6. Application of Newton’s Laws#
Mar 14, 2026 | 11490 words | 57 min read
6.1. Solving Newton’s Laws#
6.1.1. Problem-Solving Strategies#
Applying Newton’s Laws of Motion
Identify the physical principles involved by listing the givens and the quantities to calculate.
DRAW A SKETCH using arrows to represent forces.
Separate the problem into a system of interest. Each system should have a free-body diagram (FBD) that is essential to solving the problem.
Apply Newton’s 2nd law to solve the problem component-wise. If necessary apply appropriate kinematic equations (by component).
Check the solution to see whether is makes sense! Do your result have the right units?
Consider the challenge of lifting a grand piano into a second-story apartment.
Which of Newton’s laws are involved?
Fig. 6.1 (a) A piano is lifted by a crane. The original scene is redrawn in a deliberately simplified form to emphasize the essential geometry. (b) With the simplified sketch, the relevant forces are easily identified: the rope tension \(\vec{T}\), the force exerted by the piano on the rope \(\vec{F}_T\), and the weight of the piano \(\vec{w}\). (c) Defining the piano as the system of interest removes \(\vec{F}_T\), which acts on the outside world, leaving only \(\vec{T}\) and \(\vec{w}\) in the free-body diagram (FBD). (d) When the piano is stationary, these forces are equal in magnitude and opposite in direction, so \(\vec{T} = -\vec{w}\). Image Credit: modified from original on Openstax.#
Figure 6.1 shows that we can represent all forces with arrows and simplify the analysis with a more crude representation.
It is important to make a list of knowns and unknowns so that we can properly identify the system of interest. Recall that Newton’s 2nd law involves only external forces, so we
first list all the forces, and
then identify which are internal or external.
The identified forces are:
\(\vec{T}\): tension in the rope above the piano
\(\vec{F}_{T}\): force that the piano exerts on the rope, and
\(\vec{w}\): weight of the piano.
All other forces (e.g., nudge of a breeze) are assumed to be negligible.
Newton’s 3rd law can be used to identify whether forces are exerted between internal components of a system or between the system and an external component.
The \(\vec{T}\) and \(\vec{F}_T\) form a force-pair, where we only include the external component \(\vec{T}\) in our FBD (see Fig. 6.1b).
Figure 6.1c shows a FBD, where Newton’s 2nd law is applied in Fig. 6.1d.
Once external forces are clearly identified in FBDs, it should be straightforward to put them into Newton’s 2nd law and form equations that can be solved.
If the problem is 1D (i.e., forces are parallel), then the forces can be handled algebraically.
If the problem is 2D, then it must be broken into components and then, solved as a pair of 1D problems.
It is usually convenient to make one axis parallel to the direction of motion, if this is known.
Then you have
We must check the solution to tell whether it is reasonable. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exits.
In practice, intuition (i.e. knowing what is reasonable at a glance) develops through solving many problems (i.e., experience).
Ut est rerum omnium magister usus. (“Experience is the teacher of all things.”)
—Julius Caesar
Another way to check a solution is to check the units. If we are solving for force and end up with units of only velocity, then we have made a mistake.
6.1.2. Particle Equilibrium#
A particle in equilibrium is one that obey’s Newton’s 1st law (i.e., the external forces are balanced). Static equilibrium involves objects at rest, and dynamic equilibrium involves objects in motion without acceleration.
6.1.2.1. Example Problem: Different Tensions at Different Angles#
Exercise 6.1
The Problem
Consider the traffic light (mass of \(15.0\ \mathrm{kg}\)) suspended from two wires as shown in Figure 6.2. Find the tension in each wire, neglecting the masses of the wires.
Fig. 6.2 Image Credit: Openstax.#
The Model
The system of interest is the traffic light. The traffic light is treated as a point mass in static equilibrium, suspended by two massless, inextensible wires (similar to the Tension in a Tightrope). One wire makes an angle of \(30^\circ\) with the horizontal on the left, and the other makes an angle of \(45^\circ\) with the horizontal on the right. The only forces acting on the system are the tensions in the two wires and the gravitational force (weight) acting downward. Air resistance and any motion of the light are neglected, so the acceleration of the system is zero.
The Math
Because the traffic light is stationary, the net external force acting on it is zero. Newton’s 2nd law therefore applies independently along the horizontal and vertical axes.
Let \(T_1\) be the tension in the wire at \(30^\circ\) and \(T_2\) be the tension in the wire at \(45^\circ\). The weight of the traffic light is \(\vec{w} = -w\,\hat{j}\) with \(w = mg\).
First apply equilibrium along the horizontal (\(x\)) direction. The horizontal components of the two tension forces must cancel:
Using trigonometric components,
Solving for \(T_2\) in terms of \(T_1\) gives
Next apply equilibrium along the vertical (\(y\)) direction. The upward components of the tensions must balance the weight:
Substituting for the vertical components yields
Replacing \(T_2\) using the earlier relation gives
This simplifies to
Solving for \(T_1\) gives
Finally, substituting back to find \(T_2\),
The Conclusion
The tension in the wire making a \(30^\circ\) angle with the horizontal is \(T_1 = 108\ \mathrm{N}\), while the tension in the wire making a \(45^\circ\) angle is \(T_2 = 132\ \mathrm{N}\). The larger tension occurs in the wire that is more vertical, since it must provide a greater vertical force component to support the weight of the traffic light. These results are consistent with the assumption of static equilibrium and the chosen force model.
The Verification
The algebraic solution is verified by numerically recomputing the force components and confirming that the net force along both the horizontal and vertical directions is zero.
import numpy as np
m = 15.0
g = 9.80
T1 = 108.0
T2 = 132.0
# Components
T1x = -T1 * np.cos(np.deg2rad(30))
T2x = T2 * np.cos(np.deg2rad(45))
T1y = T1 * np.sin(np.deg2rad(30))
T2y = T2 * np.sin(np.deg2rad(45))
w = m * g
print(f"Net horizontal force: {(T1x + T2x):.3g} N")
print(f"Net vertical force: {(T1y + T2y - w):.3g} N")
6.1.3. Particle Acceleration#
Instead of reducing the problem to a particle in static equilibrium (i.e., \(\sum F_{\rm net} = 0\)), we can also reduce complex problems into one where the net force is nonzero, or a particle with an acceleration.
We’ll look through 4 examples that illustrate different ways we can decompose the complex motion into a simpler one:
Drag force on a barge,
Net force in an elevator,
Two attached blocks connected with a pulley,
Atwood Machine: vertical blocks with a pulley.
Note
It is useful to identify constraints that make a problem solvable. We saw this in the Tension in a Tightrope problem when identifying the tension.
In Chapter 5, we showed that the normal (a contact force) acts normal to the surface so that an object does not have an acceleration perpendicular to the surface (i.e., stays static). The bathroom scale is an excellent example of a normal force acting on a body.
A bathroom scale provides a measure of how much it must push upward to support the weight of an object. The net force in an elevator problem will examine whether we can predict what you would measure on a bathroom scale that is moving upward (accelerating or moving at constant speed).
Do you think the scale will measure a different weight when the elevator is moving at constant speed?
What if the elevator is accelerating upward?
Take a guess before looking at the example problem.
6.1.3.1. Example Problem: Drag Force on a Barge#
Exercise 6.2
The Problem
Two tugboats push on a barge at different angles as shown in Figure 6.3. The first tugboat exerts a force of \(2.7\times10^{5}\ \mathrm{N}\) in the \(x\)-direction, and the second tugboat exerts a force of \(3.6\times10^{5}\ \mathrm{N}\) in the \(y\)-direction. The mass of the barge is \(5.0\times10^{6}\ \mathrm{kg}\) and its acceleration is observed to be \(7.5\times10^{-2}\ \mathrm{m/s^2}\) in the direction shown. What is the drag force of the water on the barge resisting the motion?
Note
Drag force is a frictional force exerted by fluids (e.g., air or water). The drag force opposes the motion of the object. Since the barge is flat bottomed, we can assume that the drag force is in the direction opposite of the barge’s motion.
Fig. 6.3 Image Credit: Openstax.#
The Model
The system of interest is the barge, treated as a rigid body translating on the surface of the water. Only forces acting in the horizontal plane are included. The two tugboats apply known forces perpendicular to one another, producing a resultant applied force on the barge. The water exerts a drag force that opposes the direction of motion. Vertical forces, such as the weight of the barge and the buoyant force of the water, cancel and are excluded from the model. The barge experiences a constant acceleration, so Newton’s 2nd law applies.
The Math
The applied force from the tugboats is the vector sum of the two perpendicular forces,
Because the forces are perpendicular, the magnitude of the applied force is found using the Pythagorean theorem,
The direction of \(\vec{F}_{\text{app}}\) relative to the \(x\)-axis is
The drag force \(\vec{F}_D\) acts opposite the direction of motion and therefore opposite \(\vec{F}_{\text{app}}\). Along the direction of motion, the problem is one-dimensional. From Newton’s second law,
Taking magnitudes along the direction of motion gives
Solving for the drag force,
The Conclusion
The magnitude of the drag force exerted by the water on the barge is \(F_D = 7.5\times10^{4}\ \mathrm{N}\). This force acts in the direction opposite the barge’s motion, which is \(53^\circ\) south of west. The relatively small drag force compared to the applied force is consistent with a large barge moving at modest acceleration.
The Verification
The result is verified by numerically recomputing the applied force, the inertial term \(ma\), and confirming that their difference equals the drag force.
import numpy as np
F1 = 2.7e5
F2 = 3.6e5
m = 5.0e6
a = 7.5e-2
F_app = np.sqrt(F1**2 + F2**2)
F_D = F_app - m*a
print(f"Applied force magnitude: {F_app:.2g} N")
print(f"ma term: {(m*a):.2g} N")
print(f"Drag force magnitude: {F_D:.2g} N")
Show code cell source
import numpy as np
import matplotlib.pyplot as plt
# -----------------------------
# Global axis reference (Chapter 5 style)
# -----------------------------
theta = np.deg2rad(0.0)
x_hat = np.array([np.cos(theta), np.sin(theta)])
y_hat = np.array([-np.sin(theta), np.cos(theta)])
# -----------------------------
# draw_fbd helper (Chapter 5 style)
# -----------------------------
def draw_fbd(ax, forces, labels, note, colors=None, offsets=None, label_offsets=None):
ax.plot(0, 0, "b.", markersize=15)
if colors is None:
colors = ["black"] * len(forces)
if offsets is None:
offsets = [np.array([0.0, 0.0])] * len(forces)
if label_offsets is None:
label_offsets = [(6, 6)] * len(forces)
for F, lab, col, off, laboff in zip(forces, labels, colors, offsets, label_offsets):
Fx, Fy = F
ox, oy = off
ax.annotate("", xy=(Fx + ox, Fy + oy), xytext=(ox, oy), arrowprops=dict(arrowstyle="->", lw=2, color=col, shrinkA=0, shrinkB=0))
dx_pts, dy_pts = laboff
ax.annotate(lab, xy=(Fx + ox, Fy + oy), xytext=(dx_pts, dy_pts), textcoords="offset points", color=col, fontsize=14, ha="center", va="bottom")
mags = [np.hypot(F[0], F[1]) for F in forces]
R = 1.1 * max(mags + [1e-6])
ax.set_xlim(-0.5*R, R)
ax.set_ylim(-0.25*R, R)
ax.set_aspect("equal", adjustable="box")
ax.set_xticks([])
ax.set_yticks([])
for spine in ax.spines.values():
spine.set_visible(False)
ax.text(0.03, 0.98, note, transform=ax.transAxes, fontsize=12, va="top")
ax.plot([-1 * x_hat[0], 0], [-x_hat[1], 0], linestyle="--", color="gray", lw=1.2)
ax.text(-0.3 * x_hat[0], 0.04 * R, r"$-x$", color="gray", fontsize=11)
# -----------------------------
# Barge forces (scaled to show directions + relative magnitudes)
# -----------------------------
F1 = 2.7e5
F2 = 3.6e5
m = 5.0e6
a = 7.5e-2
F_app = np.sqrt(F1**2 + F2**2)
F_D = F_app - m * a
u_app = np.array([F1, F2]) / F_app
ratio_drag = F_D / F_app
F1_vec = np.array([F1, 0.0]) / F_app
F2_vec = np.array([0.0, F2]) / F_app
F12_vec = u_app
FD_vec = -ratio_drag * u_app
forces = [F1_vec, F2_vec, F12_vec, FD_vec]
labels = [r"$\vec{F}_1$", r"$\vec{F}_2$", r"$\vec{F}_{12}$", r"$\vec{F}_D$"]
colors = ["black", "black", "red", "black"]
label_offsets = [
(12, -6), # F1
(0, 10), # F2
(10, 6), # F12 (red)
(-10, 6), # FD
]
offsets = [np.array([0.0, 0.0])] * len(forces)
# -----------------------------
# Plot
# -----------------------------
fig, ax = plt.subplots(1, 1, figsize=(5, 4), dpi=120)
draw_fbd(ax, forces, labels, "Barge (horizontal plane)", colors=colors, offsets=offsets, label_offsets=label_offsets)
plt.tight_layout()
plt.show()
6.1.3.2. Example Problem: Bathroom Scale in an Elevator#
Exercise 6.3
The Problem
Figure 6.4 shows a \(75.0\)-\(\mathrm{kg}\) man (weight of about \(165\ \mathrm{lb}\)) standing on a bathroom scale in an elevator. Calculate the scale reading: (a) if the elevator accelerates upward at a rate of \(1.20\ \mathrm{m/s^2}\), and (b) if the elevator moves upward at a constant speed of \(1\ \mathrm{m/s}\).
