Static Equilibrium and Elasticity

12. Static Equilibrium and Elasticity#

Apr 07, 2026 | 8735 words | 44 min read

12.1. Conditions for Static Equilibrium#

A rigid body is in equilibrium when both its linear and angular acceleration are zero relative to an inertial frame of reference. This means that a body in equilibrium can be moving, where its linear and angular velocities are constant.

A rigid body is in static equilibrium when it’s at rest in our selected frame of reference. The state of rest and uniform motion is artificial, where an object may be at rest in our selected frame of reference but can appear to move with constant velocity in another frame with an observer moving at constant velocity. The laws of physics are identical for all inertial reference frames, so there is not a distinction between static equilibrium and equilibrium in an inertial frame of reference.

According to Newton’s 2nd law, the linear acceleration of a rigid body is caused by a net force acting on it, or

(12.1)#\[\begin{align} \sum_j \vec{F}_k = m\vec{a}_{\rm CoM}. \end{align}\]

The sum is of all external forces acting on the body, where $m$ is its mass and $\vec{a}_{\rm CoM}$ is the linear acceleration of its center of mass.

First Equilibrium Condition

The first equilibrium condition for the static equilibrium of a rigid body expresses translational equilibrium:

(12.2)#\[\sum_j \vec{F}_j = \vec{0}.\]

This vector equilibrium condition is equivalent to the following three scalar equations for the components of the net force:

(12.3)#\[\begin{align} \sum_j F_{jx} &= 0, \\ \sum_j F_{jy} &= 0, \\ \sum_j F_{jz} &= 0. \end{align}\]

We can state that the rotational acceleration $\vec{\alpha}$ of a rigid body about a fixed axis is caused by the net torque acting on the body, or

(12.4)#\[\begin{align} \sum_j \vec{\tau}_j = I\vec{\alpha}. \end{align}\]

The net torque depends on the moment of inertia $I$ about this axis and the summation over all torques $\vec{\tau}_j$ of external forces.

Second Equilibrium Condition

The second equilibrium for the static equilibrium of a rigid body expresses rotational equilibrium:

(12.5)#\[\begin{align} \sum_j \vec{\tau}_j = \vec{0}. \end{align}\]

This equation for equilibrium is generally valid for rotational equilibrium about any axis of rotation (fixed or otherwise). This vector equation is equivalent to three scalar equations for the vector components of the net torque:

(12.6)#\[\begin{align} \sum_j \vec{\tau}_{jx} = 0, \\ \sum_j \vec{\tau}_{jy} = 0, \\ \sum_j \vec{\tau}_{jz} = 0. \end{align}\]

The second equilibrium means that there is no net external torque to cause rotation about any axis.

The first and second equilibrium conditions are states in a particular reference frame. However, the second condition involves torque, which depends on the location of the axis in the reference frame. Recall that torque is defined through the cross product,

\[ \vec{\tau}_j = \vec{r}_j \times \vec{F}_j, \]

where $\vec{r}_j$ is the axis of rotation relative to a point in the reference frame. When rotational and translational equilibrium conditions are satisfied simultaneously in one frame of reference, then they also hold in any other inertial frame of reference and the net torque about any axis of rotation is zero.

Suppose vector $\vec{R}$ is the position of the origin of a new inertial frame of reference $S^\prime$ relative to the old inertial frame $S$. Then the position vector $\vec{r}_j^\prime$ of the point where the force $\vec{F}_j$ is applied is related to $\vec{r}_j$ by

\[ \vec{r}_j^\prime = \vec{r}_j - \vec{R}. \]

We can sum all torques $\vec{\tau}_j^\prime$ of all external forces in the new reference frame $S^\prime$:

\[\begin{align*} \sum_j \vec{\tau}_j^\prime &= \sum_j \vec{r}_j^\prime \times \vec{F}_j, \\ &= \sum_j (\vec{r}_j - \vec{R}) \times \vec{F}_j = \sum_j (\vec{r}_j \times \vec{F}_j) - (\vec{R} \times \vec{F}_j), \\ &= \underbrace{\sum_j \vec{\tau}_j}_{\rm Term\ 1} - \underbrace{\vec{R} \times \sum_j \vec{F}_j}_{\rm Term\ 2} = \vec{0}. \end{align*}\]

Term 1 vanishes because the sum of the torques goes to zero in the old reference frame $S$. Term 2 vanishes because it is ultimately a cross product with $\vec{0}$ through the first equilibrium condition (Eq. (12.2)) in the old reference frame $S$.

The practical application is that we are free to choose any point as the origin of the reference frame when applying equilibrium conditions for a rigid body. Or choice of reference frame is dictated by the physical specific of the problem we are solving. In one reference frame, the math may be quite complicated, whereas it could be much simpler in another reference frame. The origin of a selected frame of reference is called the pivot point.

In the most general case, equilibrium conditions are expressed by six scalar equations (i.e., 3 components for forces, and 3 components for torques). For planar equilibrium problems with rotation about a fixed axis, whe can reduce the number of equations to three.

The standard procedure is to choose frame of reference, where the $z$-axis aligns with the axis of rotation. Then the net torque has only a $z$ component, where all forces that have non-zero torques lie in the $xy$-plane. As a result, the contributions to the net torque only come from the $F_{jx}$ and $F_{jy}$ of external forces.

For planar problems with the axis of rotation perpendicular to the $xy$-plane, we have the following three equilibrium conditions for forces and torques:

(12.7)#\[\begin{align} F_{1x} + F_{2x} + \cdots + F_{Nx} &= 0, \\ F_{1y} + F_{2y} + \cdots + F_{Ny} &= 0, \\ \tau_1 + \tau_2 + \cdots + \tau_N &= 0, \end{align}\]

where the summation is over all $N$ external forces acting on the body and over their torques. Note, we dropped the $z$ subscript for the torques because all the forces were in the $xy$-plane and the other torques ($\tau_{jx}$ and $\tau_{jy}$) are zero in the cross product. Thus, the $z$-component of each torque $\vec{\tau}_j$ from the force $\vec{F}_j$ is

(12.8)#\[\begin{align} \tau_j &= r_jF_j\sin{\theta}, \end{align}\]

where $r_j$ is the lever arm of the force and $F_j$ is the magnitude of the force. The angle $\theta$ is the angle between vectors $\vec{r}_j$ and $\vec{F}_j$, measured starting from $\vec{r}_j$ in the counterclockwise (positive) direction using the right-hand rule.

sense of a torque — Fig. 12.1 Image Credit: Openstax.#

The magnitude of torque is computed from the sum of terms, where each one has a sense (positive or negative) depending on the direction of rotation. If the lever $\vec{r}_j$ is measured from the origin and the force $\vec{F}_j$ is applied at its tip, then we can translate the $\vec{r}_j$ radially so that it shares an origin with $\vec{F}_j$ to determine the sense (see Fig. 12.1). If $\theta >0$ (i.e., curls naturally with your right hand), then the sense of torque is positive. Otherwise, the sense of torque is negative.

Center of Gravity
In many equilibrium situations, one of the forces acting on the body is its own weight. In free-body diagrams, the weight vector is attached to the center of gravity of the body. For all practical purposes, the center of gravity is identical to the center of mass. Only in certain situations are the center of gravity and the center of mass located at different points, such as, where a body has a large spatial extension so that the gravitational field is nonuniform throughout its volume.

In most situations, the object lies in a uniform gravitational field where $g = 9.81\ {\rm m/s^2}$ and the center of gravity is the identical to the center of mass. The center of mass $\rm CoM$ is the point where the weight is attached. When an external force is applied to a body exactly at its $\rm CoM$, then the body (as a whole) undergoes translational motion and such a force, does not cause rotation.

When the $\rm CoM$ is located off the axis of rotation, a net gravitational torque occurs on an object. Gravitational torque is the torque caused by the weight. The gravitational torque may rotate the object if there is no support present to balance it. The magnitude of the gravitational torque depends on how far away from the pivot the $\rm CoM$ is located.

In the case of a tipping truck (see Fig. 12.2), the pivot is located on the line where the tires make contact with the road’s surface. if the $\rm CoM$ is located high above the road’s surface, the gravitational torque may be large enough to turn the truck over. Passenger cars with a low-lying $\rm CoM$ are more resistant to tipping over than are trucks.

tipping truck example — Fig. 12.2 Image Credit: Openstax.#

Note

A special case of static equilibrium occurs when all external forces on an object act at (or along) the axis of rotation or when the spatial extension of the object can be disregarded. Then, the object can be effectively treated like a point mass. We need not worry about the second equilibrium condition because all torques are identically zero and the first condition (for forces) is the only condition to be satisfied.

12.1.1. Example Problem: Center of Gravity#

Exercise 12.1

The Problem

A passenger car with a $2.5\ {\rm m}$ wheelbase has $52\%$ of its weight on the front wheels on level ground (see Figure 12.3). Where is the center of mass of this car located with respect to the rear axle?

center of gravity of a car — Fig. 12.3 Image Credit: Openstax.#

The Model

The car is modeled as a rigid body in static equilibrium supported by two contact forces at the front and rear axles. The entire weight $w$ of the car acts at its center-of-mass ($\rm CoM$), which lies somewhere between the two axles. The normal reaction forces at the front and rear axles support $52\%$ and $48\%$ of the total weight, respectively.

