# 9.2 The second condition for equilibrium  (Page 4/8)

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$\text{net}\phantom{\rule{0.25em}{0ex}}{F}_{y}=0$

where we again call the vertical axis the y -axis. Choosing upward to be the positive direction, and using plus and minus signs to indicate the directions of the forces, we see that

${F}_{\text{p}}–{w}_{1}–{w}_{2}=0.$

This equation yields what might have been guessed at the beginning:

${F}_{\text{p}}={w}_{1}+{w}_{2}.$

So, the pivot supplies a supporting force equal to the total weight of the system:

${F}_{\text{p}}={m}_{1}g+{m}_{2}g.$

Entering known values gives

$\begin{array}{lll}{F}_{\text{p}}& =& \left(\text{26.0 kg}\right)\left(9.80\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)+\left(\text{32.0 kg}\right)\left(9.80\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)\\ & =& \text{568 N.}\end{array}$

Discussion

The two results make intuitive sense. The heavier child sits closer to the pivot. The pivot supports the weight of the two children. Part (b) can also be solved using the second condition for equilibrium, since both distances are known, but only if the pivot point is chosen to be somewhere other than the location of the seesaw’s actual pivot!

Several aspects of the preceding example have broad implications. First, the choice of the pivot as the point around which torques are calculated simplified the problem. Since ${F}_{\text{p}}$ is exerted on the pivot point, its lever arm is zero. Hence, the torque exerted by the supporting force ${F}_{\text{p}}$ is zero relative to that pivot point. The second condition for equilibrium holds for any choice of pivot point, and so we choose the pivot point to simplify the solution of the problem.

Second, the acceleration due to gravity canceled in this problem, and we were left with a ratio of masses. This will not always be the case . Always enter the correct forces—do not jump ahead to enter some ratio of masses.

Third, the weight of each child is distributed over an area of the seesaw, yet we treated the weights as if each force were exerted at a single point. This is not an approximation—the distances ${r}_{\text{1}}$ and ${r}_{\text{2}}$ are the distances to points directly below the center of gravity    of each child. As we shall see in the next section, the mass and weight of a system can act as if they are located at a single point.

Finally, note that the concept of torque has an importance beyond static equilibrium. Torque plays the same role in rotational motion that force plays in linear motion. We will examine this in the next chapter.

## Take-home experiment

Take a piece of modeling clay and put it on a table, then mash a cylinder down into it so that a ruler can balance on the round side of the cylinder while everything remains still. Put a penny 8 cm away from the pivot. Where would you need to put two pennies to balance? Three pennies?

## Section summary

• The second condition assures those torques are also balanced. Torque is the rotational equivalent of a force in producing a rotation and is defined to be
$\tau =\text{rF}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$

where $\tau$ is torque, $r$ is the distance from the pivot point to the point where the force is applied, $F$ is the magnitude of the force, and $\theta$ is the angle between $\mathbf{\text{F}}$ and the vector directed from the point where the force acts to the pivot point. The perpendicular lever arm ${r}_{\perp }$ is defined to be

${r}_{\perp }=r\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$

so that

$\tau ={r}_{\perp }F.$
• The perpendicular lever arm ${r}_{\perp }$ is the shortest distance from the pivot point to the line along which $F$ acts. The SI unit for torque is newton-meter $\text{(N·m)}$ . The second condition necessary to achieve equilibrium is that the net external torque on a system must be zero:
$\text{net}\phantom{\rule{0.25em}{0ex}}\tau =0$

By convention, counterclockwise torques are positive, and clockwise torques are negative.

## Conceptual questions

What three factors affect the torque created by a force relative to a specific pivot point?

A wrecking ball is being used to knock down a building. One tall unsupported concrete wall remains standing. If the wrecking ball hits the wall near the top, is the wall more likely to fall over by rotating at its base or by falling straight down? Explain your answer. How is it most likely to fall if it is struck with the same force at its base? Note that this depends on how firmly the wall is attached at its base.

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? (It is also hazardous since it can break the bolt.)