Fig. 6.4 Image Credit: Openstax.#
The Model
The system of interest is the person. The person is modeled as a particle moving vertically with the elevator. The only external forces acting on the person are the weight \(\vec{w}\) (Earth’s gravitational force on the person) acting downward and the upward contact force from the scale \(\vec{F}_s\) acting upward. The scale reading is the magnitude of the force the scale exerts on the person, \(F_s\). Air resistance is neglected. Motion is one-dimensional along the vertical axis, with \(+\hat{j}\) upward. In part (a) the acceleration is upward; in part (b) the elevator’s speed is constant so the acceleration is zero.
The Math
Newton’s 2nd law for the person is
Along the vertical axis, the forces on the person are \(\vec{F}_s = +F_s\,\hat{j}\) and \(\vec{w} = -w\,\hat{j}\), so
In components along \(\hat{j}\),
Using \(w = mg\) gives a direct expression for the scale reading:
(a) First, we evaluate the case when the elevator is accelerating upward at \(a = 1.20\ \mathrm{m/s^2}\)
(b) Then, we evaluate when the elevator is moving upward at constant speed \(1\ \mathrm{m/s}\) (constant speed implies \(a = 0\)), so
The Conclusion
If the elevator accelerates upward at \(1.20\ \mathrm{m/s^2}\), the scale reads \(F_s = 825\ \mathrm{N}\) as shown in part (a). If the elevator moves upward at constant speed instead, the acceleration is zero and the scale reads \(F_s = 735\ \mathrm{N}\) as shown in part (b).
Physically, the scale reads the upward contact force. When the elevator accelerates upward, the scale must push up harder than the person’s weight to produce the upward acceleration, so the reading increases. When the elevator’s speed is constant, the net force is zero and the scale reads the person’s normal weight.
The Verification
We verify by recomputing \(F_s = m(g+a)\) for both cases and confirming that part (b) reduces to \(F_s = mg\) when \(a=0\).
import numpy as np
m = 75.0
g = 9.80
a_up = 1.20
a_const = 0.0
Fs_a = m * (g + a_up)
Fs_b = m * (g + a_const)
print(f"(a) Fs = {Fs_a:.3g} N")
print(f"(b) Fs = {Fs_b:.3g} N")
print(f"mg = {(m*g):.3g} N")
Show code cell source
import numpy as np
import matplotlib.pyplot as plt
# -----------------------------
# Global axis reference
# -----------------------------
theta = np.deg2rad(0.0)
x_hat = np.array([np.cos(theta), np.sin(theta)])
y_hat = np.array([-np.sin(theta), np.cos(theta)])
# -----------------------------
# draw_fbd helper (Chapter 5 style, adjusted limits)
# -----------------------------
def draw_fbd(ax, forces, labels, note, colors=None, offsets=None, label_offsets=None):
ax.plot(0, 0, "b.", markersize=15)
if colors is None:
colors = ["black"] * len(forces)
if offsets is None:
offsets = [np.array([0.0, 0.0])] * len(forces)
if label_offsets is None:
label_offsets = [(6, 6)] * len(forces)
for F, lab, col, off, laboff in zip(forces, labels, colors, offsets, label_offsets):
Fx, Fy = F
ox, oy = off
ax.annotate("", xy=(Fx + ox, Fy + oy), xytext=(ox, oy),
arrowprops=dict(arrowstyle="->", lw=2, color=col, shrinkA=0, shrinkB=0))
dx_pts, dy_pts = laboff
ax.annotate(lab, xy=(Fx + ox, Fy + oy), xytext=(dx_pts, dy_pts),
textcoords="offset points", color=col, fontsize=14,
ha="center", va="bottom")
mags = [np.hypot(F[0], F[1]) for F in forces]
R = max(mags + [1e-6])
ax.set_xlim(-1.65 * R, 1.65 * R)
ax.set_ylim(-1.75 * R, 1.75 * R)
ax.set_aspect("equal", adjustable="box")
ax.set_xticks([])
ax.set_yticks([])
for spine in ax.spines.values():
spine.set_visible(False)
ax.text(0.02, 0.97, note, transform=ax.transAxes, fontsize=12, va="top")
ax.plot([-x_hat[0], 0], [-x_hat[1], 0], linestyle="--", color="gray", lw=1.2)
ax.text(-0.75 * R, 0.08 * R, r"$-x$", color="gray", fontsize=11)
# -----------------------------
# Physical parameters
# -----------------------------
m = 75.0
g = 9.81
a_up = 1.20
w = m * g
Fs_up = m * (g + a_up)
Fs_const = m * g
# Normalize by w
w_vec = np.array([0.0, -1.0])
Fs_up_vec = np.array([0.0, Fs_up / w])
Fs_const_vec = np.array([0.0, Fs_const / w])
# -----------------------------
# Panel 1: All forces (expanded x offsets)
# -----------------------------
T_vec = np.array([0.0, 1.35])
N_vec = np.array([0.0, 1.10])
Fp_vec = np.array([0.0, -0.85])
Ft_vec = np.array([0.0, -1.05])
ws_vec = np.array([0.0, -0.45])
we_vec = np.array([0.0, -0.65])
forces_all = [T_vec, N_vec, Fs_const_vec, Fp_vec, Ft_vec, ws_vec, we_vec, w_vec]
labels_all = [r"$\vec{T}$", r"$\vec{N}$", r"$\vec{F}_s$", r"$\vec{F}_p$",
r"$\vec{F}_t$", r"$\vec{w}_s$", r"$\vec{w}_e$", r"$\vec{w}$"]
d = 0.18
offsets_all = [
np.array([+1.5*d, 0.0]),
np.array([-1.5*d, 0.0]),
np.array([0.0, 0.0]),
np.array([+2*d, 0.0]),
np.array([-2*d, 0.0]),
np.array([+4*d, 0.0]),
np.array([-4*d, 0.0]),
np.array([0.0, 0.0])
]
label_offsets_all = [(0, 8)] * 3 + [(0, -22)] * 5
# -----------------------------
# Panel 2 & 3: Simplified systems
# -----------------------------
forces_up = [Fs_up_vec, w_vec]
forces_const = [Fs_const_vec, w_vec]
labels_simple = [r"$\vec{F}_s$", r"$\vec{w}$"]
label_offsets_simple = [(0, 8), (0, -22)]
# -----------------------------
# Plot (tight spacing)
# -----------------------------
fig, axes = plt.subplots(1, 3, figsize=(11, 4), dpi=120, gridspec_kw={"wspace": 0.15})
draw_fbd(axes[0], forces_all, labels_all, "All forces (system not simplified)",
offsets=offsets_all, label_offsets=label_offsets_all)
draw_fbd(axes[1], forces_up, labels_simple, r"(a) Upward acceleration ($a>0$)",
label_offsets=label_offsets_simple)
draw_fbd(axes[2], forces_const, labels_simple, r"(b) Constant speed ($a=0$)",
label_offsets=label_offsets_simple)
# --- reference line at mg level (tangent to Fs when a=0) ---
# --- enforce common y-scale for comparison panels ---
R_ref = max(np.hypot(Fs_up_vec[0], Fs_up_vec[1]), np.hypot(w_vec[0], w_vec[1]))
for ax in axes[1:3]:
ax.set_ylim(-1.75 * R_ref, 1.75 * R_ref)
# --- reference line at mg level ---
y_ref = Fs_const_vec[1]
axes[1].axhline(y_ref, linestyle="--", color="gray", lw=1.2,alpha=0.5)
axes[2].axhline(y_ref, linestyle="--", color="gray", lw=1.2,alpha=0.5)
plt.subplots_adjust(left=0.03, right=0.98, top=0.92, bottom=0.08)
plt.show()
6.1.3.3. Example Problem: Two attached Blocks#
Exercise 6.4
The Problem
Figure 6.5 shows a block of mass \(m_1\) on a frictionless, horizontal surface. It is pulled by a light string that passes over a frictionless and massless pulley. The other end of the string is connected to a block of mass \(m_2\). Find the acceleration of the blocks and the tension in the string in terms of \(m_1\), \(m_2\), and \(g\).
Fig. 6.5 Image Credit: Openstax.#
The Model
We model the system as two point-like blocks connected by a light, inextensible string over a frictionless, massless pulley. Block \(m_1\) moves horizontally on a frictionless surface, so it experiences no friction force. Block \(m_2\) moves vertically. The string transmits the same tension \(T\) to both blocks.
Each block is treated as a separate system of interest. The only forces on block \(m_1\) are its weight \(\vec{w}_1\), the normal force \(\vec{N}\) from the surface, and the string tension \(\vec{T}\). The only forces on block \(m_2\) are its weight \(\vec{w}_2\) and the string tension \(\vec{T}\).
The blocks share a common acceleration magnitude because they are connected by the string. When block \(m_1\) accelerates to the right, block \(m_2\) accelerates downward by the same amount.
Before proceeding, recall the free-body-diagram workflow developed in Chapter 5 (see the example Block on the Table). You do not need to re-derive the general FBD rules here. Focus instead on identifying the forces and writing Newton’s 2nd law along the appropriate axes.
The Math
We apply Newton’s 2nd law separately to each block.
For block 1, the vertical forces balance, so we write Newton’s 2nd law only along the horizontal direction. The only horizontal force on block 1 is the tension:
For block 2, motion is vertical. Taking upward as the positive \(y\)-direction, the forces are the upward tension and the downward weight:
Because the blocks are connected by the string, their accelerations have the same magnitude but opposite directions. If block 1 accelerates to the right with acceleration \(a\), block 2 accelerates downward with acceleration \(a\), so \(a_y = -a\).
Substituting \(a_y = -a\) into the equation for block 2 gives
We now have two equations involving \(T\) and \(a\):
Subtracting the second equation from the first eliminates \(T\) and allows us to solve for the acceleration:
Substituting this result for \(a\) back into \(T = m_1 a\) gives the tension:
The Conclusion
The magnitude of the acceleration of both blocks is \(a = \frac{m_2}{m_1 + m_2}\, g.\) The tension in the string is \(T = \frac{m_1 m_2}{m_1 + m_2}\, g.\)
The blocks accelerate together because they are constrained by the string, but the acceleration depends on how the total mass is distributed between the horizontal and hanging blocks.