We choose a coordinate system along the horizontal axis of the car, with the rear axle as the origin. The front axle is located at a distance $d = 2.5\ {\rm m}$ from the rear axle. The CM is located at an unknown distance $x$ from the rear axle.

The forces acting on the car are the upward normal forces at the front and rear axles and the downward gravitational force at the CM. Because the system is in static equilibrium, both the net force and the net torque on the car must be zero.

center of gravity of a car FBD — Fig. 12.4 Image Credit: Openstax.#

The Math

The forces acting on the car are the normal force at the front axle, the normal force at the rear axle, and the weight of the car acting at the center of mass. The condition for vertical force equilibrium requires that the sum of the vertical forces is zero.

Thus, the vertical force balance is given by

\[ F_F + F_R - w = 0. \]

The problem states that the front axle supports $52\%$ of the weight and the rear axle supports $48\%$ of the weight. Therefore, the forces are given by $F_F = 0.52w$ and $F_R = 0.48w$. Substituting these values into the force balance confirms that the system is in equilibrium.

To determine the location of the center of mass, we apply the condition for rotational equilibrium. We take torques about the rear axle so that the torque due to the rear force is zero because its lever arm is zero.

The torque due to the front axle force produces a counterclockwise rotation, and the torque due to the weight produces a clockwise rotation. Setting the net torque equal to zero gives

\[ \tau_F - \tau_w = 0. \]

The magnitude of each torque is given by the product of the force and its perpendicular lever arm. The front force acts at a distance $d$ from the rear axle, and the weight acts at a distance $x$ from the rear axle. Thus, the torque balance becomes

\[ F_F d - w x = 0. \]

Substituting $F_F = 0.52w$ into this expression gives

\[ 0.52w\,d - w\,x = 0. \]

We now solve this expression for the unknown position $x$ of the center of mass. Dividing both sides of the equation by $w$ gives

\[ 0.52d - x = 0. \]

Solving for $x$ yields $x = 0.52d.$ Substituting $d = 2.5\ {\rm m}$ into this expression gives

\[ x = 0.52(2.5\ {\rm m}) = 1.3\ {\rm m}. \]

The Conclusion

The center of mass of the car is located $1.3\ {\rm m}$ from the rear axle. This means that the center of mass lies slightly closer to the front axle than the midpoint of the wheelbase, which is consistent with the greater fraction of the weight supported by the front wheels.

The Verification

The verification computes the torque balance numerically using the given force distribution and wheelbase. The calculated value of $x$ satisfies the torque equilibrium condition, confirming that the analytical solution is correct.

import numpy as np

# Given values
d = 2.5  # wheelbase (m)
w = 1.0  # normalized weight
F_F = 0.52 * w

# Solve for x from torque balance
x = F_F * d / w

print(f"The center of mass is located {x:.2f} m from the rear axle.")

12.1.2. Example Problem: A Breaking Tension#

Exercise 12.2

The Problem

A small pan of mass $42.0\ {\rm g}$ is supported by two strings, as shown in Figure 12.5. The maximum tension that the string can support is $2.80\ {\rm N}$. Mass is added gradually to the pan until one of the strings snaps. Which string is it? How much mass must be added for this to occur?

find tension with mass and two supports — Fig. 12.5 Image Credit: Openstax.#

The Model

The knot where the two upper strings and the vertical string meet can be treated as a particle in static equilibrium. The forces acting on this knot are the tension $T_1$ in the shorter $5.0\ {\rm cm}$ string, the tension $T_2$ in the longer $10.0\ {\rm cm}$ string, and the downward force due to the weight of the pan plus the added mass. If the pan has mass $M = 42.0\ {\rm g} = 0.0420\ {\rm kg}$ and the added mass is $m$, then the total downward force is $(M+m)g$.

We choose the $x$-axis horizontally to the right and the $y$-axis vertically upward. The two strings and the ceiling form a right triangle in which the string lengths are $a$ and $2a$, where $a = 5.0\ {\rm cm}$. The horizontal separation of the ceiling attachment points is therefore $\sqrt{a^2 + (2a)^2} = a\sqrt{5}$. From this geometry, the direction angles of the strings satisfy

\[\begin{align*} \cos\alpha_1 &= \frac{1}{\sqrt{5}}, \qquad &\sin\alpha_1 = \frac{2}{\sqrt{5}}, \\ \cos\alpha_2 &= \frac{2}{\sqrt{5}}, \qquad &\sin\alpha_2 = \frac{1}{\sqrt{5}}. \end{align*}\]

Because the system is in static equilibrium, the net force on the knot must be zero in both the $x$- and $y$-directions.

find tension with mass and two supports FBD — Fig. 12.6 Image Credit: Openstax.#

The Math

The forces acting on the knot are the two string tensions and the downward weight of the pan plus the added mass. Resolving the tensions into rectangular components allows us to apply the equilibrium condition separately in the horizontal and vertical directions.

The conditions for equilibrium in each direction are

\[\begin{align*} -T_{1x} + T_{2x} &= 0. \\ T_{1y} + T_{2y} - (M+m)g &= 0. \end{align*}\]

From the geometry of the strings, the components of the tensions are

\[\begin{align*} T_{1x} &= T_1\cos\alpha_1 = \frac{1}{\sqrt{5}}T_1, \qquad &T_{1y} = T_1\sin\alpha_1 = \frac{2}{\sqrt{5}}T_1, \\ T_{2x} &= T_2\cos\alpha_2 = \frac{2}{\sqrt{5}}T_2, \qquad &T_{2y} = T_2\sin\alpha_2 = \frac{1}{\sqrt{5}}T_2. \end{align*}\]

Substituting these expressions into the equilibrium equation in the $x$-direction gives

\[ -\frac{1}{\sqrt{5}}T_1 + \frac{2}{\sqrt{5}}T_2 = 0. \]

We multiply this equation by $\sqrt{5}$ to simplify it, which gives

\[ -T_1 + 2T_2 = 0.\]

Solving for $T_1$ gives $T_1 = 2T_2.$

This result shows that the tension in the shorter string is always twice the tension in the longer string. Therefore, the shorter $5.0\ {\rm cm}$ string will reach the maximum tension first and will snap first. We now substitute $T_2 = T_1/2$ into the equilibrium equation in the $y$-direction to determine the mass required to make $T_1$ reach its maximum value. The vertical equilibrium equation becomes

\[ \frac{2}{\sqrt{5}}T_1 + \frac{1}{\sqrt{5}}\left(\frac{T_1}{2}\right) - (M+m)g = 0.\]

Combining the terms involving $T_1$ gives

\[ \frac{2.5}{\sqrt{5}}T_1 = (M+m)g.\]

The string breaks when the tension in the shorter string reaches the critical value $T_1 = 2.80\ {\rm N}$. Solving this expression for the added mass $m$ gives

\[\begin{align*} m &= \frac{2.5}{\sqrt{5}\,g}T_1 - M \\ &= \frac{2.5(2.80\ {\rm N})}{\sqrt{5}(9.81\ {\rm m/s^2})} - 0.0420\ {\rm kg} \\ &= 0.277\ {\rm kg}. \end{align*}\]

Expressed in grams, this added mass is $m = 277\ {\rm g}.$

The Conclusion

The shorter $5.0\ {\rm cm}$ string snaps first because its tension is always twice the tension in the longer string. The mass that must be added to the pan to make this occur is $0.277\ {\rm kg}$.

The Verification

The verification computes the string-angle factors from the given geometry, uses horizontal equilibrium to relate the two tensions, and then uses vertical equilibrium to solve for the added mass when the shorter string reaches its maximum tension. This reproduces the analytical result and confirms that the shorter string breaks first.

import numpy as np

# Given quantities
M = 0.0420  # Mass of the pan (kg)
T1_max = 2.80  # Maximum tension in the shorter string (N)
g = 9.81  # Gravitational acceleration (m/s^2)

# Geometry factors from the right triangle
cos_alpha1 = 1 / np.sqrt(5)
sin_alpha1 = 2 / np.sqrt(5)
cos_alpha2 = 2 / np.sqrt(5)
sin_alpha2 = 1 / np.sqrt(5)

# Use horizontal equilibrium to find the tension in the longer string
T2 = T1_max / 2

# Use vertical equilibrium to find the total supported mass
total_mass = (T1_max * sin_alpha1 + T2 * sin_alpha2) / g

# Subtract the pan mass to find the added mass
m_added = total_mass - M

print(f"The tension in the shorter string is {T1_max:.2f} N when it reaches its limit.")
print(f"The tension in the longer string is {T2:.2f} N at that same instant.")
print(f"The added mass required to break the shorter string is {m_added:.3f} kg, or {m_added*1000:.0f} g.")

12.2. Examples of Static Equilibrium#

All examples in this chapter are planar problems, so we can use the simplified forms with only forces in the $xy$-plane and the corresponding torques along the $z$-axis.

