7.2 The second condition for equilibrium  (Page 4/8)

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

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

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

Entering known values gives

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.

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
  $\tau =\text{rF}\text{sin}\theta$

  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

  $r_{\perp }=r\text{sin}\theta$

  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:
  $\text{net}\tau =0$

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

Conceptual questions

Problems&Exercises