The Verification
We can verify these expressions by checking limiting cases numerically. For example, if \(m_2 \ll m_1\), the acceleration should be small and the tension should be much less than \(m_2 g\). If \(m_2 \gg m_1\), the acceleration should approach \(g\), but never exceed it.
import numpy as np
g = 9.81
# Example values
m1 = 4.0 # kg
m2 = 1.0 # kg
a = (m2 / (m1 + m2)) * g
T = (m1 * m2 / (m1 + m2)) * g
print(f"Acceleration a = {a:.3g} m/s^2")
print(f"Tension T = {T:.3g} N")
6.1.3.4. Example Problem: Atwood Machine#
Exercise 6.5
The Problem
A classic problem in physics, similar to the one we just solved, is the Atwood machine, which consists of a light rope running over a frictionless pulley, with two objects of different mass attached. It is particularly useful in understanding the connection between force and motion.
In Figure 6.6, the masses are \(m_1 = 2.00\ \mathrm{kg}\) and \(m_2 = 4.00\ \mathrm{kg}\). The pulley is frictionless. (a) If \(m_2\) is released, what will its acceleration be? (b) What is the tension in the string?
Fig. 6.6 Image Credit: Openstax.#
The Model
Each mass is treated as a point particle connected by a light, inextensible string over a frictionless, massless pulley. Because the string and pulley are ideal, the tension \(T\) is the same on both sides of the pulley.
Each block is treated as its own system of interest, just as in the example of two attached blocks. The key difference here is that both blocks move vertically rather than one moving horizontally and the other vertically.
The only forces on each block are its weight and the string tension. The blocks accelerate with the same magnitude but in opposite directions. Since \(m_2 > m_1\), block \(m_2\) accelerates downward while block \(m_1\) accelerates upward.
As in the previous problem, we do not re-derive general free-body-diagram rules. Instead, we apply the same FBD logic developed in Chapter 5 and used in the earlier attached-blocks example.
The Math
We begin by applying Newton’s second law to each block. As in the previous attached-blocks problem, the two blocks share a common acceleration magnitude \(a\) because they are connected by a light, inextensible string over a frictionless pulley.
Taking upward as the positive \(y\)-direction for both blocks, the force equations are
These two equations form the governing equations for the system and will be reused below.
(a) First, we determine the acceleration of the blocks after \(m_2\) is released. Since the blocks move together, we solve the governing equations simultaneously for \(a\).
Subtracting the equation for \(m_1\) from the equation for \(m_2\) eliminates the tension:
Solving for the acceleration gives
Substituting \(m_1 = 2.00\ \mathrm{kg}\), \(m_2 = 4.00\ \mathrm{kg}\), and \(g = 9.81\ \mathrm{m/s^2}\),
(b) Next, we find the tension in the string using the same governing equations. We substitute the acceleration found in part (a) into the equation for block \(m_1\).
From the force balance on \(m_1\),
Substituting numerical values,
The Conclusion
The acceleration of the Atwood machine is \(a = 3.27\ \mathrm{m/s^2},\) with block \(m_2\) accelerating downward and block \(m_1\) accelerating upward. The tension in the string is \(T = 26.2\ \mathrm{N}.\)
This result closely mirrors the attached-blocks problem: in both cases, the acceleration is the ratio of the net unbalanced force to the total mass of the system.
The Verification
We can verify the structure of the result by checking limiting cases. If \(m_1 = m_2\), the acceleration goes to zero, as expected. If one mass is much larger than the other, the acceleration approaches \(g\) but never exceeds it.
import numpy as np
g = 9.81
m1 = 2.00
m2 = 4.00
a = (m2 - m1) / (m1 + m2) * g
T = m1 * (g + a)
print(f"Acceleration a = {a:.3g} m/s^2")
print(f"Tension T = {T:.3g} N")
6.1.4. Newton’s Laws of Motion and Kinematics#
Newton’s laws of motion can also be integrated with other concepts to solve problems of motion. For example, forces produce accelerations, which was the main topic when discussing kinematics.
For problems that involve various types of forces, acceleration, velocity and/or position, it helps to identify the principles involved when listing the givens and quantities to be calculated. The following examples illustrate how to apply the problem-solving strategies. These example problems are
Soccer player at top speed,
Force on a model helicopter,
Force on a baggage tractor, and
The motion of vertical projectile (Bonus).
Recall that \( v = \frac{ds}{dt}\) and \(a = \frac{dv}{dt}\). This means that we can use the calculus forms when acceleration is not constant (i.e., \(a = f(s,v,t)\)).
First we solve for \(dt\) in each equation, then set them equal to each other, or
When \(a = f(v)\), we can apply the following integrals
6.1.4.1. Example Problem: Soccer Player Top Speed#
Exercise 6.6
The Problem
A soccer player starts at rest and accelerates forward, reaching a velocity of \(8.00\ \mathrm{m/s}\) in \(2.50\ \mathrm{s}\). (a) What is her average acceleration? (b) What average force does the ground exert forward on the runner so that she achieves this acceleration? The player’s mass is \(70.0\ \mathrm{kg}\), and air resistance is negligible.
The Model
The player is modeled as a particle moving in one dimension along a horizontal surface. Air resistance is neglected, so the net external horizontal force on the player is due entirely to the interaction with the ground.
For part (a), the motion is treated kinematically, assuming constant average acceleration over the time interval. For part (b), the player is the system of interest, and Newton’s second law is applied in the horizontal direction. Vertical forces are not relevant because they balance and do not affect the horizontal motion.
The Math
We first determine the player’s average acceleration from the definition of acceleration. That result is then reused to find the average horizontal force using Newton’s second law.
(a) First, we evaluate the player’s average acceleration during the sprint. The player starts from rest and reaches a final speed of \(8.00\ \mathrm{m/s}\) in \(2.50\ \mathrm{s}\), so the change in velocity and the elapsed time are known.
The definition of average acceleration is
Substituting the given values,
(b) Next, we determine the average force exerted by the ground on the runner. Since air resistance is negligible, this force is equal in magnitude to the net external horizontal force on the player.
We reuse the acceleration found in part (a) and apply Newton’s second law in the horizontal direction:
Substituting the mass of the player and the acceleration,
The Conclusion
The player’s average acceleration is \(a = 3.20\ \mathrm{m/s^2}.\) The average forward force exerted by the ground on the runner is \(F_{\text{net}} = 224\ \mathrm{N}. \)
This force is reasonable for an athlete accelerating rapidly over a short time interval and illustrates how kinematics and dynamics are combined to analyze real motion.
The Verification
We can verify the result by checking units and magnitudes. The acceleration has units of \(\mathrm{m/s^2}\), and multiplying by mass correctly produces force units of newtons. The force corresponds to a moderate fraction of the runner’s weight, which is consistent with a realistic sprint acceleration.
m = 70.0 # kg
v = 8.00 # m/s
t = 2.50 # s
a = v / t
F = m * a
print(f"Average acceleration = {a:.3g} m/s^2")
print(f"Average force = {F:.3g} N")
6.1.4.2. Example Problem: Force on a Model Helicopter#
Exercise 6.7
The Problem
A \(1.50\ \mathrm{kg}\) model helicopter has a velocity of \(5.00\,\hat{j}\ \mathrm{m/s}\) at \(t = 0\). It is accelerated at a constant rate for \(2.00\ \mathrm{s}\), after which it has a velocity of \((6.00\,\hat{i} + 12.00\,\hat{j})\ \mathrm{m/s}\). What is the magnitude of the resultant force acting on the helicopter during this time interval?
The Model
The helicopter is modeled as a particle moving in two dimensions. The acceleration is assumed to be constant over the \(2.00\ \mathrm{s}\) interval. External forces other than those producing the observed acceleration are not considered explicitly; instead, we determine the net external force using Newton’s second law.
A Cartesian coordinate system is used with \(\hat{i}\) horizontal and \(\hat{j}\) vertical. The system of interest is the helicopter itself.
The Math
We first determine the helicopter’s acceleration from the change in velocity over the given time interval. We then apply Newton’s second law in vector form to find the net force and finally compute its magnitude.
The initial and final velocities are
The change in velocity is therefore
The acceleration follows from its definition:
Applying Newton’s second law in vector form,
Substituting the mass of the helicopter,
The magnitude of the resultant force is
The Conclusion
The magnitude of the resultant force acting on the helicopter during the \(2.00\ \mathrm{s}\) interval is \(F = 6.91\ \mathrm{N}.\)
This result follows directly from the constant acceleration inferred from the change in velocity and illustrates the use of Newton’s second law in vector form.
The Verification
We can verify the calculation by recomputing the acceleration and force components numerically and confirming the force magnitude.
import numpy as np
m = 1.50
vi = np.array([0.0, 5.00])
vf = np.array([6.00, 12.00])
dt = 2.00
a = (vf - vi) / dt
F = m * a
F_mag = np.linalg.norm(F)
print(f"Acceleration vector = {a} m/s^2")
print(f"Force vector = {F} N")
print(f"Force magnitude = {F_mag:.3g} N")
6.1.4.3. Example Problem: Baggage Tractor#
Exercise 6.8
The Problem
Figure 6.7a shows a baggage tractor pulling luggage carts from an airplane. The tractor has mass \(650.0\ \mathrm{kg}\), while cart A has mass \(250.0\ \mathrm{kg}\) and cart B has mass \(150.0\ \mathrm{kg}\). The driving force acting for a brief period of time accelerates the system from rest and acts for \(3.00\ \mathrm{s}\). (a) If this driving force is given by \(\vec{F} = (820.0\,t)\,\hat{i}\ \mathrm{N}\), find the speed after \(3.00\ \mathrm{s}\). (b) What is the horizontal force acting on the connecting cable between the tractor and cart A at this instant?
Fig. 6.7 Image Credit: Openstax.#
The Model
The tractor and both carts move together in a straight line along a horizontal surface. The string connections are ideal, and friction is neglected. Vertical forces balance and do not affect the motion, so only horizontal forces are considered.
For part (a), the entire system (tractor + cart A + cart B) is treated as a single object acted on by the external driving force. For part (b), the tractor alone is isolated to determine the internal coupling force in the cable to cart A.
The motion starts from rest, and the applied force varies linearly with time.
The Math
The governing principle for both parts is Newton’s 2nd law applied in the horizontal direction,
For the full system, the total mass is
(a) First, we consider the motion of the entire system while the driving force is applied.
Applying Newton’s 2nd law to the system gives
Solving for the acceleration as a function of time,
Because acceleration varies with time, we determine the velocity using
With the initial condition \(v_o = 0\) at \(t = 0\), we integrate from \(t = 0\) to \(t = 3.00\ \mathrm{s}\):
(b) Next, we isolate the tractor to determine the force in the connecting cable to cart A.
At \(t = 3.00\ \mathrm{s}\), the acceleration is
Applying Newton’s 2nd law to the tractor alone,
Solving for the cable force,
The Conclusion
After \(3.00\ \mathrm{s}\), the tractor and carts reach a speed of \(3.51\ \mathrm{m/s}\). At this instant, the horizontal force in the cable connecting the tractor to cart A is \(938\ \mathrm{N}\).
The solution highlights two important modeling choices: treating multiple objects as a single system when external forces are applied, and isolating individual components when internal forces are required.
The Verification
The acceleration and velocity are recomputed numerically by integrating the time-dependent force, and the cable force is checked by applying Newton’s second law to the tractor at \(t = 3.00\ \mathrm{s}\).
import numpy as np
# Given values
m_system = 650.0 + 250.0 + 150.0
m_tractor = 650.0
t = 3.00
a = 0.7809 * t
v = 0.3905 * t**2
T = 820.0 * t - m_tractor * a
print(f"Speed at 3.00 s: {v:.3g} m/s")
print(f"Cable force at 3.00 s: {T:.3g} N")
6.1.4.4. Example Problem: Projectile Vertical Motion (Bonus)#
Exercise 6.9
The Problem
A \(10.0\)-kg mortar shell is fired vertically upward from the ground, with an initial velocity of \(50.0\ \mathrm{m/s}\) (see Figure 6.8). Determine the maximum height it will travel if atmospheric resistance is measured as \(F_D = (0.010v^2)\ \mathrm{N}\), where \(v\) is the speed at any instant.