Problem Solving Strategy

Static Equilibrium

Identify all forces acting on each object in the system.
Set up a free-body diagram (FBD) for each object.
- (a) Choose the $xy$-reference frame for the problem. Draw the corresponding FBDs, representing the components of the forces in the chosen reference frame. Label all forces.
- (b) Choose the location of the rotation axis (i.e., the pivot point) where you will compute torques of the acting forces. On the FBD, indicate the location of the pivot and lever arms of acting forces. Note: choose a pivot point that simplifies the calculation of the torques.
Set up the equations of equilibrium for the object.
- (a) Use the FBD to write a correct equilibrium (force) condition in the $x$ and $y$ directions.
- (b) Use the FBD to write a correct equilibrium (torque) condition along the axis of rotation. Be sure to evaluate the torque magnitudes and senses.
Simplify and solve the system of equations for equilibrium to obtain unknown quantities. Keep in mind that the number of equations must match the number of unknowns to be solved.
Evaluate the expressions for the unknown quantities. Your final answers should have the correct sig figs and physical units. You may independently check your numerical answers by shifting the pivot to a different location and solving again.

Setting up a free-body diagram (FBD) for a rigid-body equilibrium problem is the most important component in the solution process. Without the correct setup and diagram, you will not have the correct conditions for equilibrium.
A FBD for an extended rigid body that may undergo rotational motion is different than a FBD for body that experiences only translational motion.

In translational dynamics, a body is represented as its center-of-mass $\rm CoM$, where all forces on the body are attached and no torques appear. This does not hold true in rotational dynamics, where an extended rigid body cannot be represented by one point alone. The torques acting on the body depend on both the acting force and its lever arm. The FBD for an extended rigid body helps us to identify the external torques.

12.2.1. Example Problem: Torque Balance#

Exercise 12.3

The Problem

Three masses are attached to a uniform meter stick, as shown in Figure 12.7. The mass of the meter stick is $150\ {\rm g}$ and the masses to the left of the fulcrum are $m_1 = 50\ {\rm g}$ and $m_2 = 75\ {\rm g}$. Find the mass $m_3$ that balances the system when it is attached at the right end of the stick, and the normal reaction force at the fulcrum when the system is balanced.

torque balance problem — Fig. 12.7 Image Credit: Openstax.#

The Model

The meter stick is modeled as a rigid body in static equilibrium about the fulcrum at point $S$. The forces acting on the stick are the downward weights of the three attached masses, the downward weight of the uniform meter stick acting at its center of mass, and the upward normal reaction force $F_S$ at the fulcrum.

We choose the $x$-axis along the stick, the $y$-axis vertically upward, and the axis of rotation perpendicular to the page through the fulcrum. Because the stick is uniform, its center of mass is at the $50\ {\rm cm}$ mark. From Figure 12.7, the fulcrum is at the $70\ {\rm cm}$ mark, so the lever arms measured from the fulcrum are $70\ {\rm cm}$ for $m_1$, $40\ {\rm cm}$ for $m_2$, $20\ {\rm cm}$ for the stick’s weight, and $30\ {\rm cm}$ for $m_3$.

The system is balanced, so both the net torque about the fulcrum and the net vertical force on the stick must be zero.

torque balance fbd — Fig. 12.8 Image Credit: Openstax.#

The Math

The forces acting on the stick are the weights $m_1g$, $m_2g$, $mg$, and $m_3g$, together with the upward normal reaction force $F_S$ at the fulcrum. We first apply the condition for rotational equilibrium about the fulcrum so that the torque due to $F_S$ is zero because its lever arm is zero.

The masses to the left of the fulcrum produce counterclockwise torques, and the mass $m_3$ to the right of the fulcrum produces a clockwise torque. Therefore, the torque balance is

\[ r_1 m_1 g + r_2 m_2 g + r m g - r_3 m_3 g = 0.\]

The lever arms are $r_1 = 70\ {\rm cm}$, $r_2 = 40\ {\rm cm}$, $r = 20\ {\rm cm}$, and $r_3 = 30\ {\rm cm}$. Substituting these values into the torque equation gives

\[ (70.0\ {\rm cm})m_1 g + (40.0\ {\rm cm})m_2 g + (20.0\ {\rm cm})m g - (30.0\ {\rm cm})m_3 g = 0.\]

We divide this equation by $g$ to eliminate the common factor of gravitational acceleration, which gives

\[ r_3 m_3 = r_1 m_1 + r_2 m_2 + r m. \]

We now solve this expression for the unknown mass $m_3$, which gives

\[\begin{align*} m_3 &= \frac{r_1}{r_3}m_1 + \frac{r_2}{r_3}m_2 + \frac{r}{r_3}m \\ &= \frac{70}{30}(50\ {\rm g}) + \frac{40}{30}(75\ {\rm g}) + \frac{20}{30}(150\ {\rm g}) \\ &= 316.7\ {\rm g}. \end{align*}\]

The condition for vertical force equilibrium requires that the sum of the vertical forces is zero. Taking upward as positive, the vertical force balance is

\[ F_S - m_1 g - m_2 g - m g - m_3 g = 0. \]

We solve this expression for the normal reaction force, which gives

\[ F_S = (m_1 + m_2 + m + m_3)g. \]

Substituting $m_1 = 0.05\ {\rm kg}$, $m_2 = 0.075\ {\rm kg}$, $m = 0.15\ {\rm kg}$, $m_3 = 0.3167\ {\rm kg}$, and $g = 9.81\ {\rm m/s^2}$ into this expression gives

\[\begin{align*} F_S &= (0.05 + 0.075 + 0.15 + 0.3167)\ {\rm kg}\,(9.81\ {\rm m/s^2}) \\ &= 5.8\ {\rm N}. \end{align*}\]

The Conclusion

The mass required at the right end of the stick to balance the system is $m_3 = 317\ {\rm g}$. The normal reaction force at the fulcrum is $F_S = 5.8\ {\rm N}$. This result shows that the torque balance depends only on the mass distribution and lever arms, while the support force depends on the total weight of the balanced system.

The Verification

The verification reproduces the torque balance about the fulcrum to compute the unknown mass $m_3$, and then applies vertical force equilibrium to compute the support force $F_S$. This confirms the analytical result and shows that the balanced mass is determined independently of the value of $g$, while the support force is not.

import numpy as np

# Given masses (kg)
m1 = 0.05  # Mass on the left end
m2 = 0.075 # Second mass on the left
m_stick = 0.15  # Mass of the uniform meter stick

# Lever arms measured from the fulcrum (m)
r1 = 0.7  # Lever arm of m1
r2 = 0.4  # Lever arm of m2
r_stick = 0.2  # Lever arm of the stick's center of mass
r3 = 0.3  # Lever arm of m3

# Gravitational acceleration (m/s^2)
g = 9.81

# Use torque balance about the fulcrum to find the balancing mass
m3 = (r1 * m1 + r2 * m2 + r_stick * m_stick) / r3

# Use vertical force equilibrium to find the support force at the fulcrum
F_S = (m1 + m2 + m_stick + m3) * g

print(f"The mass required to balance the meter stick is {m3*1000:.1f} g.")
print(f"The normal reaction force at the fulcrum is {F_S:.2f} N.")

12.2.2. Example Problem: Forces in the Forearm#

Exercise 12.4

The Problem

A weightlifter is holding a $50.0\ {\rm lb}$ weight (equivalent to $222.4\ {\rm N}$) with his forearm, as shown in Figure 12.9. His forearm is positioned at $\beta = 60^\circ$ with respect to his upper arm. The forearm is supported by a contraction of the biceps muscle, which causes a torque around the elbow. Assuming that the tension in the biceps acts along the vertical direction given by gravity, what tension must the muscle exert to hold the forearm at the position shown? What is the force on the elbow joint? Assume that the forearm’s weight is negligible. Give your final answers in SI units.

forces in the forearm problem — Fig. 12.9 Image Credit: Openstax.#

The Model

The forearm is modeled as a rigid body in static equilibrium about the elbow joint. The forces acting on the forearm are the downward force of the weight in the hand, the upward tension exerted by the biceps muscle, and the force exerted by the elbow joint. The forearm’s own weight is neglected, as stated in the problem.

We choose the elbow as the pivot point so that the torque due to the elbow force is zero. The forearm makes an angle $\beta = 60^\circ$ with the vertical. The tension in the biceps and the weight both act vertically, so each makes the same angle $\beta$ with the forearm. The lever arm of the biceps tension is $r_T = 1.5\ {\rm in}$, and the lever arm of the weight is $r_w = 13.0\ {\rm in}$. Because these distances appear only as a ratio in the torque equation, they do not need to be converted into SI units until the final force calculation.

The system is in static equilibrium, so the net torque about the elbow and the net vertical force on the forearm must both be zero.

forces in the forearm fbd — Fig. 12.10 Image Credit: Openstax.#

The Math

The forces acting on the forearm are the elbow force $F$, the biceps tension $T$, and the weight $w$ of the object in the hand. Because the biceps tension and the weight both act vertically, and because the elbow force must balance their combined effect, all three forces are vertical. Therefore, the force balance may be written directly along the vertical direction.

