A woman of mass m = 51.2 kg sits on the left end of a seesaw—a plank of length L

Question

A woman of mass m = 51.2 kg sits on the left end of a seesaw—a plank of length L = 3.73 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 = 78.2 kg sit if the system (seesaw plus man and woman) is to be balanced?
1.22 m (CORRECT ANSWER)

(b) Find the normal force exerted by the pivot if the plank has a mass of mpl = 11.3 kg.
1.38*10^3 N (CORRECT ANSWER)

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

Suppose a 27.8 kg child sits 1.23 m to the left of center on the same seesaw as the problem you just solved. A second child sits at the end on the opposite side, and the system is balanced.

(a) Find the mass of the second child.
mchild 2 = ________kg ?????

(b) Find the normal force acting at the pivot point.
Fn = _________ N??????

Explanation / Answer

part a

let counter clockwise is positive

when only lady sit on plank torque about pivot = 51.2*9.81*3.73/2 =936.73N-m

now for balnace of plank torque shoud be equal

so

51.2*9.81*3.73/2= 78.2*9.81*x

x= 1.22m from pivot

part b

normal force on pivot = weight by lady + weight by man + weight of plank =51.2*9.91+78.2*9.81+11.3*9.81 = 1385.3875 N = 1.3865*103KN

part c

torque about left corner if only lady sit there is zero because axis is passes through the force so distance is zero

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