Fig. 6.8 Image Credit: Openstax.#
The Model
We model the mortar shell as a particle moving only in the vertical direction. Take \(+y\) upward with the shell launched from \(y_o = 0\). The forces acting during the upward motion are:
Weight \(\vec{w} = -mg\,\hat{j}\) (downward).
Drag force \(\vec{F}_D\) (downward during ascent because it opposes the upward velocity).
The drag magnitude depends on the instantaneous speed: \(F_D = kv^2\) with \(k = 0.010\ \mathrm{N\,s^2/m^2}\). We neglect any horizontal motion and treat \(g\) as constant with \(g = 9.81\ \mathrm{m/s^2}\).
The Math
We want the maximum height \(h\), which occurs at the instant the upward speed reaches zero. During ascent, the acceleration is not constant because drag depends on \(v\).
Newton’s second law in the vertical direction is
Using \(F_D = kv^2\) gives the governing acceleration as a function of speed:
To connect acceleration to position, we use
Substitute \(a = v\frac{dv}{dy}\) into the governing equation:
At launch, \(y=0\) and \(v=v_o\). At the top, \(y=h\) and \(v=0\). Integrate:
The integral \(\int \frac{du}{u}\)
In this example the integral becomes
This can simplified via a method called u-substitution.
The denominator contains \(v^2\) and the numerator contains \(v\,dv\). That is a strong hint that we should choose
If we differentiate both sides with respect to \(v\):
Then we solve this for \(v\,dv\):
Now the original integral becomes
Then we can use the identity \(\int \frac{du}{u}=\ln|u|+C\)
A basic derivative fact is:
So the matching antiderivative is
Applying our limits
When
So
That is the expression used in the solution.
Let \(u = \frac{k}{m}v^2 + g\), so \(du = 2\frac{k}{m}v\,dv\) and \(v\,dv = \frac{m}{2k}du\). Then
Numerical evaluation using \(m = 10.0\ \mathrm{kg}\), \(k = 0.01\ \mathrm{N\,s^2/m^2}\), \(v_o = 50.0\ \mathrm{m/s}\), and \(g = 9.81\ \mathrm{m/s^2}\):
The Conclusion
The shell reaches a maximum height of \(h \approx 110\ \mathrm{m}.\) This is lower than the no-drag result because both \(\vec{w}\) and \(\vec{F}_D\) act downward during ascent, producing a larger downward acceleration than gravity alone. The model assumes a purely vertical launch and a drag force of the form \(kv^2\) that always opposes the motion.
The Verification
We verify the analytic height by numerically integrating the ascent using \(a(v) = -\frac{k}{m}v^2 - g\) and the identity \(a = v\,dv/dy\), which gives \(dy = \dfrac{v\,dv}{-\frac{k}{m}v^2 - g}\). We also include a simple free-body diagram.
Show code cell source
import numpy as np
import matplotlib.pyplot as plt
# -----------------------------
# Global axis reference (match Chapter 5 style)
# -----------------------------
theta = np.deg2rad(0.0)
x_hat = np.array([np.cos(theta), np.sin(theta)])
y_hat = np.array([-np.sin(theta), np.cos(theta)])
# -----------------------------
# draw_fbd helper (Chapter 5 style)
# -----------------------------
def draw_fbd(ax, forces, labels, note, colors=None, offsets=None, label_offsets=None):
ax.plot(0, 0, "b.", markersize=15)
if colors is None:
colors = ["black"] * len(forces)
if offsets is None:
offsets = [np.array([0.0, 0.0])] * len(forces)
if label_offsets is None:
label_offsets = [(6, 6)] * len(forces)
for F, lab, col, off, laboff in zip(forces, labels, colors, offsets, label_offsets):
Fx, Fy = F
ox, oy = off
ax.annotate("", xy=(Fx + ox, Fy + oy), xytext=(ox, oy), arrowprops=dict(arrowstyle="->", lw=2, color=col, shrinkA=0, shrinkB=0))
dx_pts, dy_pts = laboff
ax.annotate(lab, xy=(Fx + ox, Fy + oy), xytext=(dx_pts, dy_pts), textcoords="offset points", color=col, fontsize=14, ha="center", va="bottom")
mags = [np.hypot(F[0], F[1]) for F in forces]
R = 1.25 * max(mags + [1e-6])
ax.set_xlim(-1.20 * R, 1.20 * R)
ax.set_ylim(-1.20 * R, 1.20 * R)
ax.set_aspect("equal", adjustable="box")
ax.set_xticks([])
ax.set_yticks([])
for spine in ax.spines.values():
spine.set_visible(False)
ax.text(0.03, 0.95, note, transform=ax.transAxes, fontsize=12, va="top")
ax.plot([-1 * x_hat[0], 0], [-x_hat[1], 0], linestyle="--", color="gray", lw=1.2)
ax.text(-0.9 * x_hat[0], -1.1* x_hat[1], r"$-x$", color="gray", fontsize=11)
# -----------------------------
# Given values
# -----------------------------
m = 10.0
v0 = 50.0
k = 0.01
g = 9.81
# -----------------------------
# Analytic result
# -----------------------------
h_analytic = (m / (2.0 * k)) * np.log(((k / m) * v0**2 + g) / g)
# -----------------------------
# Numerical check via integrating dy = v dv / (-(k/m)v^2 - g)
# -----------------------------
v = np.arange(v0, 0.0, -0.01)
a = -(k / m) * v**2 - g
integrand = v / a
h_numeric = np.trapz(integrand, v)
print(f"h_analytic = {h_analytic:.3g} m")
print(f"h_numeric = {h_numeric:.3g} m")
print(f"percent difference = {100.0 * abs(h_numeric - h_analytic) / h_analytic:.3g} %")
# -----------------------------
# Minimal FBD (during ascent: both forces downward)
# -----------------------------
w_vec = np.array([0.0, -1.0])
Fd_vec = np.array([0.0, -0.7])
forces = [w_vec, Fd_vec]
labels = [r"$\vec{w}$", r"$\vec{F}_D$"]
colors = ["black", "black"]
label_offsets = [(0, -20), (10, -20)]
offsets = [np.zeros(2),np.array([0.1,0])]
fig, ax = plt.subplots(1, 1, figsize=(4.5, 4), dpi=120)
draw_fbd(ax, forces, labels, "Shell during ascent", colors=colors, offsets=offsets, label_offsets=label_offsets)
plt.tight_layout()
plt.show()
h_analytic = 114 m
h_numeric = 114 m
percent difference = 4.88e-06 %
6.2. Friction#
Real objects interact with their surroundings, which means the environment can cause a resistance (a force of friction).
Friction opposed relative motion between systems in contact, but also allows us to move.
Try walking on ice. You need boots that grip (i.e., increase friction).
6.2.1. Static and Kinetic Friction#
Friction
Friction is a force that opposes relative motion between systems in contact.
There are several forms of friction.
Sliding friction \(f_s\) is parallel to the contact surface between systems and is always in a direction that opposes motion (or attempted motion of the systems relative to each other).
Kinetic friction \(f_k\) is when two systems are in contact and moving relative to one another.
For example, friction slows a hockey puck sliding on ice.
When the static friction is greater than the kinetic friction (\(f_s >f_k\)), the objects are stationary.
Static and Kinetic Friction
If two systems are in contact and stationary relative to one another, then the friction between them is called static friction. If two systems are in contact and moving relative to one another, then the friction between them is called kinetic friction.
Consider trying to slide a heavy crate across a concrete floor. You might push very hard and not move the crate at all. This means the static friction responds to what you do; it increases (in the opposite direction) to resist your push.
Once in motion, it is easier to keep it in motion than it was to get it started, which indicates that the kinetic friction is less than the static friction. If you add mass to the crate, you need to push even harder to get it started and also to keep it moving. If you oiled the concrete you would find it easier to get the crate started and keep it going.
Figure 6.9 is a pictorial representation of how friction occurs at the interface between two objects. Close-up inspection of these surfaces reveals them to be rough. Thus, when you push to get an object moving, you must raise the object until it can skip along with just the tips of the surface hitting, breaking off the points or both.
Fig. 6.9 Image Credit: Openstax.#
A considerable force can be resisted by friction with no apparent motion. The harder the surfaces are pushed together, the more force is needed to move them. Part of the friction is due to adhesive forces between surface molecules of the two objects, which explains the dependence on the surface composition.
Adhesion varies with substances in contact and is a complicated aspect of surface physics. Once an object is moving, there are fewer points of contact (i.e., fewer molecules adhering), so less force is required to keep the object moving. At small but nonzero speeds, friction is nearly independent of speed.
The magnitude of the frictional force has two forms: static and kinetic. The models of each are empirical (experimentally determined) and they are not vector equations.
Magnitude of Static and Kinetic Friction
The magnitude of static friction \(f_s\) is
where \(\mu_s\) (or \mu_s) is the coefficient of static friction and \(N\) is the magnitude fo the normal force. Static friction is a responsive force that increases to be equal and opposite to whatever force is exerted. Once the applied force exceeds \(f_s(\text{max})\), the object moves, and
The magnitude of kinetic friction \(f_k\) is given by
where \(\mu_k\) (or \mu_k) is the coefficient of kinetic friction. Figure 6.10 illustrates the transition from static friction (stationary in Fig. 6.10a) to kinetic friction (moving in Fig. 6.10b).
Fig. 6.10 Image Credit: Openstax.#
Table 6.1 show the coefficients of kinetic friction are less than their static counterparts. The values of \(\mu\) provide only an approximate description of the friction given.
System |
Static Friction \(\mu_s\) |
Kinetic Friction \(\mu_k\) |
|---|---|---|
Rubber on dry concrete |
1.0 |
0.7 |
Rubber on wet concrete |
0.5–0.7 |
0.3–0.5 |
Wood on wood |
0.5 |
0.3 |
Waxed wood on wet snow |
0.14 |
0.10 |
Metal on wood |
0.5 |
0.3 |
Steel on steel (dry) |
0.6 |
0.3 |
Steel on steel (oiled) |
0.05 |
0.03 |
Teflon on steel |
0.04 |
0.04 |
Bone lubricated by synovial fluid |
0.016 |
0.015 |
Shoes on wood |
0.9 |
0.7 |
Shoes on ice |
0.10 |
0.05 |
Ice on ice |
0.10 |
0.03 |
Steel on ice |
0.04 |
0.02 |
In both static and kinetic frictions, the equation depends on the materials (through \(\mu_s\) or \(\mu_k\)) and the normal force. The direction of friction is always
opposite to the direction of motion,
parallel to the surface between objects, and
perpendicular to the normal force.
For example if the crate you try to push has a mass of \(100\ {\rm kg}\), then the normal force is equal to its weight,
perpendicular to the floor. If the coefficient of static friction is \(0.45\), you would have to exert a force parallel to the floor that is greater than
to move the crate. Once there is motion, friction is less and the coefficient of kinetic friction might be \(0.30\), so that a force of only
keeps it moving at a constant speed. If the floor is lubricated, both coefficients are considerably less than otherwise. The coefficient of friction is a unitless quantity with a magnitude between \(0\) and \(1.0\). The actual value depends on the two surfaces in contact.
The equations for static and kinetic friction are empirical laws that describe the behavior of the frictional force. While these formulas are useful for practical purposes, they are not general principles (e.g., Newton’s 2nd law). There are cases for which these equations are not even good approximations. For example, neither formula is accurate for lubricated surfaces or surfaces sliding at high speeds.
6.2.1.1. Example Problem: Static and Kinetic Friction#
Exercise 6.10
The Problem
A \(20.0\text{-kg}\) crate is at rest on a floor as shown in Figure 6.11. The coefficient of static friction between the crate and floor is \(0.70\) and the coefficient of kinetic friction is \(0.60\). A horizontal force \(\vec{P}\) is applied to the crate. Find the force of friction if (a) \(\vec{P}=20\,\mathrm{N}\,\hat{i}\), (b) \(\vec{P}=30\,\mathrm{N}\,\hat{i}\), (c) \(\vec{P}=120\,\mathrm{N}\,\hat{i}\), and (d) \(\vec{P}=180,\mathrm{N},\hat{i}\).
