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## Q. 8.8

BALANCING ACT

GOAL Apply the conditions of equilibrium and illustrate the use of different axes for calculating the net torque on an object.

PROBLEM A woman of mass $m=55.0 \mathrm{~kg}$ sits on the left end of a seesaw—a plank of length $L=4.00 \mathrm{~m}$, pivoted in the middle as in Figure 8.15. (a) First compute the torques on the seesaw about an axis that passes through the pivot point. Where should a man of mass $M=75.0 \mathrm{~kg}$ sit if the system (seesaw plus man and woman) is to be balanced? (b) Find the normal force exerted by the pivot if the plank has a mass of $m_{\mathrm{pl}}=12.0 \mathrm{~kg}$. (c) Repeat part (a), but this time compute the torques about an axis through the left end of the plank.

STRATEGY In part (a), apply the second condition of equilibrium, $\Sigma \tau=0$, computing torques around the pivot point. The mass of the plank forming the seesaw is distributed evenly on either side of the pivot point, so the torque exerted by gravity on the plank, $\tau_{\text {plank }}$, can be computed as if all the plank’s mass is concentrated at the pivot point. Then $\tau_{\text {plank }}$ is zero, as is the torque exerted by the pivot, because their lever arms are zero. In part (b) the first condition of equilibrium, $\sum \overrightarrow{\mathbf{F}}=0$, must be applied. Part (c) is a repeat of part (a) showing that choice of a different axis yields the same answer. ## Verified Solution

(a) Where should the man sit to balance the seesaw?

Apply the second condition of equilibrium to the plank by setting the sum of the torques equal to zero:

$\tau_{\text {pivot }}+\tau_{\text {plank }}+\tau_{\text {man }}+\tau_{\text {woman }}=0$

The first two torques are zero. Let $x$ represent the man’s distance from the pivot. The woman is at a distance $\ell=L / 2$ from the pivot.

$0+0-M g x+m g(L / 2)=0$

Solve this equation for $x$ and evaluate it:

$x=\frac{m(L / 2)}{M}=\frac{(55.0 \mathrm{~kg})(2.00 \mathrm{~m})}{75.0 \mathrm{~kg}}=1.47 \mathrm{~m}$

(b) Find the normal force $n$ exerted by the pivot on the seesaw.

Apply for first condition of equilibrium to the plank, solving the resulting equation for the unknown normal force, $n$ :

\begin{aligned}& -M g-m g-m_{\mathrm{pl}} g+n=0 \\n &=\left(M+m+m_{\mathrm{pl}}\right) g \\&=(75.0 \mathrm{~kg}+55.0 \mathrm{~kg}+12.0 \mathrm{~kg})\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right) \\n &=1.39 \times 10^{3} \mathrm{~N}\end{aligned}

(c) Repeat part (a), choosing a new axis through the left end of the plank.

Compute the torques using this axis, and set their sum equal to zero. Now the pivot and gravity forces on the plank result in nonzero torques.

\begin{aligned}&\tau_{\text {man }}+\tau_{\text {woman }}+\tau_{\text {plank }}+\tau_{\text {pivot }}=0 \\&-M g(L / 2+x)+m g(0)-m_{\text {pl }} g(L / 2)+n(L / 2)=0\end{aligned}

Substitute all known quantities:

\begin{aligned}&-(75.0 \mathrm{~kg})\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)(2.00 \mathrm{~m}+x)+0 \\&-(12.0 \mathrm{~kg})\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)(2.00 \mathrm{~m})+n(2.00 \mathrm{~m})=0 \\&-\left(1.47 \times 10^{3} \mathrm{~N} \cdot \mathrm{m}\right)-(735 \mathrm{~N}) x-(235 \mathrm{~N} \cdot \mathrm{m}) \\&\quad+(2.00 \mathrm{~m}) n=0\end{aligned}

Solve for $x$, substituting the normal force found in part (b):

$x=1.46 \mathrm{~m}$

REMARKS The answers for $x$ in parts (a) and (c) agree except for a small rounding discrepancy. That illustrates how choosing a different axis leads to the same solution.