The condition for rotational equilibrium about the elbow requires that the counterclockwise torque due to the biceps tension balance the clockwise torque due to the weight. Thus, the torque balance is

\[ r_T T\sin\beta - r_w w\sin\beta = 0. \]

Because both terms contain the factor $\sin\beta$, this factor cancels, and the torque equation becomes

\[ r_T T - r_w w = 0. \]

We solve this expression for the biceps tension, which gives

\[ T = \frac{r_w}{r_T}w. \]

Substituting $r_w = 13.0\ {\rm in}$, $r_T = 1.5\ {\rm in}$, and $w = 50.0\ {\rm lb}$ into this expression gives

\[ T = \frac{13.0}{1.5}(50.0\ {\rm lb}) = 433.3\ {\rm lb}. \]

The condition for force equilibrium requires that the net vertical force on the forearm be zero. Taking upward as positive, the force balance is

\[ -F + T - w = 0. \]

We solve this expression for the elbow force, which gives

\[ F = T - w. \]

Substituting the known values into this expression gives

\[ F = 433.3\ {\rm lb} - 50.0\ {\rm lb} = 383.3\ {\rm lb}. \]

The grouped final calculations in English units are

\[\begin{align*} T &= \frac{r_w}{r_T}w = \frac{13.0}{1.5}(50.0\ {\rm lb}) = 433.3\ {\rm lb}, \\ F &= T - w = 433.3\ {\rm lb} - 50.0\ {\rm lb} = 383.3\ {\rm lb}. \end{align*}\]

We now convert these forces into SI units using $1\ {\rm lb} = 4.448\ {\rm N}$. The final values are

\[\begin{align*} T &= 433.3\ {\rm lb}(4.448\ {\rm N/lb}) = 1.93 \times 10^3\ {\rm N}, \\ F &= 383.3\ {\rm lb}(4.448\ {\rm N/lb}) = 1.71 \times 10^3\ {\rm N}. \end{align*}\]

The biceps tension acts upward, while the elbow joint force acts downward on the forearm.

The Conclusion

The biceps muscle must exert an upward tension of $1.93 \times 10^3\ {\rm N}$ to hold the forearm in equilibrium. The force exerted by the elbow joint on the forearm has magnitude $1.71 \times 10^3\ {\rm N}$ and acts downward. This large muscle force shows that even a moderate weight in the hand can require a much larger internal force because the biceps acts with a very short lever arm.

The Verification

The verification uses the torque balance about the elbow to compute the biceps tension from the ratio of lever arms, and then uses force balance to compute the elbow force. The resulting values reproduce the analytical solution and confirm that the elbow force must act downward on the forearm.

import numpy as np

# Given quantities
w_lb = 50.0  # Weight held in the hand (lb)
lb_to_N = 4.448  # Conversion factor from pounds to newtons
r_T = 1.5  # Lever arm of the biceps tension (in)
r_w = 13.0  # Lever arm of the weight (in)

# Use torque balance about the elbow to find the biceps tension
T_lb = (r_w / r_T) * w_lb

# Use vertical force balance to find the elbow force magnitude
F_lb = T_lb - w_lb

# Convert the forces to SI units
T_N = T_lb * lb_to_N
F_N = F_lb * lb_to_N

print(f"The biceps tension is {T_N:.0f} N upward.")
print(f"The force on the elbow joint is {F_N:.0f} N downward.")

12.2.3. Example Problem: Ladder Resting Against a Wall#

Exercise 12.5

The Problem

A uniform ladder is $L = 5.0\ {\rm m}$ long and weighs $400.0\ {\rm N}$. The ladder rests against a slippery vertical wall, as shown in Figure 12.11. The minimum inclination angle between the ladder and the rough floor below which the ladder slips is $\beta = 53^\circ$. Find the reaction forces from the floor and from the wall on the ladder and the coefficient of static friction $\mu_s$ at the interface of the ladder with the floor that prevents the ladder from slipping.

ladder resting problem — Fig. 12.11 Image Credit: Openstax.#

The Model

The ladder is modeled as a rigid body in static equilibrium. Because the wall is slippery, the wall exerts only a horizontal normal reaction force on the ladder. At the floor, the ladder experiences an upward normal force and a horizontal static friction force. The ladder’s weight acts downward at its center of mass, which is located at the midpoint of the ladder because the ladder is uniform.

We choose the coordinate system so that the $x$-axis is horizontal and the $y$-axis is vertical. We select the contact point with the floor as the pivot. With this choice, both the normal force from the floor and the friction force from the floor produce zero torque because their lever arms are zero. The only torques about the pivot come from the ladder’s weight and the normal reaction force from the wall.

The system is at the threshold of slipping when $\beta = 53^\circ$, so the static friction force has reached its maximum value. That allows us to determine the coefficient of static friction from the ratio of the friction force to the normal force at the floor.

ladder resting fbd — Fig. 12.12 Image Credit: Openstax.#

The Math

The forces acting on the ladder are the friction force $f$ at the floor, the normal force $N$ from the floor, the weight $w$ of the ladder, and the horizontal normal reaction force $F$ from the wall. The condition for translational equilibrium requires that the net force on the ladder vanish in both the horizontal and vertical directions.

The force balance is

\[\begin{align*} f - F &= 0,\\ N - w = 0. \end{align*}\]

The condition for rotational equilibrium about the contact point with the floor requires that the net torque about that point be zero. The wall force produces a counterclockwise torque, and the weight produces a clockwise torque. The lever arm of the wall force is $L$, and the lever arm of the weight is $L/2$.

Thus, the torque balance about the floor contact point is

\[ r_F F\sin\theta_F + r_w w\sin\theta_w = 0. \]

For the wall force, the angle between the lever arm and the force is $\theta_F = 180^\circ - \beta$, so $\sin\theta_F = \sin\beta$. For the weight, the angle between the lever arm and the force is $\theta_w = 270^\circ - \beta$, so $\sin\theta_w = -\cos\beta$. Substituting these into the torque equation gives

\[ LF\sin\beta - \frac{L}{2}w\cos\beta = 0. \]

We solve this expression for the wall force, which gives

\[ F = \frac{w}{2}\cot\beta.\]

Substituting $w = 400.0\ {\rm N}$ and $\beta = 53^\circ$ into this expression gives

\[ F = \frac{400.0\ {\rm N}}{2}\cot 53^\circ = 150.7\ {\rm N}. \]

The horizontal force balance requires $f = F$, so the friction force at the floor is $f = 150.7\ {\rm N}.$ The vertical force balance requires $N = w$, so the normal force from the floor is $N = 400.0\ {\rm N}.$ The coefficient of static friction at the threshold of slipping is the ratio of the friction force to the normal force. Therefore,

\[ \mu_s = \frac{f}{N} = \frac{150.7}{400.0} = 0.377. \]

The reaction force from the floor is the vector sum of the floor’s normal force and the floor’s friction force. Writing this vector in unit-vector form gives

\[ \vec{F}_{\rm floor} = (-150.7\,\hat{i} + 400.0\,\hat{j})\ {\rm N}. \]

Its magnitude is $F_{\rm floor} = \sqrt{f^2 + N^2} = \sqrt{(150.7)^2 + (400.0)^2}\ {\rm N} = 427.4\ {\rm N}. $

Its direction above the floor is $ \phi = \tan^{-1}\left(\frac{N}{f}\right) = \tan^{-1}\left(\frac{400.0}{150.7}\right) = 69.3^\circ. $

Because the wall is frictionless, the reaction force from the wall is purely horizontal and has magnitude $F_{\rm wall} = 150.7\ {\rm N}.$

The Conclusion

The wall exerts a horizontal reaction force of $150.7\ {\rm N}$ on the ladder. The floor exerts a normal force of $400.0\ {\rm N}$ upward and a friction force of $150.7\ {\rm N}$ toward the wall, so the total reaction force from the floor has magnitude $427.4\ {\rm N}$ and points $69.3^\circ$ above the floor. The coefficient of static friction required to prevent slipping at the limiting angle is $\mu_s = 0.377$.

The Verification

The verification uses torque balance about the floor contact point to compute the wall force, then uses horizontal and vertical force balance to compute the floor friction and floor normal force. The resulting values reproduce the analytical solution and confirm the required coefficient of static friction at the threshold of slipping.

import numpy as np

# Given quantities
w = 400.0  # Weight of the ladder (N)
beta_deg = 53.0  # Inclination angle (degrees)
beta = np.deg2rad(beta_deg)

# Use torque balance about the floor contact point to find the wall force
F_wall = 0.5 * w * (np.cos(beta) / np.sin(beta))

# Use force balance to find the floor friction and floor normal force
f_floor = F_wall
N_floor = w

# Compute the coefficient of static friction
mu_s = f_floor / N_floor

# Compute the total reaction force from the floor
F_floor_mag = np.sqrt(f_floor**2 + N_floor**2)
phi_deg = np.degrees(np.arctan2(N_floor, f_floor))

print(f"The reaction force from the wall is {F_wall:.1f} N.")
print(f"The floor exerts a friction force of {f_floor:.1f} N toward the wall.")
print(f"The floor exerts a normal force of {N_floor:.1f} N upward.")
print(f"The coefficient of static friction is {mu_s:.3f}.")
print(f"The magnitude of the total reaction force from the floor is {F_floor_mag:.1f} N.")
print(f"The direction of the floor reaction force is {phi_deg:.1f} degrees above the floor.")