Fig. 6.11 Image Credit: Openstax.#
The Model
We model the crate as a particle on a horizontal floor. Choose axes so \(+\hat{i}\) is to the right and \(+\hat{j}\) is upward. The forces on the crate are the applied force \(\vec{P}\) (horizontal), the friction force \(\vec{f}\) (horizontal and opposite the impending or actual motion), the normal force \(\vec{N}\) from the floor (upward), and the weight \(\vec{W}\) (downward). The floor is level so \(a_y=0\) and \(N=W\). Static friction adjusts in magnitude up to \(f_{s,\max}=\mu_s N\). If the applied force exceeds \(f_{s,\max}\), the crate slides and kinetic friction acts with magnitude \(f_k=\mu_k N\).
The Math
We apply Newton’s 2nd law to the crate. Let \(\vec{P}=P,\hat{i}\) and \(\vec{f}=-f,\hat{i}\) because friction opposes the direction of \(\vec{P}\). The weight is \(\vec{W}=-W,\hat{j}\) with \(W=mg\).
Substituting the forces gives
From the vertical equation, \(N=W=mg\), so the maximum static friction is
We evaluate the shared quantities once:
If \(P \le f_{s,\max}\), the crate remains at rest so \(a_x=0\) and \(P-f=0\), which gives \(f_s=P\). If \(P>f_{s,\max}\), the crate slides and \(f=f_k=\mu_k N\).
(a) First, we evaluate \(\vec{P}=20\,\mathrm{N}\,\hat{i}\). Since \(20\,\mathrm{N} < 140\,\mathrm{N}\), the crate remains at rest and
(b) Next, we evaluate \(\vec{P}=30\,\mathrm{N}\,\hat{i}\). Since \(30\,\mathrm{N} < 140,\mathrm{N}\), the crate remains at rest and
(c) Then, we evaluate \(\vec{P}=120\,\mathrm{N},\hat{i}\). Since \(120\,\mathrm{N} < 140\,\mathrm{N}\), the crate remains at rest and
(d) Finally, we evaluate \(\vec{P}=180\,\mathrm{N}\,\hat{i}\). Since \(180\,\mathrm{N} > 140\,\mathrm{N}\), static friction is not sufficient and the crate slides, so kinetic friction acts:
and the acceleration is
The Conclusion
The maximum static friction is \(f_{s,\max}=140\,\mathrm{N}\), so for (a)–(c) the crate can remain at rest and the friction magnitude equals the applied force: \(f=20\,\mathrm{N}\), \(30\,\mathrm{N}\), and \(120\,\mathrm{N}\). In (d), the applied force exceeds \(f_{s,\max}\) so the crate slides and the friction magnitude is the kinetic value \(f=120\,\mathrm{N}\). In every case the friction force points opposite \(\vec{P}\), so \(\vec{f}=-f\,\hat{i}\).
The Verification
We recompute \(N\), \(f_{s,\max}\), and \(f_k\) using \(g=9.81,\mathrm{m/s^2}\), then apply the static-versus-kinetic decision rule for each given value of \(P\).
import numpy as np
import matplotlib.pyplot as plt
m = 20.0
mu_s = 0.70
mu_k = 0.60
g = 9.81
P_vals = np.array([20.0, 30.0, 120.0, 180.0])
N = m * g
f_s_max = np.round(mu_s * N,-1) #round to 10s place for sig figs
f_k = np.round(mu_k * N,-1)
print(f"N = {N:.3g} N")
print(f"f_(s,max) = {f_s_max:.3g} N")
print(f"f_k = {f_k:.3g} N")
for P in P_vals:
f = P if P <= f_s_max else f_k
regime = "static" if P <= f_s_max else "kinetic"
print(f"P = {P:.3g} N -> f = {f:.3g} N ({regime})")
# Piecewise plot: friction magnitude vs applied force
P = np.arange(0.0, 200, 0.1)
f = np.where(P <= f_s_max, P, f_k)
fs = 'large'
fig = plt.figure(figsize=(6, 4), dpi=140)
ax = fig.add_subplot(111)
ax.plot(P, f,'k-', lw=2)
ax.axvline(f_s_max, linestyle="--", lw=1.5)
ax.axhline(f_k, linestyle="--", lw=1.5)
ax.set_xlabel(r"Applied force $P$ (N)")
ax.set_ylabel(r"Friction force magnitude $f$ (N)")
ax.text(0.02, 0.95, rf"$f_{{s,\max}}=\mu_s N=%2d\ \mathrm{{N}}$" % f_s_max, transform=ax.transAxes, va="top")
ax.text(0.02, 0.88, rf"$f_k=\mu_k N=%2i\ \mathrm{{N}}$" % f_k, transform=ax.transAxes, va="top")
ax.set_xlim(0, P.max())
ax.set_ylim(0, max(f_s_max, f_k) * 1.15)
ax.set_xticks(np.arange(0,220,20))
# ax.set_yticks([])
ax.minorticks_on()
ax.tick_params()
ax.tick_params(which='major',direction='out',length=8.0,width=2.0,labelsize=fs)
ax.tick_params(which='minor',direction='out',length=4.0,width=2.0,labelsize=fs)
plt.tight_layout()
plt.show()
N = 196 N
f_(s,max) = 140 N
f_k = 120 N
P = 20 N -> f = 20 N (static)
P = 30 N -> f = 30 N (static)
P = 120 N -> f = 120 N (static)
P = 180 N -> f = 120 N (kinetic)
6.2.2. Friction and the Inclined Plane#
Friction plays an obvious role when there is an object on a slope. It could be a crate pushed up a ramp or a skateboarder coasting down a mountain, but the basic physics is the same.
We generalize the sloping surface and call it an inclined plane. Recall the Weight on an incline problem, where we orient our system axes so that the motion is along the \(x\)-axis only (i.e., flat).
When an object rests on a horizontal surface, the normal force supporting it is equal in magnitude to its weight. Additionally, simple friction is always proportional to the normal force. However with the inclined plane, we must find the force acting on the object that is directed perpendicular to the surface (i.e., it is a component of the weight).
Consider an object that slides down an inclined plane at constant velocity (i.e., the net force is zero). Such conditions can be used to measure the coefficient of kinetic friction between two objects.
Recall the free-body diagram from Chapter 5 (Fig. 5.18), where we found that \(w_x = w\sin{\theta}\) and the friction opposes the motion along the \(x\)-axis. The normal force is opposing the weight along the \(y\)-axis, or \(N = w\cos{\theta}\).
Therefore, the kinetic friction on a slope is \(f_k = \mu_k N = \mu_k mg\cos{\theta}\). Writing this out in terms of forces, we have
Now we can solve for \(\mu_k\) to get
6.2.2.1. Example Problem: Downhill Skier#
Exercise 6.11
The Problem
A skier with a mass of \(62\,\mathrm{kg}\) is sliding down a snowy slope at a constant acceleration. Find the coefficient of kinetic friction for the skier if the friction force is known to be \(45.0\,\mathrm{N}\).
Fig. 6.12 Image Credit: Openstax.#
The Model
The skier is modeled as a point mass sliding down a rigid incline at a fixed angle of \(25^\circ\) relative to the horizontal. Air resistance is neglected. The skier remains in contact with the slope at all times, so there is no motion perpendicular to the surface.
A coordinate system is chosen such that:
One axis is parallel to the slope.
One axis is perpendicular to the slope.
Forces acting on the skier are:
Weight \(\vec{W}\) acting vertically downward.
Normal force \(\vec{N}\) perpendicular to the slope.
Kinetic friction \(\vec{f}_k\) acting up the slope, opposing motion.
See the Weight on an incline problem for a free-body diagram.
The Math
The coefficient of kinetic friction is defined as the ratio of the kinetic friction force to the normal force,
Because there is no acceleration perpendicular to the slope, the normal force equals the component of the skier’s weight perpendicular to the surface.
Substituting this into the definition of kinetic friction gives the governing equation,
(a) The skier is sliding down the slope, so kinetic friction applies. We now evaluate the coefficient using the given values.
The Conclusion
The coefficient of kinetic friction for the skier is \(\mu_k = 0.082.\) This value is reasonable for skis on snow and is consistent with the assumption that the skier experiences kinetic friction while sliding down the slope (see Table 6.1 for Waxed wood on wet snow where \(\mu_k \approx 0.1\)).
The Verification
We verify the result by recomputing the normal force and confirming that the ratio \(f_k/N\) matches the reported coefficient.
import numpy as np
m = 62 # kg
g = 9.81 # m/s^2
theta = np.deg2rad(25)
f_k = 45.0 # N
N = m * g * np.cos(theta)
mu_k = f_k / N
print(f"Normal force N = {N:.3g} N")
print(f"Coefficient of kinetic friction μ_k = {mu_k:.3g}")
6.2.3. Atomic-scale Explanations of Friction#
The simpler aspects of friction rely on its macroscopic (large-scale) characteristics. Researchers are finding that the atomic nature of friction also has several fundamental microscopic (small-scale) characteristics. These characteristics hold the potential for the development of nearly frictionless environments that helps reduce mechanical wear and energy loss due to heat.
So far we have noted that friction is proportional to the normal force, but not to the amount of area in contact. When two rough surfaces are in contact, the actual contact area is a tiny fraction of the total area because only high spots touch. When a greater normal force is exerted, the actual contact area increases, and we find that the friction is proportional to this area, which is illustrated in Fig. 6.13.
Fig. 6.13 Image Credit: Openstax.#
The atomic scale seeks to explain, “why do surfaces get warmer when rubbed?”
Essentially, atoms are linked with one another to form lattices. When surfaces rub, the surface atoms adhere and cause atomic lattices to vibrate (i.e., create sound waves). The sound waves diminish with distance, and their energy is converted to heat.
Fig. 6.14 Image Credit: Openstax.#
Chemical reactions related to frictional wear can also occur between atoms and surfaces. Figure 6.14 shows how the tip of a probe drawn across another material is deformed by atomic-scale friction. The force needed to drag the tip can be measured (and is related to shear stress). The variation in shear stress is more than a factor of \(10^{12}\) and difficult to predict theoretically.
6.2.3.1. Example Problem: Sliding Blocks#
Exercise 6.12
The Problem
The two blocks of Figure 6.15 are attached to each other by a massless string that is wrapped around a frictionless pulley. When the bottom \(4.0\ \mathrm{kg}\) block is pulled to the left by the constant force \(\vec{P}\), the top \(2.0\ \mathrm{kg}\) block slides across it to the right. Find the magnitude of the force necessary to move the blocks at constant speed. Assume that the coefficient of kinetic friction between all surfaces is \(0.40\).
Fig. 6.15 Image Credit: Openstax.#
The Model
The system consists of two blocks connected by a massless string over a frictionless pulley. Both blocks move with constant speed, so their accelerations are zero. Motion occurs only in the horizontal direction. Vertical forces balance for each block. Kinetic friction acts at both contact surfaces: between the two blocks and between the lower block and the floor. The normal forces at each contact determine the magnitudes of the kinetic friction forces. The string tension is the same throughout.
The Math
We apply Newton’s 2nd law to each block separately. Since the blocks move at constant speed, the acceleration of each block is zero in both the horizontal and vertical directions.
(a) Starting with the top block (\(m_1 = 2.0\ \mathrm{kg}\)), which slides to the right at constant speed. The horizontal forces on it are the string tension \(T\) to the right and the kinetic friction force \(\mu_k N_1\) to the left. Vertically, the normal force \(N_1\) balances the weight.
The weight of the top block is \(w_1 = m_1 g\), where we solve the vertical equation to get \(N_1 = m_1 g.\) Substituting this into the horizontal equation yields
(b) For the bottom block (\(m_2 = 4.0\ \mathrm{kg}\)), it moves to the left at constant speed. Horizontally, it experiences the applied force \(P\) to the left, the string tension \(T\) to the right, kinetic friction from the top block of magnitude \(\mu_k N_1\) to the right, and kinetic friction from the floor of magnitude \(\mu_k N_2\) to the right. Vertically, the normal force from the floor balances the block’s weight and the downward normal force from the top block.
