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 9.9 ?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 ? p , the supporting force exerted by the pivot?
Solution 5PE Step-by-step solution Step 1 of 2 Net applied torque is zero here. So, Here, is the torque on right side from pivot, is the torque on the left side of pivot, w1 is the weight of fir st child and 1 is the distance of first child from p ivot point, s denotes the mass o f seesaw and r s is the distance of center of mass of seesaw from the pivot point, is the distance at which second child whose mass 32 kg sits and w2 is the weight of second child.