12.2.4. Example Problem: Forces on Door Hinges#

Exercise 12.6

The Problem

A swinging door that weighs $w = 400\ {\rm N}$ is supported by hinges $A$ and $B$ so that the door can swing about a vertical axis passing through the hinges. The door has a width of $b = 1.00\ {\rm m}$, and the door slab has a uniform mass density. The hinges are placed symmetrically at the door’s edge in such a way that the door’s weight is evenly distributed between them. The hinges are separated by distance $a = 2.00\ {\rm m}$. Find the forces on the hinges when the door rests half-open.

force on door hinges — Fig. 12.13 Image Credit: Openstax.#

The Model

The door is treated as a rigid body in static equilibrium. The forces acting on the door are the hinge forces at $A$ and $B$, together with the weight $w$ acting downward at the center of mass. Because the door has uniform mass density, its center of mass is at its geometric center.

We resolve each hinge force into horizontal and vertical components and adopt a coordinate system with $x$ horizontal and $y$ vertical. The problem states that the weight is evenly distributed between the hinges, so the vertical components satisfy $A_y = B_y$. We choose the upper hinge $A$ as the pivot so that its torque contribution is zero.

force on door hinges fbd — Fig. 12.14 Image Credit: Openstax.#

The Math

We apply the equilibrium conditions for forces in both directions simultaneously:

\[\begin{align*} \sum F_x &= -A_x + B_x = 0, \\ \sum F_y &= A_y + B_y - w = 0. \end{align*}\]

From the first equation, $A_x = B_x$. From the second equation together with $A_y = B_y$, we obtain

\[\begin{align*} 2A_y &= w \\ A_y &= B_y = \frac{w}{2} = 200\ {\rm N}. \end{align*}\]

Next, we apply the torque equilibrium condition about hinge $A$. Only the weight and the horizontal component $B_x$ contribute:

\[\begin{align*} \sum \tau_A &= -w\frac{b}{2} + a B_x = 0. \end{align*}\]

Solving for $B_x$ gives

\[\begin{align*} B_x &= \frac{w b}{2a} = \frac{(400)(1.00)}{2(2.00)} = 100\ {\rm N}. \end{align*}\]

Thus, $A_x = B_x = 100\ {\rm N}$. The forces exerted by the hinges on the door are therefore

\[\begin{align*} \vec{F}_{A\,{\rm on\ door}} &= -100\,\hat{i} + 200\,\hat{j}\ {\rm N}, \\ \vec{F}_{B\,{\rm on\ door}} &= +100\,\hat{i} + 200\,\hat{j}\ {\rm N}. \end{align*}\]

By Newton’s third law, the forces on the hinges are

\[\begin{align*} \vec{F}_{{\rm door\ on}\ A} &= +100\,\hat{i} - 200\,\hat{j}\ {\rm N}, \\ \vec{F}_{{\rm door\ on}\ B} &= -100\,\hat{i} - 200\,\hat{j}\ {\rm N}. \end{align*}\]

The Conclusion

Each hinge supports half the weight of the door, $200\ {\rm N}$ downward, while the horizontal components of $100\ {\rm N}$ form a couple that balances the torque produced by the door’s center of mass being offset from the hinge axis.

The Verification

The code verifies the vertical forces from force balance and equal load sharing, then computes the horizontal force using torque balance about the upper hinge.

import numpy as np 

# Given quantities
w = 400.0  # Weight of the door (N)
b = 1.00   # Width of the door (m)
a = 2.00   # Separation between hinges (m)

# Vertical force balance with equal distribution
A_y = w / 2
B_y = w / 2

# Torque balance about upper hinge
B_x = w * b / (2 * a)
A_x = B_x

print(f"Each hinge supports {A_y:.1f} N vertically.")
print(f"The horizontal hinge forces have magnitude {A_x:.1f} N.")

print(f"Force of door on upper hinge: ({A_x:.1f} i - {A_y:.1f} j) N.")
print(f"Force of door on lower hinge: (-{B_x:.1f} i - {B_y:.1f} j) N.")

12.3. Stress, Strain, and Elastic Modulus#

A model of a rigid body is an idealized example of an object that does not deform under the actions of external forces. When analyzing mechanical systems, many physical objects are rigid to a great extent, which depends on the material properties of the object. For example, a ping-pong ball made of plastic is brittle, where a tennis ball (made of rubber) is elastic when acted upon by squashing forces. Under some circumstances, both may bounce well as rigid bodies.

A change in shape due to the application of forces is called deformation, where even small forces are known to cause some deformation. Deformation is the reaction of materials under the action of external forces (e.g., squashing, squeezing, ripping, twisting, shearing, or pulling the objects apart). In the language of physics, two terms describe the forces causing deformation: stress and strain.

Stress is a quantity that describes the magnitude of forces that cause deformation and is described as a force per unit area. There are four broad types of stress:

When forces pull on an object and cause its elongation (e.g., stretching an elastic band) is called tensile stress.
When forces compress an object, we call it a compressive stress.
When an object is squeezed from all sides (e.g., a submarine in the depths of the ocean), this is called a *bulk (or volume) stress.
When deforming forces act tangentially to the object’s surface, the forces are shear forces and the stress is called a shear stress.

The SI unit of stress is the pascal ($\rm Pa$). When one newton of force presses on a unit surface area ($1\ {\rm m^2}$), the resulting stress is one pascal:

\[ \text{one pascal} = 1.0\ {\rm Pa} = \frac{1\ {\rm N}}{1\ {\rm m^2}}. \]

In the Imperial system of units, the unit of stress is psi (i.e., pound per square inch, $\rm lb/in^2$). Another unit that is often used for bulk stress is the atmosphere $\rm atm$. Conversion factors between these units are:

\[\begin{align*} 1\ {\rm psi} &= 6895\ {\rm Pa},\\ 1\ {\rm Pa} &= 1.450 \times 10^{-4}\ {\rm psi}, \\ 1\ {\rm atm} &= 14.7\ {\rm psi} = 1.013 \times 10^5\ {\rm Pa}. \end{align*}\]

Note

The bulk stress and pressure both exert a force over an area and have the same units. However, a pressure characterizes an external force uniformly over a surface, while the bulk stress is a restorative force resisting the pressure from within. Another notable difference is that pressure is a scalar, while stress is a tensor that can be complex and directional.

An object or medium under stress becomes deformed, where we quantify the deformation using strain. Strain is a fractional change in either length (under tensile stress), volume (under bulk stress), or geometry (under shear stress). Strain is a dimensionless number that is proportional to the strain.

The greater the stress, the greater the strain. This relationship need not be linear. Only when the stress is sufficiently low is the deformation in direct proportion to the stress value. The proportionality constant is called the elastic modulus. In the linear limit of low stress, the general relation between stress and strain is given by

(12.9)#\[\begin{align} \text{stress} = (\text{elastic modulus}) \times \text{strain}. \end{align}\]

Since the strain is dimensionless, this means that the elastic modulus must have the same units as the stress. When an object’s elastic modulus is large, the effect of stress is small. Also, a small elastic modulus means that stress produces large strain and a noticeable deformation. For example, the stress on a rubber band produces a larger strain (deformation) than the same stress on a steel band of the same dimensions because the elastic modulus for rubber is about $100\times$ smaller than the elastic modulus for steel.

tensile stress: the elastic modulus is called Young’s modulus.
bulk stress: the elastic modulus is called the bulk modulus.
shear stress: the elastic modulus is called the shear modulus.

The relation between stress and strain is an observed (or empirical) relation that is measured in a laboratory. Elastic moduli for various materials are measured under various physical conditions (e.g., varying temperature) and collected in engineering data tables for reference.

Table 12.1 Approximate Elastic Moduli for Selected Materials#
Material	Young’s modulus ($\times 10^{10}\ {\rm Pa}$)	Bulk modulus ($\times 10^{10}\ {\rm Pa}$)	Shear modulus ($\times 10^{10}\ {\rm Pa}$)
Aluminum	7.0	7.5	2.5
Bone (tension)	1.6	0.8	8.0
Bone (compression)	0.9
Brass	9.0	6.0	3.5
Brick	1.5
Concrete	2.0
Copper	11.0	14.0	4.4
Crown glass	6.0	5.0	2.5
Granite	4.5	4.5	2.0
Hair (human)	1.0
Hardwood	1.5		1.0
Iron	21.0	16.0	7.7
Lead	1.6	4.1	0.6
Marble	6.0	7.0	2.0
Nickel	21.0	17.0	7.8
Polystyrene	3.0
Silk	6.0
Spider thread	3.0
Steel	20.0	16.0	7.5
Acetone		0.07
Ethanol		0.09
Glycerin		0.45
Mercury		2.5
Water		0.22

12.3.1. Tensile or Compressive Stress, Strain, and Young’s Modulus#

Tension or compression occurs when two antiparallel forces of equal magnitude act on an object along only one of its dimensions and the object does not move. Consider a rod segment that is either stretched or squeezed by a pair of forces acting along its length and perpendicular to its cross-section (see Fig. 12.15).

rod compression example — Fig. 12.15 Image Credit: Openstax.#

The rod changes its length under the action of the forces, where this change in length $\Delta L = L- L_o$ may be either elongation $(L>L_o)$ or contraction $(L<L_o)$.