The weight of the bottom block is \(w_2 = m_2 g.\) Solving the vertical equation gives
We can substitute \(N_1 = m_1 g\) and \(N_2\) into the horizontal force balance, and use the previously found expression for \(T\) to obtain
Using \(g = 9.81\ \mathrm{m/s^2}\) and \(\mu_k = 0.40\), we have
The Conclusion
The magnitude of the applied force required to move the blocks at constant speed is \(P = 39\ \mathrm{N}.\) This force exactly balances the total kinetic friction acting on the system, consistent with the assumption of zero acceleration.
The Verification
We verify the result by numerically computing all friction forces and confirming that their sum equals the applied force when acceleration is zero.
import numpy as np
g = 9.81
mu_k = 0.40
m1 = 2.0
m2 = 4.0
N1 = m1 * g
N2 = m2 * g + N1
T = mu_k * N1
P = T + mu_k * N1 + mu_k * N2
print(f"N1 = {N1:.3g} N")
print(f"N2 = {N2:.3g} N")
print(f"T = {T:.3g} N")
print(f"P = {P:.3g} N")
6.2.3.2. Example Problem: Crate on an Accelerating Truck#
Exercise 6.13
The Problem
A \(50.0\ \mathrm{kg}\) crate rests on the bed of a truck as shown in Figure 6.16. The coefficients of friction between the surfaces are \(\mu_k = 0.30\) and \(\mu_s = 0.40\). Find the frictional force on the crate when the truck is accelerating forward relative to the ground at (a) \(2.0\ \mathrm{m/s^2}\) and (b) \(5.0\ \mathrm{m/s^2}\).
Fig. 6.16 Image Credit: Openstax.#
The Model
The crate is modeled as a particle in contact with the horizontal truck bed. The reference frame is inertial and fixed to the ground. Forces acting on the crate are its weight \(\vec{W}\), the normal force \(\vec{N}\) exerted by the truck bed, and the frictional force \(\vec{f}\) exerted by the truck bed. Air resistance is neglected. The crate may either remain at rest relative to the truck bed, in which case static friction acts, or slip relative to the bed, in which case kinetic friction acts. The vertical acceleration of the crate is zero.
The Math
Newton’s 2nd law is applied independently in the horizontal and vertical directions. The vertical force balance determines the normal force, which sets the maximum possible static friction and the kinetic friction magnitude.
The governing equations are
From the vertical force balance,
The maximum static friction is \(f_{s,\max} = \mu_s N,\) and the kinetic friction magnitude is \(f_k = \mu_k N.\)
(a) Starting with the truck acceleration \(a = 2.00\ \mathrm{m/s^2}\). If the crate does not slip, static friction must provide the horizontal acceleration of the crate. Applying Newton’s second law in the horizontal direction gives \(f_s = m a.\)
Using the given values,
The normal force is
so the maximum static friction is
Since \(f_s < f_{s,max}\), the crate does not slip and the frictional force is \(f = 100\ \mathrm{N}.\)
(b) Now, we evaluate the truck acceleration \(a = 5.00\ \mathrm{m/s^2}\). If the crate were to remain at rest relative to the truck, the required static friction would be
This exceeds the maximum static friction of \(200\ \mathrm{N}\), so the crate must slip. Kinetic friction therefore acts, with magnitude
Thus, the frictional force on the crate is \(f = 150\ \mathrm{N}.\)
The horizontal acceleration of the crate relative to the ground is now found from
The Conclusion
When the truck accelerates at \(2.0\ \mathrm{m/s^2}\), static friction of magnitude \(100\ \mathrm{N}\) acts on the crate and prevents slipping. When the truck accelerates at \(5.0\ \mathrm{m/s^2}\), the required static friction exceeds its maximum value, the crate slips, and kinetic friction of magnitude \(150\ \mathrm{N}\) acts on the crate. These results follow directly from comparing the required horizontal force to the maximum possible static friction set by the normal force.
Relative to the ground, the truck is accelerating forward at \(5.0\ {\rm m/s^2}\) and the crate is accelerating forward at \(2.9\ {\rm m/s^2}.\) The crate is sliding backward relative to the bed of the truck with an acceleration \(2.9\ {\rm m/s^2} - 5.0\ {\rm m/s^2} = -2.1\ {\rm m/s^2}.\)
The Verification
We verify the friction regimes by numerically computing the required static friction and comparing it to the maximum static friction, then computing the kinetic friction magnitude.
import numpy as np
m = 50.0
g = 9.81
mu_s = 0.40
mu_k = 0.30
N = m * g
fs_max = mu_s * N
fk = mu_k * N
a_vals = np.array([2.0, 5.0])
fs_required = m * a_vals
print(f"Normal force N = {N:.3g} N")
print(f"Maximum static friction = {fs_max:.3g} N")
print(f"Kinetic friction = {fk:.3g} N")
print(f"Required static friction at 2.00 m/s^2 = {fs_required[0]:.3g} N")
print(f"Required static friction at 5.00 m/s^2 = {fs_required[1]:.3g} N")
6.2.3.3. Example Problem: Snowboarding#
Exercise 6.14
The Problem
Earlier, we analyzed the situation of a downhill skier moving at constant velocity to determine the coefficient of kinetic friction. Now let’s do a similar analysis to determine acceleration. The snowboarder of Figure 6.17 glides down a slope that is inclined at \(13^\circ\) to the horizontal. The coefficient of kinetic friction between the board and the snow is \(\mu_k = 0.20\). What is the acceleration of the snowboarder?
Fig. 6.17 Image Credit: Openstax.#
The Model
The snowboarder is treated as a particle sliding on a rigid inclined plane. Air resistance is neglected. The snowboarder remains in continuous contact with the slope, so a normal force acts perpendicular to the surface and kinetic friction acts parallel to the surface, opposing the motion. The slope is fixed at an angle \(\theta = 13^\circ\) relative to the horizontal.
A coordinate system is chosen with the \(x\)-axis along the slope, positive downhill, and the \(y\)-axis perpendicular to the slope. Motion occurs only along the slope, so the acceleration perpendicular to the slope is zero.
The Math
We apply Newton’s 2nd law in component form along and perpendicular to the slope. The forces acting on the snowboarder are the weight \(\vec{W}\), the normal force \(\vec{N}\), and the kinetic friction force \(\vec{f}_k\).
Resolving the weight into components relative to the incline gives a component \(w_x = mg\sin\theta\) along the slope and \(w_y = mg\cos\theta\) perpendicular to the slope. Kinetic friction has magnitude \(f_k = \mu_k N\) and acts up the slope.
Applying Newton’s 2nd law in the two coordinate directions gives
Using the force components, these become
From the second equation, the normal force is \(N = mg\cos\theta.\) Substituting this expression for \(N\) into the friction force and then into the \(x\)-equation yields the general result
We now evaluate this expression numerically using \(g = 9.81\ \mathrm{m/s^2}\) and \(\theta = 13^\circ\):
The Conclusion
The snowboarder accelerates downhill with a magnitude of \(0.29\ \mathrm{m/s^2}\). This acceleration is smaller than the component of gravity alone (\(a_{gx} = g\sin{\theta} = 2.2\ {\rm m/s^2}\)) because kinetic friction opposes the motion along the slope. The result follows directly from balancing gravity and friction along the incline under the assumption of steady sliding.
If \(a_x <0\), that would mean \(\sin\theta - \mu_k \cos\theta<0\), or \( \tan{\theta}<\mu_k. \) For the snowboarder to slow down, the \(\mu_k\) needs to be sufficiently large, or the \(\theta\) needs to be small so that the above is satisfied.
The Verification
We verify the result by directly evaluating the expression \(a_x = g(\sin\theta - \mu_k\cos\theta)\) numerically using Python.
import numpy as np
g = 9.81
theta = np.deg2rad(13.0)
mu_k = 0.20
a = g * (np.sin(theta) - mu_k * np.cos(theta))
print(f"Acceleration = {a:.2f} m/s^2")
This computation confirms an acceleration of \(0.29\ \mathrm{m/s^2}\), consistent with the analytical result.
6.3. Centripetal Force#
Recall that an object undergoing circular motion must be accelerating because the direction of the object’s velocity is changing. This centripetal acceleration is given by:
where the magnitude of \(\vec{\rm v}\) is \(v\) (or its speed) and \(r\) is the radius of the curve. The velocity \(\vec{\rm v}\) is directed along a tangent line to the curve at any instant, which is represented by the unit vector \(\hat{\theta}\) and is perpendicular to the radial unit vector \(\hat{r}\).
Note
The velocity can be represented as \(\vec{\rm v} = v\, \hat{\theta}\), where \(v^2\) is determined through the dot product or
If we also know the angular speed \(\omega\) (measured in \(\text{rad/s}\)), then we can write the magnitude of the centripetal acceleration \(a_c\) in two ways:
where this also referred to as a radial acceleration.
An acceleration must be produced by a force, where any force (or combination of forces) can cause a centripetal acceleration. For example,
the rope on a tether ball,
Earth’s gravity on the Moon,
friction between roller skates and the rink floor,
a banked roadway’s force on a car, and
forces on the tube of a spinning centrifuge.
Any net force causing uniform circular motion is called a centripetal force. The direction of the centripetal force is radial, meaning it is directed in the \(\hat{r}\) that is aligned with the circle’s radius. The word centripetal means ‘center-seeking’, so the direction is toward the center of curvature.
According to Newton’s 2nd law, the net force is mass times acceleration or \(F_{\rm net} = m a\). For uniform circular motion, the acceleration is the centripetal acceleration and the magnitude of the centripetal force is
Because we had 2 expressions for the centripetal acceleration, we therefore have two expressions for the centripetal force:
When solving problems, you may use whichever expression is more convenient (i.e., whichever has the proper balance of knowns vs. unknowns). Moreover, you may need to rearrange the equation to fit the given problem. Suppose that you know the mass, speed, and centripetal force needed to keep an object fixed to the ground, then you could solve the first expression for \(r\):
to determine the radius required to keep the system in balance. This implies that for a large centripetal force \(F_c\) requires a small \(r\) for an object with a given mass and speed (see Fig. {numref}`{number}
Fig. 6.18 Image Credit: Openstax#
6.3.1. Example Problem: Friction for Car on a Flat Curve#
Exercise 6.15
The Problem
What Coefficient of Friction Do Cars Need on a Flat Curve? (a) Calculate the centripetal force exerted on a \(900\ \text{kg}\) car that negotiates a \(500\ \text{m}\) radius curve at \(25\ \mathrm{m/s}\). (b) Assuming an unbanked curve, find the minimum static coefficient of friction between the tires and the road, static friction being the reason that keeps the car from slipping (Figure 6.19).
Fig. 6.19 Image Credit: Openstax#
The Model
We model the car as a particle moving in a horizontal circle of radius \(r\). The road is level (unbanked), so the vertical forces are the normal force \(\vec{N}\) upward and the weight \(\vec{W}\) downward, where they balance. The only horizontal force is static friction \(\vec{f}_s\), which points toward the center of the circular path and provides the required centripetal acceleration. We assume rolling without slipping, neglect air resistance and, neglect any drivetrain details. We use \(g = 9.81\ \mathrm{m/s^2}\).
The Math
The car’s inward (radial) acceleration has magnitude \(a_c = v^2/r\), so the required centripetal force is
For a level road, the vertical forces balance so the normal force magnitude is
Static friction can adjust up to a maximum magnitude \(f_{s,\max}=\mu_s N\). To avoid slipping on an unbanked curve, static friction must be at least the required centripetal force:
Canceling \(m\) and solving for the minimum coefficient gives
(a) The car moves at speed \(v\) around a curve of radius \(r\), so the required centripetal force is
(b) The curve is unbanked, so static friction must supply the centripetal force. The minimum static coefficient is
The Conclusion
For a \(900\ \mathrm{kg}\) car traveling at \(25\ \mathrm{m/s}\) around a level curve of radius \(500\ \mathrm{m}\), the required inward (centripetal) force is \(F_c = 1100\ \mathrm{N}\). On an unbanked road, that inward force must come from static friction, so the road-tire system must provide at least \(\mu_s = 0.13\) to prevent slipping. This result follows from the level-road assumption that \(N=mg\) and from requiring that static friction be able to match the needed centripetal force.