## Problems&Exercises

(a) When opening a door, you push on it perpendicularly with a force of 55.0 N at a distance of 0.850m from the hinges. What torque are you exerting relative to the hinges? (b) Does it matter if you push at the same height as the hinges?

a) $\text{46.8 N·m}$

b) It does not matter at what height you push. The torque depends on only the magnitude of the force applied and the perpendicular distance of the force’s application from the hinges. (Children don’t have a tougher time opening a door because they push lower than adults, they have a tougher time because they don’t push far enough from the hinges.)

When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of the bolt. (a) How much torque are you exerting in newton × meters (relative to the center of the bolt)? (b) Convert this torque to footpounds.

Two children push on opposite sides of a door during play. Both push horizontally and perpendicular to the door. One child pushes with a force of 17.5 N at a distance of 0.600 m from the hinges, and the second child pushes at a distance of 0.450 m. What force must the second child exert to keep the door from moving? Assume friction is negligible.

23.3 N

Use the second condition for equilibrium $\text{(net τ = 0)}$ to calculate ${F}_{\text{p}}$ in [link] , employing any data given or solved for in part (a) of the example.

Repeat the seesaw problem in [link] with the center of mass of the seesaw 0.160 m to the left of the pivot (on the side of the lighter child) and assuming a mass of 12.0 kg for the seesaw. The other data given in the example remain unchanged. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium.

Given:

$\begin{array}{ccc}{m}_{1}& =& \text{26.0 kg,}\phantom{\rule{0.25em}{0ex}}{m}_{2}=\text{32.0 kg,}\phantom{\rule{0.25em}{0ex}}{m}_{\text{s}}=\text{12.0 kg,}\phantom{\rule{0.25em}{0ex}}\\ {r}_{1}& =& \text{1.60 m,}\phantom{\rule{0.25em}{0ex}}{r}_{\text{s}}=\text{0.160 m, find (a)}\phantom{\rule{0.25em}{0ex}}{r}_{2,\phantom{\rule{0.25em}{0ex}}}\text{(b)}\phantom{\rule{0.25em}{0ex}}{F}_{\text{p}}\end{array}$

a) Since children are balancing:

$\begin{array}{}\text{net}\phantom{\rule{0.25em}{0ex}}{\tau }_{\text{cw}}=–\text{net}\phantom{\rule{0.25em}{0ex}}{\tau }_{\text{ccw}}\\ ⇒{w}_{1}{r}_{1}+{m}_{\text{s}}{\mathit{gr}}_{\text{s}}={w}_{2}{r}_{2}\\ \end{array}$

So, solving for ${r}_{2}$ gives:

$\begin{array}{lll}{r}_{2}& =& \frac{{w}_{1}{r}_{1}+{m}_{\text{s}}{\mathit{gr}}_{\text{s}}}{{w}_{2}}=\frac{{m}_{1}{\mathit{gr}}_{1}+{m}_{\text{s}}{\mathit{gr}}_{\text{s}}}{{m}_{2}g}=\frac{{m}_{1}{r}_{1}+{m}_{\text{s}}{r}_{\text{s}}}{{m}_{2}}\\ & =& \frac{\left(\text{26.0 kg}\right)\left(\text{1.60 m}\right)+\left(\text{12.0 kg}\right)\left(\text{0.160 m}\right)}{\text{32.0 kg}}\\ & =& \text{1.36 m}\end{array}$

b) Since the children are not moving:

$\begin{array}{}\text{net}\phantom{\rule{0.25em}{0ex}}F=0={F}_{\text{p}}–{w}_{1}–{w}_{2}–{w}_{\text{s}}\\ ⇒{F}_{\text{p}}={w}_{1}+{w}_{2}+{w}_{s}\end{array}$

So that

$\begin{array}{lll}{F}_{\text{p}}& =& \left(\text{26.0 kg}+\text{32.0 kg}+\text{12.0 kg}\right)\left(\text{9.80}\phantom{\rule{0.25em}{0ex}}\text{m}/{\text{s}}^{2}\right)\\ & =& \text{686 N}\end{array}$

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