Tensile stress and strain occur when the forces are stretching the object causing its elongation $(\Delta L > 0)$.
Compressive stress and strain occur when the forces are contracting the objection causing its shortening $(\Delta L <0)$.

In either of these situations, we define stress as the ratio of the deforming force $F_\perp$ to the cross-sectional area $A$ of the object. The deforming force acts perpendicularly to the cross-section of the object. Forces that act parallel to the cross-section do not change the length of the object. The definition of the tensile stress is

(12.10)#\[\begin{align} \text{tensile stress} &= \frac{F_\perp}{A}. \end{align}\]

Tensile strain is the measure of the deformation of an object under tensile stress and is defined as the fractional change of the object’s length under the stress, or

(12.11)#\[\begin{align} \text{tensile strain} &= \frac{\Delta L}{L_o}. \end{align}\]

Compressive stress and strain are defined by the same formulae, where we take the absolute values on the right-hand side.

Young’s modulus is the elastic modulus when deformation is caused by either tensile or compressive stress. The expression for the Young’s modulus is the ratio of the tensile stress to the strain, or

(12.12)#\[\begin{align} Y &= \frac{\text{tensile stress}}{\text{tensile strain}}= \left(\frac{F_\perp}{A}\right)\left(\frac{L_o}{\Delta L}\right). \end{align}\]

Objects can often experience both compressive and tensile stress simultaneously (see Fig. 12.16). For example, a long shelf loaded with heavy book sags between the end supports under the weight of the books. The top surface of the shelf is in compressive stress, while the bottom surface is in tensile stress. Similarly, long and heavy beams sag under their own weight, where such bending strains can be almost eliminated with the use of I-beams.

tensile and compressive stress example — Fig. 12.16 Image Credit: Openstax.#

12.3.1.1. Example Problem: Compressive Stress in a Pillar#

Exercise 12.7

The Problem

A sculpture weighing $10{,}000\ {\rm N}$ rests on a horizontal surface at the top of a $6.0\ {\rm m}$-tall vertical pillar (Figure 12.17). The pillar’s cross-sectional area is $0.20\ {\rm m}^2$ and it is made of granite with a mass density of $2700\ {\rm kg/m}^3$. Find the compressive stress at the cross-section located $3.0\ {\rm m}$ below the top of the pillar and the value of the compressive strain of the top $3.0\ {\rm m}$ segment of the pillar.

12.3.1.2. Example Problem: Stretching a Rod#

Exercise 12.8

The Problem

A $2.0\ {\rm m}$-long steel rod has a cross-sectional area of $0.30\ {\rm cm}^2$. The rod is a part of a vertical support that holds a heavy $550\ {\rm kg}$ platform that hangs attached to the rod’s lower end. Ignoring the weight of the rod, what is the tensile stress in the rod and the elongation of the rod under the stress?

The Model

The rod is modeled as a uniform elastic member under tension. Because the weight of the rod is neglected, the tensile force is simply the weight of the attached platform. The tensile stress is the applied force divided by the cross-sectional area, and the elongation follows from the definition of Young’s modulus, $Y = \dfrac{\text{stress}}{\text{strain}}$.

The cross-sectional area is given in ${\rm cm}^2$, so it must be converted to ${\rm m}^2$ before substitution into the stress formula. The elongation is then found from the strain relation $\dfrac{\Delta L}{L_0} = \dfrac{\text{stress}}{Y}$ using Young’s modulus for steel.

The Math

The tensile force in the rod is the weight of the platform. Therefore,

\[ F = mg = (550\ {\rm kg})(9.81\ {\rm m/s^2}) = 5395.5\ {\rm N}. \]

The cross-sectional area must be written in SI units. Since $1\ {\rm cm}^2 = 10^{-4}\ {\rm m}^2$, the given area becomes

\[ A = 0.30\ {\rm cm}^2 = 3.0 \times 10^{-5}\ {\rm m}^2.\]

The tensile stress is the applied force divided by the area, so

\[\begin{align*} \text{stress} &= \frac{F}{A} = \frac{5395.5\ {\rm N}}{3.0 \times 10^{-5}\ {\rm m}^2} = 1.80 \times 10^8\ {\rm Pa}. \end{align*}\]

From Table 12.1, Young’s modulus for steel is $Y = 2.0 \times 10^{11}\ {\rm Pa}$. The strain is therefore

\[ \text{strain} = \frac{\text{stress}}{Y} = \frac{1.80 \times 10^8}{2.0 \times 10^{11}} = 9.0 \times 10^{-4}. \]

The elongation is related to the strain by $\Delta L = (\text{strain})L_0$, so

\[\begin{align*} \Delta L &= (9.0 \times 10^{-4})(2.0\ {\rm m}) = 1.8 \times 10^{-3}\ {\rm m}. \end{align*}\]

Because this value is between $0.01$ and $10^4$ when expressed in millimeters, we convert it to a more readable unit:

\[ \Delta L = 1.8\ {\rm mm}.\]

The Conclusion

The tensile stress in the steel rod is $1.80 \times 10^8\ {\rm Pa}$, and the corresponding elongation is $1.8\ {\rm mm}$. Although the platform is quite massive, the resulting extension is small because steel has a very large Young’s modulus.

The Verification

The verification computes the tensile force from the platform’s weight, then uses the definitions of stress and Young’s modulus to reproduce the stress and elongation. This confirms the analytical result and shows that the rod stretches only a small amount under the applied load.

import numpy as np

# Given quantities
m = 550.0  # Mass of the platform (kg)
g = 9.81  # Gravitational acceleration (m/s^2)
A = 0.30e-4  # Cross-sectional area (m^2)
L0 = 2.0  # Original length of the rod (m)
Y = 2.0e11  # Young's modulus for steel (Pa)

# Compute the tensile force from the weight of the platform
F = m * g

# Compute the tensile stress in the rod
stress = F / A

# Compute the strain and elongation
strain = stress / Y
delta_L = strain * L0

print(f"The tensile stress in the rod is {stress:.2e} Pa.")
print(f"The elongation of the rod is {delta_L*1000:.1f} mm.")

12.3.2. Bulk Stress, Strain, and Modulus#

When you dive into water, you feel a force pressing on every part of your body from all directions that is a bulk stress, or pressure. Bulk stress always tends to decrease the volume enclosed by the surface of a submerged object. The forces of this “squeezing” are always perpendicular to the submerged surface.

bulk stress example — Fig. 12.18 Image Credit: Openstax.#

These forces decrease the volume of the submerged object by an amount $\Delta V$. This kind of deformation is called a bulk strain and is described by

(12.13)#\[\begin{align} \text{bulk strain} &= \frac{\Delta V}{V_o}. \end{align}\]

The bulk strain results from the bulk stress, which is a perpendicular force $F_\perp$ that presses on the unit surface area $A$. This pressure $p$ is defined as

(12.14)#\[\begin{align} p \equiv \frac{F_\perp}{A}. \end{align}\]

Pressure is a scalar quantity and does not have any particular direction, where it acts equally in all possible directions. When you submerge your hand in water, you sense the same amount of pressure action the the top, bottom, and side surfaces as you do on the surface of skin between your fingers. The increase in pressure $\Delta p$ that you feel is relative to the normal pressure $p_o$ of $1\ {\rm atm}$ you feel from the atmosphere. The bulk stress is the change in pressure $\Delta p$ relative to the normal level $p_o$, or

(12.15)#\[\begin{align} B &= \frac{\text{bulk stress}}{\text{bulk strain}}, \\[6pt] &= -\frac{\Delta p}{\Delta V/V_o} = -\frac{\Delta p}{\Delta V}V_o. \end{align}\]

When the bulk stress increases, the bulk strain increases in response. The minus sign appears to ensure that $B$ is a positive quantity. An increase in $\Delta p$ always causes a decrease $\Delta V$ in volume so that $\frac{\Delta p}{\Delta V} <0$.

The reciprocal of the bulk modulus is called compressibility, or

(12.16)#\[\begin{align} k &= \frac{1}{B} &= -\frac{\Delta V}{\Delta p}\frac{1}{V_o}. \end{align}\]

The term ‘compressibility’ is used in relation to fluids (i.e., gases and liquids). Compressibility describes the change in the volume of a fluid per unit increase in pressure. Fluids that are easy to compress are characterized by a large compressibility. For example, the compressibility water and acetone are

\[\begin{align*} k_{\text{water}} = 4.64 \times 10^{-5}\ {\rm atm^{-1}}, \\ k_{\text{acetone}} = 1.45 \times 10^{-4}\ {\rm atm^{-1}}, \end{align*}\]

which means that under normal pressure ($1\ {\rm atm}$) the relative decrease in volume for acetone is approximately $3\times$ larger than for water.

12.3.2.1. Example Problem: Hydraulic Press#

Exercise 12.9

The Problem

In a hydraulic press (Figure 12.19), a $250\ {\rm L}$ volume of oil is subjected to a $2300\ {\rm psi}$ pressure increase. If the compressibility of oil is $2.0 \times 10^{-5}/{\rm atm}$, find the bulk strain and the absolute decrease in the volume of oil when the press is operating.

hydraulic press example — Fig. 12.19 Image Credit: Openstax.#

The Model

The oil is modeled as a compressible fluid undergoing a uniform pressure increase. The relevant material property is the compressibility $k$, which relates the fractional volume change to the pressure increase through the relation for bulk deformation. Because the problem asks for the absolute decrease in volume, we first determine the bulk strain $\Delta V/V_0$ and then multiply by the initial volume.