The Verification
We verify the arithmetic for \(F_c\) and \(\mu_{s,\min}\) using direct numerical evaluation.
import numpy as np
m = 900.0 # kg
r = 500.0 # m
v = 25.00 # m/s
g = 9.81 # m/s^2
Fc = m * v**2 / r
mu_min = v**2 / (r * g)
print(f"F_c = {Fc:.2g} N")
print(f"mu_s,min = {mu_min:.2f}")
6.3.2. Banked Curves#
For banked curves, the slope of the road helps you make the curve (see Fig. 6.20); that’s why they have them in Nascar. The greater the angle \(\theta\) (relative to flat ground), the faster you can take the curve. In an ideally banked curve, the angle \(\theta\) the angle \(\theta\) allows you to traverse the curve at a certain speed without the aid of friction between the tires and the road.
Fig. 6.20 Image Credit: Openstax#
For ideal banking, the net external force equals the horizontal centripetal force in the absence of friction (i.e., \(F_{\rm net} = ma_{c,x}\)). The components of the normal force \(\vec{N}\) in the horizontal and vertical directions must equal the centripetal force and the weight of the car, respectively.
Figure 6.20 shows a free-body diagram for a car on a frictionless banked curve. However, you should recognize that the setup is very similar to the sliding block on an inclined plane. If the angle \(\theta\) is ideal for the speed and radius, then the net external force equals the necessary centripetal force.
The only two external forces acting on the car are: (1) its weight \(\vec{w}\) and (2) the normal force \(\vec{N}\) of the road. These two forces must add to give a net external force that is horizontal toward the center of curvature (to be centripetal) and has magnitude \(mv^2/r\). Only the normal force has a horizontal component (recall this from Common Forces), so we have
Because the car does not leave the surface of the road (i.e., no flying cars–yet), the net vertical force must be zero. The only vertical forces are the vertical component of the normal force and the car’s weight. Thus, we have
We can combine these two equations to eliminate the normal force \(N\) and get an expression for \(\theta\). We have
Taking the inverse tangent gives
This expression shows that a large \(\theta\) is obtained if either \(v\) is larger or \(r\) is small. Roads must be steeply banked for high speeds and sharp curves. Friction allows you to take the curve at greater or lower speed than if the curve were frictionless. Note that \(\theta\) does not depend on the mass of the vehicle.
Airplanes also make turns by banking, where the lift force acts at right angles to the wing. When the airplane banks, there is greater lift than required for level flight. The vertical component of the lift balances the airplanes weight (like the Normal force), and the horizontal component accelerates the plane.
6.3.2.1. Example Problem: Ideal Speed Banked Tight Curve#
Exercise 6.16
The Problem
Curves on some test tracks and race courses, such as Daytona International Speedway, are very steeply banked. This banking allows the curves to be taken at high speed even without tire friction. Calculate the speed at which a \(100\ \mathrm{m}\) radius curve banked at \(31^\circ\) should be driven if the road is frictionless.
The Model
The car is modeled as a particle moving in uniform circular motion on a rigid, frictionless, banked surface. The road is inclined at a fixed angle \(\theta\) relative to the horizontal, and the car follows a circular path of radius \(r\). The only forces acting on the car are its weight \(\vec{W}\) and the normal force \(\vec{N}\) from the road. Air resistance and tire friction are neglected. The vertical and horizontal axes are chosen so that vertical is aligned with gravity and horizontal points toward the center of the circular path (i.e., a banked curve, see Fig. 6.20). We use \(g = 9.81\ \mathrm{m/s^2}\).
The Math
For uniform circular motion, the car requires a centripetal acceleration of magnitude \(a_c = v^2/r\) directed horizontally toward the center of the curve.
The forces acting on the car are its weight \(\vec{W}\) downward and the normal force \(\vec{N}\) perpendicular to the road surface. Resolving the normal force into horizontal and vertical components gives
We divide the horizontal equation by the vertical equation to eliminate \(N\) and \(m\),
Then, we solve for the speed \(v\), which yields
Substituting the given values,
The Conclusion
For a frictionless road banked at \(31^\circ\) with a curve radius of \(100\ \mathrm{m}\), the ideal speed at which a car can travel without slipping up or down the bank is \(24.3\ \mathrm{m/s}\) (or \(54.4\ \text{mph}\)). At this speed, the horizontal component of the normal force provides exactly the required centripetal force while the vertical component balances the car’s weight.
The Verification
The result is verified by directly evaluating the expression \(v=\sqrt{rg\tan\theta}\) numerically using the given parameters.
import numpy as np
r = 100 # m
g = 9.81 # m/s^2
theta = np.deg2rad(31)
v = np.sqrt(r * g * np.tan(theta))
print(f"Ideal speed = {v:.3g} m/s")
mi2meter = 1609.34 #meters in 1 mile
hr2sec = 3600 #seconds in an hour
v *=hr2sec/mi2meter
print(f"Ideal speed = {v:.3g} mph")
6.3.3. Inertial Forces and Noninertial Frames: The Coriolis Force#
Consider the following:
taking off in a jet airplane,
turning a corner in a car,
riding a merry-go-round, and
the circular motion of a tropical cyclone.
What do they all have in common?
Each exhibits fictitious (non-inertial) forces that merely seem to arise from motion because the observer’s frame of reference is accelerating or rotating.
When taking off in a jet, it feels as if you are being pushed back into the seat as the airplane accelerates. A physicist would say that you tend to remain stationary while the seat pushes you forward on you.
When you make a tight curve in your car to the right, you feel as if you’re thrown (or forced) to the left. However, you are actually going in a straight line (i.e., Newton’s 1st law), but the car moves to the right.
We can reconcile these points of view by examining the frames of reference. Passengers instinctively use the car as the frame of reference, whereas physicists would use the Earth instead. The Earth is nearly an inertial frame fo reference, where all forces have an identifiable physical origin. Newton’s laws behave just as expected.
Fig. 6.21 Image Credit: Openstax#
The car is a noninertial frame of reference because it is accelerated to the side, where the force to the left sensed by the passengers is an inertial force having no physical origin. It is purely due to the inertia of the passenger (i.e., moving in a straight line) rather than some physical cause (e.g., tension, friction, or gravitation).
Consider a rapidly rotating merry-go-round, where you take the merry-go-round as your frame of reference because you rotate together. When rotating in that noninertial frame of reference, you feel an inertial force that tends to throw you off. This is often referred to as a centrifugal force because your moving outward.
Fig. 6.22 Image Credit: Openstax#
The centrifugal force does not actually exist, where you must hang on tightly to counteract your inertia (i.e., tendency to move in a straight line). You must hang on to make yourself go in a circle because Newton’s 1st law would have you go off in a straight line. The force you exert acts toward the center of the circle.
This inertial effect is put to good use in centrifuges. A centrifuge spins a sample very rapidly, where the inertial force throws particles outward and hastens their sedimentation. The greater the angular speed \(\omega\), the greater the centrifugal force. What really happens is the inertia of the particles carries them along a line tangent to the circle while the test tube is forced in a circular path by a centripetal force.
Fig. 6.23 Image Credit: Openstax#
Consider what happens if something moves in a rotating frame of reference.
What if you slide a ball directly away from the center of the merry-go-round?
The ball follows a straight path relative to Earth and a path curved to the right on the merry-go-round’s surface. A person standing next to the merry-go-round sees the ball moving straight and the merry-go-round rotating underneath it. The apparent curve to the right is caused by an inertial force called the Coriolis force.
6.4. Drag Force and Terminal Speed#
You feel the drag force when you move your hand through water or during a strong wind (i.e., outside the car window).
The faster you move your hand, the harder it is to move.
You feel a smaller drag force when you tilt your hand so only the side goes through the air (i.e., decreased your surface area in the direction of motion).
6.4.1. Drag Forces#
The drag force always opposes the motion of an object (like friction). Unlike simple friction, the drag force is proportional to some function of the velocity of the object in that fluid. Which function to choose is complicated and depends on several factors:
shape of the object (its aerodynamics),
size,
velocity, and
the fluid it is in.
For most large objects (e.g., cyclists, cars, baseballs) not moving too slowly, the magnitude of the drag force \(F_{\rm D}\) is proportional to the square of the speed of the object (i.e., \(F_{\rm D}\propto v^2\)). The full relationship becomes
where \(C\) is the drag coefficient, \(A\) is the cross-sectional area (i.e., side facing the fluid), and \(\rho\) is the density of the fluid. Recall that density is mass per unit volume (\(\rho = m/{\rm V}\)).
The more generalized form is \(F_{\rm D} = bv^n\), where the constant \(b = 0.5C \rho A\) bundles all the constants together.
The exponent \(n=2\) comes from our assumption about moving at high speed. If the object is moving a low speed then \(n=1\).
Fig. 6.24 Image Credit: OpenStax#
Athletes (as well as car designers) seek to reduce the drag force to lower their race times. Aerodynamic shaping of an automobile can reduce the drag force, and thus increase a car’s gas mileage. The value of the drag coefficient is determined empirically either by using a wind tunnel or through a complex numerical simulation on a supercomputer.
The drag coefficient can depend on velocity, but we assume that it is constant here. See the table (Typical Values of Drag Coefficient) below that lists some typical drag coefficients for a variety of objects, while Drag Coefficients by Geometric Shape provides values by the geometric shape. The drag coefficient is a dimensionless quantity.
Object |
\(C\) |
|---|---|
Airfoil |
0.05 |
Toyota Camry |
0.28 |
Ford Focus |
0.32 |
Honda Civic |
0.36 |
Ferrari Testarossa |
0.37 |
Dodge Ram Pickup |
0.43 |
Sphere |
0.45 |
Hummer H2 SUV |
0.64 |
Skydiver (feet first) |
0.70 |
Bicycle |
0.90 |
Skydiver (horizontal) |
1.0 |
Circular flat plate |
1.12 |
Fig. 6.25 Image Credit: Wikipedia:drag_coefficient#
At highway speeds, over \(50\%\) of the power of the car is used to overcome air drag. For average conditions (e.g., altitude and temperature), the mos fuel-efficient cruising speed is about \(45\)-\(50\ {\rm mph}\) (or about \(70\)-\(80\ {\rm km/h}\)).
Substantial research is undertaken to minimize drag in racing sports as well as specific sporting instruments. For example, the dimples on golf balls are being redesigned as well as the clothes athletes wear. Most elite swimmers and cyclists shave their body hair to reduce drag and can slice milliseconds of a race time, which can be the difference between a gold or silver medal.
6.4.2. Terminal Velocity#
Consider a skydiver falling under the influence of gravity. The 2 forces acting on him are the force of gravity \(F_g\) and the drag force \(F_{\rm D}\) (ignoring the small buoyant force caused by displacing the air). The downward force of gravity remains roughly constant (i.e., the altitude above Earth’s surface is much less than Earth’s radius) and does not depend on his velocity.
However, the magnitude of the drag force increases with velocity squared (assuming his descent is sufficiently fast) until it equals the gravitational force, thus producing a net force of zero and stops accelerating. At this point the person’s velocity remains constant and he has reached his terminal velocity \(v_{\rm T}\).
At the terminal velocity,
Using Equation (6.5), we solve for \(v_{\rm T}\)
Assume the density of air is \(\rho = 1.21\ {\rm kg/m^3}\), and a skydiver with a mass of \(75\ {\rm kg}\) whose cross-sectional area is \(A=0.18\ {\rm m^2}\). He has a drag coefficient \(C = 0.70\), which means we can calculate his terminal velocity \(v_{\rm T}\) as,
This means a skydiver with a with a mass of \(75\ {\rm kg}\) achieves a terminal velocity of
\(350\ {\rm km/h}\) in a headfirst position (minimizing the area and his drag),
\(200\ {\rm km/h}\) in a spread-eagle position (area has increased).
