MODEL Model the arm as two rigid rods connected by a hinge. We’ll ignore the arm’s weight because it is so much less than that of the barbell. Although the tendon pulls at a slight angle, it is close enough to vertical that we’ll treat it as such.

VISUALIZE FIGURE 12.32 shows the forces acting on our simplified model of the forearm. The biceps pulls the forearm up against the upper arm at the elbow, so the force $\vec{F}{}_{\text{elbow}}$ on the forearm at the elbow—a force due to the upper arm—is a downward force.

FIGURE 12.32 A pictorial representation of the forces involved.

SOLVE Static equilibrium requires both the net force and the net torque on the forearm to be zero. Only the y-component of force is relevant, and setting it to zero gives a first equation:

Because each arm supports half the weight of the barbell, $F_{\text{barbell}}=450\text{ N}$. We don’t know either $F_{\text{tendon}}$ or $F_{\text{elbow}}$, nor does the force equation give us enough information to find them. But the fact that the net torque also must be zero gives us that extra information. The torque is zero about every point, so we can choose any point we wish to calculate the torque. The elbow joint is a convenient point because force $\vec{F}{}_{\text{elbow}}$ exerts no torque about this point; its moment arm is zero. Thus the torque equation is

The tension in the tendon tries to rotate the arm ccw, so it produces a positive torque. Similarly, the torque due to the barbell is negative. We can solve the torque equation for $F_{\text{tendon}}$ to find

REVIEW The short distance $d_{\text{tendon}}$ from the tendon to the elbow joint means that the force supplied by the biceps has to be very large to counter the torque generated by a force applied at the opposite end of the forearm. Although we ended up not needing the force equation in this problem, we could now use it to calculate that the force exerted at the elbow is $F_{\text{elbow}}=3450\text{N}$. These large forces can easily damage the tendon or the elbow.

MODEL Model the plank as a uniform rigid body with its center of mass at the center.

VISUALIZE FIGURE 12.34 shows the forces acting on the plank. Both supports exert upward forces. $\vec{n}{}_{\text{A}}$ and $\vec{n}{}_{\text{B}}$ are the normal forces of Adrienne’s and Bo’s feet pushing down on the board.

FIGURE 12.34 A pictorial representation of the forces on the plank.

SOLVE Because the plank is resting on the supports, not held down, forces $\vec{n}{}_1$ and $\vec{n}{}_2$ must point upward. (The supports could pull down if the plank were nailed to them, but that’s not the case here.) Force $\vec{n}{}_1$ will decrease as Bo moves to the right, and the tipping point occurs when $n_1=0$. The plank remains in static equilibrium right up to the tipping point, so both the net force and the net torque on it are zero. The force equation is

Adrienne is at rest, with zero net force, so her downward force on the board, an action/reaction pair with the upward normal force of the board on her, equals her weight: $n_{\text{A}}=m_{\text{A}}g$. Bo’s center of mass oscillates up and down as he walks, so he’s not in equilibrium and, strictly speaking, $n_{\text{B}}\ne m_{\text{B}}g$. But we’ll assume that he edges out onto the board slowly, with minimal bouncing, in which case $n_{\text{B}}=m_{\text{B}}g$ is a reasonable approximation.

We can again choose any point we wish for calculating torque. Let’s use the support on the left. Adrienne and the support on the right exert positive torques about this point; the other forces exert negative torques. Force $\vec{n}{}_1$ exerts no torque because it acts at the pivot point. Thus the torque equation is

At the tipping point, where $n_1=0$, the force equation gives $n_2=$ $(m_{\text{A}}+m_{\text{B}}+M)g$. Substituting this into the torque equation and then solving for Bo’s position give

Bo doesn’t quite make it to the end. The plank tips when he’s 6.3 m past the left support, our pivot point, and thus 3.3 m past the support on the right.

REVIEW We could have solved this problem somewhat more simply had we chosen the support on the right for calculating the torques. However, you might not recognize the “best” point for calculating the torques in a problem. The point of this example is that it doesn’t matter which point you choose.

MODEL The ladder is a rigid rod of length L. To not slip, it must be in both translational equilibrium $(\vec{F}{}_{\text{net}}=\vec{0})$ and rotational equilibrium $({\tau }_{\text{net}}=0)$.

VISUALIZE FIGURE 12.35 shows the ladder and the forces acting on it.

FIGURE 12.35 A ladder in total equilibrium.

SOLVE The x- and y-components of $\vec{F}{}_{\text{net}}=\vec{0}$ are

The net torque is zero about any point, so which should we choose? The bottom corner of the ladder is a good choice because two forces pass through this point and have no torque about it. The torque about the bottom corner is

The signs are based on the observation that $\vec{F}{}_{\text{G}}$ would cause the ladder to rotate ccw while $\vec{n}{}_2$ would cause it to rotate cw. All together, we have three equations in the three unknowns $n_1$, $n_2$, and $f_{\text{s}}$. If we solve the third for $n_2$,

we can then substitute this into the first to find

Our model of friction is $f_{\text{s}}\le f_{\text{s max}}={\mu }_{\text{s}}{n}_1$. We can find $n_1$ from the second equation: $n_1=Mg$. Using this, the model of static friction tells us that

Comparing these two expressions for $f_{\text{s}}$, we see that ${\mu }_{\text{s}}$ must obey

Thus the minimum value of the coefficient of static friction is 0.29.

REVIEW You know from experience that you can lean a ladder or other object against a wall if the ground is “rough,” but it slips if the surface is too smooth. 0.29 is a “medium” value for the coefficient of static friction, which is reasonable.