The compressibility is given in ${\rm atm}^{-1}$, so the pressure increase must be converted from psi to atm before substitution. Since the oil is being compressed, the actual change in volume is negative, but the problem asks for the absolute decrease, so we report the magnitude of the volume change.

The Math

The bulk strain is related to the compressibility and the pressure increase by

\[ \frac{\Delta V}{V_0} = k\,\Delta p. \]

We first convert the pressure increase from psi to atm:

\[\begin{align*} \Delta p = 2300\ {\rm psi}\left(\frac{1\ {\rm atm}}{14.7\ {\rm psi}}\right) = 156\ {\rm atm}. \end{align*}\]

Substituting $k = 2.0 \times 10^{-5}/{\rm atm}$ and $\Delta p = 156\ {\rm atm}$ into the bulk-strain expression gives

\[\begin{align*} \frac{\Delta V}{V_0} &= \left(2.0 \times 10^{-5}/{\rm atm}\right)(156\ {\rm atm}) \\ &= 3.1 \times 10^{-3}. \end{align*}\]

This is the magnitude of the bulk strain. The absolute decrease in volume is then

\[\begin{align*} |\Delta V| &= \left(\frac{\Delta V}{V_0}\right)V_0 \\ &= \left(3.1 \times 10^{-3}\right)(250\ {\rm L}) \\ &= 0.78\ {\rm L}. \end{align*}\]

The Conclusion

The bulk strain of the oil is $3.1 \times 10^{-3}$, and the absolute decrease in the oil’s volume is $0.78\ {\rm L}$. This result shows that even under a large pressure increase, the oil undergoes only a small fractional compression.

The Verification

The verification converts the applied pressure from psi to atm, computes the bulk strain from the compressibility, and then computes the corresponding volume decrease from the initial volume. This reproduces the analytical result and confirms the small but nonzero compression of the oil.

import numpy as np

# Given quantities
V0 = 250.0  # Initial volume of oil (L)
dp_psi = 2300.0  # Pressure increase (psi)
psi_per_atm = 14.7  # Conversion from atm to psi
k = 2.0e-5  # Compressibility of oil (1/atm)

# Convert the pressure increase to atmospheres
dp_atm = dp_psi / psi_per_atm

# Compute the bulk strain
bulk_strain = k * dp_atm

# Compute the absolute decrease in volume
delta_V = bulk_strain * V0

print(f"The bulk strain is {bulk_strain:.2e}.")
print(f"The absolute decrease in volume is {delta_V:.2f} L.")

12.3.3. Shear Stress, Strain, and Modulus#

The concepts of shear stress and strain concern only solid objects or materials. Buildings and tectonic planets are examples that may be subjected to shear stresses. In general, these concepts do not apply to fluids.

Shear deformation occurs when two antiparallel forces of equal magnitude are applied tangentially to opposite surfaces of a solid object. This does not cause deformation in the transverse direction to the line of force (see Fig. 12.20).

shear stress example — Fig. 12.20 Image Credit: Openstax.#

Shear deformation is characterized by a gradual shift $\Delta x$ of layers in the direction tangent to the acting forces. This gradation in $\Delta x$ occurs in the transverse direction along some distance $L_o$. Shear strain is defined by the ratio of the largest displacement $\Delta x$ to the transverse distance $L_o$, or

(12.17)#\[\begin{align} \text{shear strain} = \frac{\Delta x}{L_o}. \end{align}\]

Shear stress is due to forces that act parallel to the surface, where we use $F_\parallel$ (or $F_\parallel$ ) for such forces. The magnitude $F_\parallel$ per unit surface area $A$ where the shearing force is applied is the measure of shear stress, or

(12.18)#\[\begin{align} \text{shear stress} &= \frac{F_\parallel}{A}. \end{align}\]

The shear modulus is the proportionality constant and is defined by the ratio of stress to strain. The shear modulus is commonly denoted by $S$ and defined as:

(12.19)#\[\begin{align} S &= \frac{\text{shear stress}}{\text{shear strain}}, \\[6pt] &= \frac{F_\parallel /A}{\Delta x/L_o} = \left(\frac{F_\parallel}{A}\right)\left(\frac{L_o}{\Delta x} \right). \end{align}\]

12.3.3.1. Example Problem: An Old Bookshelf#

Exercise 12.10

The Problem

A cleaning person tries to move a heavy, old bookcase on a carpeted floor by pushing tangentially on the surface of the very top shelf. However, the only noticeable effect of this effort is similar to that seen in Figure 12.20, and it disappears when the person stops pushing. The bookcase is $180\ {\rm cm}$ tall and $90.0\ {\rm cm}$ wide with four $30.0\ {\rm cm}$-deep shelves, all partially loaded with books. The total weight of the bookcase and books is $600\ {\rm N}$. If the person gives the top shelf a $50.0\ {\rm N}$ push that displaces the top shelf horizontally by $15.0\ {\rm cm}$ relative to the motionless bottom shelf, find the shear modulus of the bookcase.

The Model

The bookcase is modeled as a deformable solid undergoing shear. The applied tangential force produces a shear stress over the side face of the bookcase, and the horizontal displacement of the top relative to the bottom produces a shear strain. The shear modulus is the ratio of the shear stress to the shear strain.

The relevant dimensions are the height of the bookcase, which sets the original length $L_0$ over which the shear displacement occurs, and the area of the face parallel to the applied force. Because the push is applied tangentially to the top shelf, the resisting area is the rectangular side face with dimensions $30.0\ {\rm cm}$ by $90.0\ {\rm cm}$.

The Math

The shear modulus is given by the ratio of shear stress to shear strain:

\[ S = \frac{F_{\parallel}/A}{\Delta x/L_0}.\]

The area of the sheared face is $A = (30.0\ {\rm cm})(90.0\ {\rm cm}) = 2700\ {\rm cm}^2.$

The shear strain is determined from the horizontal displacement of the top relative to the bottom:

\[ \frac{\Delta x}{L_0} = \frac{15.0\ {\rm cm}}{180\ {\rm cm}} = \frac{1}{12} = 0.0833.\]

The shear stress is

\[\begin{align*} \frac{F_{\parallel}}{A} &= \frac{50.0\ {\rm N}}{2700\ {\rm cm}^2} = \frac{1}{54}\ {\rm N/cm}^2. \end{align*}\]

Using these results in the definition of shear modulus gives

\[\begin{align*} S &= \frac{F_{\parallel}}{A}\frac{L_0}{\Delta x} \\ &= \left(\frac{50.0\ {\rm N}}{2700\ {\rm cm}^2}\right)\left(\frac{180\ {\rm cm}}{15.0\ {\rm cm}}\right) \\ &= \frac{2}{9}\ {\rm N/cm}^2. \end{align*}\]

To express this in SI units, we use $1\ {\rm N/cm}^2 = 10^4\ {\rm N/m}^2$. Therefore,

\[\begin{align*} S &= \frac{2}{9}\times 10^4\ {\rm N/m}^2 \\ &= 2.22 \times 10^3\ {\rm N/m}^2. \end{align*}\]

For reference, the individual shear stress and shear strain are

\[\begin{align*} \frac{F_{\parallel}}{A} &= \frac{50.0\ {\rm N}}{2700\ {\rm cm}^2} = 185.2\ {\rm Pa}, \\ \frac{\Delta x}{L_0} &= \frac{15.0\ {\rm cm}}{180\ {\rm cm}} = 0.0833. \end{align*}\]

The Conclusion

The shear modulus of the bookcase is $2.22 \times 10^3\ {\rm N/m}^2$. This value is quite small compared with ordinary structural solids, which is consistent with the visibly noticeable but reversible deformation of the old bookcase under a modest sideways push.

The Verification

The verification computes the sheared area, the shear stress, and the shear strain directly from the given geometry and force, then forms their ratio to obtain the shear modulus. This reproduces the analytical result and confirms the very small rigidity of the bookcase.

import numpy as np

# Given quantities
F_parallel = 50.0  # Tangential force (N)
dx_cm = 15.0  # Horizontal displacement (cm)
L0_cm = 180.0  # Height of the bookcase (cm)
depth_cm = 30.0  # Shelf depth (cm)
width_cm = 90.0  # Bookcase width (cm)

# Compute the sheared area in cm^2 and convert to m^2
A_cm2 = depth_cm * width_cm
A_m2 = A_cm2 * 1e-4

# Compute the shear stress
shear_stress = F_parallel / A_m2

# Compute the shear strain
shear_strain = dx_cm / L0_cm

# Compute the shear modulus
S = shear_stress / shear_strain

print(f"The sheared area is {A_cm2:.0f} cm^2.")
print(f"The shear stress is {shear_stress:.1f} Pa.")
print(f"The shear strain is {shear_strain:.4f}.")
print(f"The shear modulus of the bookcase is {S:.2e} N/m^2.")

12.4. Elasticity and Plasticity#

The proportionality constant between stress and strain is the elastic modulus. This leads to a few questions:

Why is that?
What does it mean for an object to be elastic?
How do we describe the object’s behavior?

