5. Repeat the seesaw problem in Example 9.1 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.
EXAMPLE 9.1
She Saw Torques On A Seesaw
The two children shown in Figure 9.8 are balanced on a seesaw of negligible mass. (This assumption is made to keep the example simple—more involved examples will follow.) The first child has a mass of 26.0 kg and sits 1.60 m from the pivot. (a) If the second child has a mass of 32.0 kg, how far is she from the pivot? (b) What is Fp?
Strategy
Both conditions for equilibrium must be satisfied. In part (a), we are asked for a distance; thus, the second condition (regarding torques) must be used, since the first (regarding only forces) has no distances in it. To apply the second condition for equilibrium, we first identify the system of interest to be the seesaw plus the two children. We take the supporting pivot to be the point about which the torques are calculated. We then identify all external forces acting on the system.
Solution (a)
The three external forces acting on the system are the weights of the two children and the supporting force of the pivot. Let us examine the torque produced by each. Torque is defined to be τ = rFsinθ.
Here θ = 90º, sinθ = 1 for all three forces. That means r⊥ = r for all three. The torques exerted by the three forces are first,
τ1 = r1w1
second,
τ2 = –r2w2
and third,
τp = rpFp = 0⋅Fp = 0.
Note that a minus sign has been inserted into the second equation because this torque is clockwise and is therefore negative by convention. Since Fp acts directly on the pivot point, the distance rp is zero. A force acting on the pivot cannot cause a rotation, just as pushing directly on the hinges of a door will not cause it to rotate. Now, the second condition for equilibrium is that the sum of the torques on both children is zero. Therefore,
τ2 = –τ1,
or
r2w2 = r1w1.
Weight is mass times the acceleration due to gravity. Entering mg for w,
r2m2g = r1m1g.
Solve this for the unknown r2:
r2 = r1m1/m2.
The quantities on the right side of the equation are known; thus, r2 is
r2 = (1.60 m)(26.0 kg)/(32.0 kg) = 1.30 m.
As expected, the heavier child must sit closer to the pivot (1.30 m versus 1.60 m) to balance the seesaw.
Solution (b)
This part asks for a force Fp.
netF = 0.
The forces are all vertical, so that we are dealing with a one-dimensional problem along the vertical axis; hence, the condition can be written as
netFy = 0.
Choosing upward to be the positive direction, and using plus and minus signs to indicate the directions of the forces, we see that
Fp – w1 – w2 = 0.
This equation yields what might have been guessed at the beginning:
Fp = w1 + w2.
So, the pivot supplies a supporting force equal to the total weight of the system:
Fp = m1g + m2g.
Entering known values gives
Fp = (26.0 kg)(9.80 m/s^2) + (32.0 kg)(9.80 m/s^2) = 568 N.