a woman of mass m=55.0 kg sits on the left end of a seesaw plank of length L=4.0
ID: 1468486 • Letter: A
Question
a woman of mass m=55.0 kg sits on the left end of a seesaw plank of length L=4.00m, pivoted in the middle as in the figure. a) where should a man of mass M= 75.0 kg sit if the system (seesaw plus man and woman) is to be balanced? (compute the torques on the seesaw about an axis that passes through the pivot point) b) find the normal force exerted by the pivot if the plank has a mass of mpl= 12.0kg c) repeat part (a) , but this time compute the torques about an axis through the left end of the plank
Explanation / Answer
Suppose a woman sits on the left of center on the same seesaw . A man sits at the end on the opposite side, and the system is balanced.
plank length L = 4m, pivot point at 4 ÷ 2 =2 m
The right end is 2m from the pivot point
Torque caused by man = m * 9.8 * 2 =
Torque caused by woman left of center = 55* 9.8 * 4
m * 9.8 * 2 =55 * 9.8 * 4
m = (55 * 4) ÷ 2 = 110 kg
The normal force acting at the pivot point supports the weight of both children
Weight =Total Weight
woman W = 9.8 * 55 = 539
man W = 9.8 * 110 = 1078
Fn = 1078+539
F = 1617 N
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