Elasticity is the tendency of solid objects (and materials) to return to their original shape after the external forces (i.e., load) causing deformation are removed. An object is elastic whin it comes back to its original size and shape when the load is no longer present. For example, the elasticity of polymers and rubbers is caused by stretching polymer chains under an applied force. In contrast, the elasticity of metals is caused by the deformation (resizing and reshaping) of the lattices under the action of externally applied forces.

The two parameters that determine the elasticity of a material are its (1) elastic modulus and (2) elastic limit.

A high elastic modulus is typical for materials that are hard to deform, or materials that require a high load to achieve a significant strain. For example a steel band.
A low elastic modulus is typical for materials that are easy to deform under a load (e.g., a rubber band).

If the stress under a load becomes too high, then when the load is removed, the material no longer comes back to its original shape and size, but relaxes to a different shape and size (e.g., a deformed Slinky) and the material becomes permanently deformed. The elastic limit is the stress value beyond which the material no longer behaves elastically but becomes permanently deformed.

Our perception of an elastic material depends on both its elastic limit and elastic modulus. For example, all rubbers are characterized by a low elastic modulus and a high elastic limit. As a result, they are easy to stretch and the stretch is noticeable.

When the load increases from zero, the resulting stress is in direct proportion to strain, but only when stress does not exceed some limiting value. For stress values near this linear limit, we describe the elastic behavior in an analogous way as a spring using Hooke’s law.

According to Hooke’s law, the stretch value of a spring under an applied force is directly proportional to the magnitude of the force.
Conversely, the response force from the spring to an applied stretch is directly proportional to the stretch.

The deformation of a material under a load is directly proportional to the load, and the resulting stress is directly proportional to the strain. The linearity limit (or the proportionality limit) is the largest stress value beyond which stress is no longer proportional to strain. Beyond the linearity limit, the relation between stress and strain is on longer linear. When stress becomes larger than the linearity limit but still within the elasticity limit, the behavior is still elastic. But the relation between stress and strains becomes nonlinear.

For stresses beyond the elastic limit, a material exhibits plastic behavior. This means the material deforms irreversibly and does not return to its original shape and size (even when the load is removed). When stress is gradually increased beyond the elastic limit, the material undergoes plastic deformation.

Rubber-like materials show an increase in stress with the increasing strain, which means they become more difficult to stretch and eventually reach a fracture point where they break.
Ductile materials (e.g., metals) show a gradual decrease in stress with increasing strain, which means they become easier to deform as stress-strain values approach the breaking point. For example, bending a paper clip is initially difficult until you’ve bent it enough times.

stress-strain plot — Fig. 12.21 Image Credit: Openstax.#

We can graph the stress-strain relationship (see Figure 12.21), where each material has its own characteristic stress-strain curve.

Strain is a fractional elongation. When the load is gradually increased, the linear behavior (e.g., red line) ends at the linearity limit (point $H$).
For further load increases beyond point $H$, the stress-strain relation is nonlinear but still elastic (i.e., from point $H$ to $E$). Ever larger loads take the stress to the elasticity limit $E$.
Beyond the elasticity limit, it enters plastic deformation and when the load is removed, the material relaxes to a new shape and size along the green line. The material becomes permanently deformed and does not return to its initial shape and size when stress is zero.
A material can plastically deform until the stress reaches the fracture (or breaking) point. Beyond this point, we no longer have one sample of material, so the diagram ends.

Note

The linear, elastic, and plasticity limits denote a range of values rather than a single sharp point.

The value of stress at the fracture point is called breaking stress (or ultimate stress). Materials with similar elastic properties (e.g., metals) may have very different breaking stresses. For example, the ultimate stress for aluminum is $2.2\times 10^8\ {\rm Pa}$ and for steel, it may be as high as $20\times 10^8\ {\rm Pa}$ (depending on the kind of steel).

12.5. In-Class Problems#

12.5.1. Part I#

Problem 1

Find the magnitude of the tension in each supporting cable shown below. In each case, the weight of the suspended body is $100.0\ \text{N}$ and the masses of the cables are negligible.

Image Credit: Openstax.

Problem 2

A uniform plank rests on a level surface as shown below. The plank has a mass of $30\ \text{kg}$ and is $6.0\ \text{m}$ long. How much mass can be placed at its right end before it tips? Hint: When the board is about to tip over, it makes contact with the surface only along the edge that becomes a momentary axis of rotation.

Image Credit: Openstax.

Problem 3

A uniform horizontal strut weighs $400.0\ \text{N}$. One end of the strut is attached to a hinged support at the wall, and the other end of the strut is attached to a sign that weighs $200.0\ \text{N}$. The strut is also supported by a cable attached between the end of the strut and the wall. Assuming that the entire weight of the sign is attached at the very end of the strut, find the tension in the cable and the force at the hinge of the strut.

Image Credit: Openstax.

12.5.2. Part II#

Problem 4

TV broadcast antennas are the tallest artificial structures on Earth. In 1987, a $72.0\ \text{kg}$ physicist placed himself and $400\ \text{kg}$ of equipment at the top of a $610\ \text{m}$-high antenna to perform gravity experiments. By how much was the antenna compressed, if we consider it to be equivalent to a steel cylinder $0.150\ \text{m}$ in radius?

Problem 5

By how much does a $65.0\ \text{kg}$ mountain climber stretch her $0.800\ \text{cm}$ diameter nylon rope when she hangs $35.0\ \text{m}$ below a rock outcropping? (For nylon, $Y = 1.35 \times 10^9\ \text{Pa}$.)

Problem 6

When water freezes, its volume increases by $9.05\%$. What force per unit area is water capable of exerting on a container when it freezes?

Problem 7

An aluminum $(\rho = 2.7\ \text{g/cm}^3)$ wire is suspended from the ceiling and hangs vertically. How long must the wire be before the stress at its upper end reaches the proportionality limit, which is $8.0 \times 10^7\ \text{N/m}^2$?

12.6. Homework#

12.6.1. Conceptual Problems#

Problem 1

Mechanics sometimes put a length of pipe over the handle of a wrench when trying to remove a very tight bolt. How does this help?

Problem 2

A painter climbs a ladder. Is the ladder more likely to slip when the painter is near the bottom or near the top? Explain your reasoning using concepts of torque and equilibrium.

12.6.2. Quantitative Problems#

Problem 3

The uniform seesaw is balanced at its center of mass, as seen below. The smaller boy on the right has a mass of $40.0\ \text{kg}$. What is the mass of his friend?

Image Credit: Openstax.

Problem 4

The forearm shown below is positioned at an angle $\theta$ with respect to the upper arm, and a $5.0\ \text{kg}$ mass is held in the hand. The total mass of the forearm and hand is $3.0\ \text{kg}$, and their center of mass is $15.0\ \text{cm}$ from the elbow.

(a) What is the magnitude of the force that the biceps muscle exerts on the forearm for $\theta = 60^\circ$?
(b) What is the magnitude of the force on the elbow joint for the same angle?
(c) How do these forces depend on the angle $\theta$?

Image Credit: Openstax.

Problem 5

As an oil well is drilled, each new section of drill pipe supports its own weight and the weight of the pipe and the drill bit beneath it. Calculate the stretch in a new $6.00\ \text{m}$-long steel pipe that supports a $100\ \text{kg}$ drill bit and a $3.00\ \text{km}$ length of pipe with a linear mass density of $20.0\ \text{kg/m}$. Treat the pipe as a solid cylinder with a $5.00\ \text{cm}$ diameter.

Problem 6

A uniform rope of cross-sectional area $0.50\ \text{cm}^2$ breaks when the tensile stress in it reaches $6.00 \times 10^6\ \text{N/m}^2$.

(a) What is the maximum load that can be lifted slowly at a constant speed by the rope?
(b) What is the maximum load that can be lifted by the rope with an acceleration of $4.00\ \text{m/s}^2$?

Static Equilibrium and Elasticity

Contents

12. Static Equilibrium and Elasticity#

12.1. Conditions for Static Equilibrium#

12.1.1. Example Problem: Center of Gravity#

12.1.2. Example Problem: A Breaking Tension#

12.2. Examples of Static Equilibrium#

12.2.1. Example Problem: Torque Balance#

12.2.2. Example Problem: Forces in the Forearm#

12.2.3. Example Problem: Ladder Resting Against a Wall#

12.2.4. Example Problem: Forces on Door Hinges#

12.3. Stress, Strain, and Elastic Modulus#

12.3.1. Tensile or Compressive Stress, Strain, and Young’s Modulus#

12.3.1.1. Example Problem: Compressive Stress in a Pillar#

12.3.1.2. Example Problem: Stretching a Rod#

12.3.2. Bulk Stress, Strain, and Modulus#

12.3.2.1. Example Problem: Hydraulic Press#

12.3.3. Shear Stress, Strain, and Modulus#

12.3.3.1. Example Problem: An Old Bookshelf#

12.4. Elasticity and Plasticity#

12.5. In-Class Problems#

12.5.1. Part I#

12.5.2. Part II#

12.6. Homework#

12.6.1. Conceptual Problems#

12.6.2. Quantitative Problems#