A woman of mass m = 51.9 kg sits on the left end of a seesaw—a plank of length L

Question

A woman of mass m = 51.9 kg sits on the left end of a seesaw—a plank of length L = 4.29 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 = 65.3 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 = 11.1 kg.
  N

(c) Repeat part (a), but this time compute the torques about an axis through the left end of the plank.

Explanation / Answer

here,

mass of the woman , mw = 51.9 kg

length of plank , l = 4.29 m

(a)

mass of man , M = 65.3 kg

let the distance of man be x

taking moment of force about the pivot be zero

mw*l/2 - M*x =0

51.9 * 4.29/2 - 65.3 * x = 0

x = 1.7 m

the man should sit at 1.7 m to the right

(b)

the normal force exerted , N = M*g + mw*g + mpl*g

N = 1257.34 N

the normal force exerted by the pivot is 1257.34 N

(c)

for the axis through the left end of the plank

mpl*l/2 + x*M = 0

11.1 * 4.29/2 + x*65.3 = 0

x = 0.36 m

the distance from the left end for the man is 0.36 m

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