Equation (6.5) does not hold if the object is very small, is going very slow, or is in a denser medium than air. For these cases, we find the drag force is proportional to just the velocity (i.e., \(F_{\rm D} \propto v\)), This relation is given by Stokes’ law.
Stokes’ Law
For a spherical object falling in a medium, the drag force is
where \(r\) is the radius of the sphere, \(\eta\) is the viscosity of the fluid, and \(\vec{\rm v}\) is the object’s velocity.
Good examples of Stokes’ law are provided by microorganisms, pollen, and dust particles. Each of these objects is small, where we find that many of these objects travel unaided only at a constant (terminal) velocity.
6.4.2.1. Example Problem: Terminal Velocity of a Skydiver#
Exercise 6.17
The Problem
Find the terminal velocity of an \(85\ \mathrm{kg}\) skydiver falling in a spread-eagle position. Assume that in the spread-eagle position, the diver has a cross-sectional area of \(0.70\ \mathrm{m^2}\).
The Model
The skydiver is modeled as a particle moving vertically through air under the influence of gravity and air drag. The motion is one-dimensional and downward. At terminal velocity, the skydiver moves at constant speed, so the acceleration is zero. The forces acting on the skydiver are the weight \(\vec{w}\) acting downward and the drag force \(\vec{F}_D\) acting upward, opposite the direction of motion. The drag force is modeled using the quadratic drag law appropriate for motion through air at high speed. Air density is assumed constant, and lift forces are neglected.
The Math
Using Newton’s second law in the vertical direction gives
This force balance implies
The drag force for motion through air at speed \(v\) is given by
where \(\rho\) is the air density, \(C\) is the drag coefficient, and \(A\) is the cross-sectional area.
Equating drag and weight yields
We solve this expression for the terminal speed \(v_T\), which gives
Substituting the given values,
The Conclusion
The terminal velocity of the \(85\ \mathrm{kg}\) skydiver in a spread-eagle position is \(\boxed{44\ \mathrm{m/s}}\). At this speed, the upward drag force exactly balances the downward force of gravity, resulting in zero acceleration and constant downward motion.
The Verification
The result is verified by numerically evaluating the terminal velocity formula using the given parameters.
import numpy as np
m = 85.0 # kg
g = 9.81 # m/s^2
rho = 1.21 # kg/m^3
C = 1.0
A = 0.70 # m^2
v_T = np.sqrt((2*m*g)/(rho*C*A))
print(f"Terminal velocity = {v_T:.3g} m/s")
6.4.3. The Calculus of Velocity-Dependent Frictional Forces#
When a body slides across a surface, the frictional force is approximately constant and given by \(\mu_k N\). The drag force through air is generally a complicated function of the body’s velocity. However, for a body moving in a straight line at moderate speeds through a liquid (e.g., water), the frictional force can be approximated by
where \(b\) is a constant whose value depends on the dimensions (and shape) of the body and the properties of the liquid.
Consider an object falling through a liquid. Newton’s 2nd law in the vertical direction gives the differential equation
where we have written the acceleration as \(dv/dt\). As \(v\) increases, the frictional force increase until it matches the weight \(mg\). At this point there is no acceleration and the velocity remains constant at the terminal velocity \(v_{\rm T}\). Thus, we have
We can find the object’s velocity by integrating the differential equation for \(v\). First, we rearrange the equation to obtain
Assuming that \(v=0\) at \(t=0\), integration of the equation yields
where \(v^\prime\) and \(t^\prime\) are dummy variables of integration. Apply the limits of integration, we find
Note
We can use the properties of logarithms to simplify expressions using the following identities:
Using the properties of logarithms, we find
Solving for \(v\), we find
Notice that as \(t\rightarrow \infty,\ v\rightarrow mg/b = v_{\rm T}\), which is the terminal velocity. The position at any time can be found by integration the equation for \(v\). with \(v = dy/dt\), we have
Assuming that \(y=0\), when \(t=0\),
which integrates to
6.4.3.1. Example Problem: Resistive Force on a Motorboat#
Exercise 6.18
The Problem
A motorboat is moving across a lake at a speed \(v_o\) when its motor suddenly freezes up and stops. The boat then slows down under the frictional force \(\vec{f}_R = -bv\,\hat{i}\). (a) What are the velocity and position of the boat as functions of time? (b) If the boat slows down from \(4.0\ \mathrm{m/s}\) to \(1.0\ \mathrm{m/s}\) in \(10\ \mathrm{s}\), how far does it travel before stopping?
The Model
The boat is modeled as a particle moving in one dimension along the horizontal surface of the lake. After the motor stops, the only horizontal force acting on the boat is a resistive force proportional to the boat’s speed and opposite the direction of motion. Vertical forces are neglected because they balance and do not affect the horizontal motion. The resistive force is assumed to be linear in speed, with proportionality constant \(b\). The reference frame is inertial and fixed to the shore.
The Math
The force acting on the boat after the motor stops is a resistive force opposite the motion. Applying Newton’s second law along the direction of motion gives
We rearrange this equation to separate variables,
(a) To determine the velocity as a function of time, we integrate from \(v_o\) at \(t=0\) to \(v\) at time \(t\),
Evaluating the integrals gives
which we solve for the velocity,
The position is obtained from the definition of velocity. Since \(v = dx/dt\), we write
Integrating from \(x=0\) at \(t=0\) to \(x\) at time \(t\) gives
As time increases, the exponential term approaches zero, and the position approaches a limiting value
(b) To determine the numerical distance traveled, we use the given change in speed. The velocity expression gives
Substituting \(v = 1.0\ \mathrm{m/s}\), \(v_o = 4.0\ \mathrm{m/s}\), and \(t = 10\ \mathrm{s}\),
Taking the natural logarithm and rearranging (via algebra),
The limiting distance traveled is then
The Conclusion
After the motor stops, the boat’s velocity decreases exponentially according to \(v(t) = v_o e^{-bt/m}\), and its position approaches a limiting distance \(x_{\max} = m v_o/b\). For the given numerical values, the boat travels approximately \(29\ \mathrm{m}\) before effectively coming to rest. Although the model predicts that the boat never fully stops in finite time, the speed becomes negligibly small after a short interval.
The Verification
The result is verified by numerically evaluating the exponential decay of velocity and confirming that the position approaches the limiting distance \(x_{\max}\).
import numpy as np
v0 = 4.0 # m/s
b_over_m = 0.14 # 1/s
t = np.linspace(0, 60, 500)
v = v0 * np.exp(-b_over_m * t)
x = (v0 / b_over_m) * (1 - np.exp(-b_over_m * t))
print(f"The boat's limiting distance is {v0 / b_over_m:.2g} m".)
print(f"The boat's speed at t = 10 s is {v0*np.exp(-b_over_m*10):.2g} m/s.")
6.5. In-class Problems#
6.5.1. Part I#
Problem 1
Find the tension in each of the three cables supporting the traffic light if it weighs \(2.00 \times 10^{2}\ \mathrm{N}\).
Fig. 6.26 Image Credit: Openstax.#
Problem 2
A large rocket has a mass of \(2.00 \times 10^{6}\ \mathrm{kg}\) at takeoff, and its engines produce a thrust of \(3.50 \times 10^{7}\ \mathrm{N}\).
(a) Find its initial acceleration if it takes off vertically.
(b) How long does it take to reach a velocity of \(120\ \mathrm{km/h}\) straight up, assuming constant mass and thrust?
Problem 3
An elevator filled with passengers has a mass of \(1.70 \times 10^{3}\ \mathrm{kg}\).
(a) The elevator accelerates upward from rest at a rate of \(1.20\ \mathrm{m/s^2}\) for \(1.50\ \mathrm{s}\). Calculate the tension in the cable supporting the elevator.
(b) The elevator continues upward at constant velocity for \(8.50\ \mathrm{s}\). What is the tension in the cable during this time?
(c) The elevator accelerates opposite to the motion at a rate of \(0.600\ \mathrm{m/s^2}\) for \(3.00\ \mathrm{s}\). What is the tension in the cable during acceleration opposite to the motion?
Problem 4
Two blocks are connected by a massless rope as shown below. The mass of the block on the table is \(4.0\ \mathrm{kg}\) and the hanging mass is \(1.0\ \mathrm{kg}\). The table and the pulley are frictionless.
(a) Find the acceleration of the system.
(b) Find the tension in the rope.
(c) Find the speed with which the hanging mass hits the floor if it starts from rest and is initially located \(1.0\ \mathrm{m}\) from the floor.
Fig. 6.27 Image Credit: Openstax.#
6.5.2. Part II#
Problem 5
A \(120\ \mathrm{kg}\) wooden crate rests on a wood floor. The coefficient of static friction between the wood surfaces is \(\mu_s = 0.50\).
(a) What maximum horizontal force can you exert on the crate without causing it to move?
(b) If you continue to exert this force once the crate starts to slip, what will its acceleration be? The coefficient of kinetic friction is \(\mu_k = 0.300\).
Problem 6
Calculate the maximum acceleration of a car that is heading down a \(6.00^\circ\) slope (measured from the horizontal) under the following road conditions. You may assume that the weight of the car is evenly distributed on all four tires and that static friction is involved; that is, the tires do not slip during the acceleration opposite the direction of motion. Ignore rolling resistance.
(a) On dry concrete.
(b) On ice, assuming \(\mu_s = 0.100\).
Problem 7
What is the ideal banking angle for a gentle turn of radius \(1.20\ \mathrm{km}\) on a highway with a speed limit of \(105\ \mathrm{km/h}\) (about \(65\ \mathrm{mi/h}\)), assuming everyone travels at the speed limit?
Problem 8
Modern roller coasters have vertical loops like the one shown. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top is greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop if the radius of curvature there is \(15\ \mathrm{m}\) and the downward acceleration of the car is \(1.50\,g\)?
Fig. 6.28 Image Credit: Openstax.#
6.6. Homework#
6.6.1. Conceptual Problems#
Problem 1
When you learn to drive, you discover that you need to let up slightly on the brake pedal as you come to a stop or the car will stop with a jerk. Explain this in terms of the relationship between static and kinetic friction.
Problem 2
Define centripetal force. Can any type of force (e.g., tension, gravitational force, friction, etc.) be a centripetal force? Can any combination of forces be a centripetal force?
6.6.2. Numerical Problems#
Problem 3
A \(2.5\text{-kg}\) fireworks shell is fired straight up from a mortar and reaches a height of \(110\ \mathrm{m}\).
(a) Neglecting air resistance (a poor assumption, but we will make it for this example), calculate the shell’s velocity when it leaves the mortar.
(b) The mortar itself is a tube \(0.45\ \mathrm{m}\) long. Calculate the average acceleration of the shell in the tube as it goes from zero to the velocity found in (a).
(c) What is the average force on the shell in the mortar? Express your answer in newtons and as a ratio to the weight of the shell.
Problem 4
A team of eight dogs pulls a sled with waxed wood runners on wet snow (mush!). The dogs have average masses of \(19\ \mathrm{kg}\), and the loaded sled with its rider has a mass of \(210\ \mathrm{kg}\).
(a) Calculate the acceleration of the dogs starting from rest if each dog exerts an average force of \(185\ \mathrm{N}\) backward on the snow.
(b) Calculate the force in the coupling between the dogs and the sled.
Problem 5
(a) What is the radius of a bobsled turn banked at \(75^\circ\) and taken at \(30\ \mathrm{m/s}\), assuming it is ideally banked?
(b) Calculate the centripetal acceleration.
(c) Does this acceleration seem large to you?
Problem 6
A \(560\text{-g}\) squirrel with a surface area of \(930\ \mathrm{cm^2}\) falls from a \(5.0\ \mathrm{m}\) tree to the ground. Estimate its terminal velocity.
(Use a drag coefficient for a skydiver falling feet first.)
What will be the velocity of a \(56\text{-kg}\) person hitting the ground, assuming no drag contribution in such a short distance?