A woman of mass m 561 kg sits on the left end of a seesawa

A woman of mass m = 56.1 kg sits on the left end of a seesaw—a plank of length L = 3.67 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 = 65.5 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 = 12.1 kg.

N

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

m

Solution

a)

Torque = mg(L/2) = 56.1 x 9.8 x 3.67 /2 = 1008.85 Nm

d = distance of man from pivot

Torque = Mgd

1008.85 = 65.5 x 9.8 d

d = 1.6 m

b)

F_n = Mg + mg + m_plg = (56.1 + 65.5 + 12.1) (9.8) = 1310.3 N

c)

T_woman = 0

T_pl = m_pl g (L/2) = 12.1 x 9.8 (3.67/2) = 217.6 Nm

T_man = Mg(d + L/2) = 65.5 x 9.8 (1.6 + 1.8) = 2182.5 Nm