A woman of mass m = 51.3 kg sits on the left end of a seesaw—a plank of length L

Question

A woman of mass m = 51.3 kg sits on the left end of a seesaw—a plank of length L = 3.83 m, pivoted in the middle as shown in the figure. (a) First compute the torques on the seesaw about an axis that passes through the pivot point. Where should a man of mass M = 76.5 kg sit if the system (seesaw plus man and woman) is to be balanced? m (b) Find the normal force exerted by the pivot if the plank has a mass of mpl = 13.6 kg. N (c) Repeat part (a), but this time compute the torques about an axis through the left end of the plank. m

Explanation / Answer

a) Torque due to gravity + Torque due to pivot + Torque due to man + Torque due to Woman = 0

0 + 0 - (M*g*x) + (m*g*(L/2)) = 0

M*g*x = m*g*(L/2)

x = (m/M)*(L/2) = (51.3/76.5)*(3.83/2) = 1.284 m

b) Using the equilibrium conditions

Mg + mg + (Mpl*g) = N

N = (M+m+Mpl)*g = (76.5+51.3+13.6)*9.8

N = 1385.72 newton

C) Torque due to Planck + Torque due to man + Torque due to pivot = 0

-Mpl*g*(L/2) -Mg*((L/2)+x) + N*(L/2) = 0

-(13.6*9.8*(3.83/2)) - (76.5*9.8*((3.83/2)+x)) + 1385.72*(3.83/2) = 0

x = 1.28 m

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