A woman of mass m = 53.8 kg sits on the left end of a seesaw—a plank of length L

Question

A woman of mass m = 53.8 kg sits on the left end of a seesaw—a plank of length L = 3.95 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 = 75.2 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 = 11.3 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) A women is sit on the left end of the seesaw therefore her distance from the pivot is
(3.95/2) = 1.975 m
Therefore the torque about the pivot is = force*distance = (53.8*9.81)*(1.975) = 1042.36 Nm
So to balance the seesaw , man should sit at a distance that his torque about the pivot should be equal to the
women torque
that is
(75.2*9.81) *X = 1042.36
X = 1.413 m
therefore the distance of the man from the pivot is 1.413 m
(b) If the plank mass is 11.3 kg therefore the net mass acting downward is of women + men + planck mass
Now ,
Let us consider that the reaction of the plank is R.
since the Centre of mass of the plank has no acceleration therefore
R = (53.8+75.2+11.3)*9.81 = 1376.34 N
(c) If we consider that both men and women are on the seesaw therefore the torque about the left hand will be due to the weight of the man and the normal reaction at the pivot .
Man is at distance of = 1.975 + 1.413 = 3.385 m
Therefore
= R*(1.975)-(75.2*9.81 )*(3.385)
= 221.11 N m

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