Question

4) The two children shown in the are balanced on a seesaw of negligible mass.
(This assumption is made to keep the example simple-more involved
examples will follow.) The first child has a mass of 26.0 kg and sits 160 m
from the pivot. (a) If the second child has a mass of 32.0 kg, how far is she
from the pivot? (b) What is Fl, the supporting force exerted by the pivot
using the second condition for equilibrium (net t = o), employing any data
given or solved for in part (a).​

Answer 1

Bhoot Bhoot Bhoot Bhoot Bhoot

Answer 2

Given:

Mass of first child = 26 kg

Mass of second child = 32 kg

Distance of first child from the pivot = 160 m

To find:

How far is the second child from the pivot.

Solution:

a) When the center of mass of the seesaw and the two children passes through the pivot, the seesaw is in the static equilibrium. In the balanced position, the net torque about the pivot is zero .

To maintain rotational equilibrium: Torque about pivot = 0

m1 l1 - m2 l2 = 0

m1 l1 = m2 l2

26 * 160 = 32 * l2

l2 = 130 m

Therefore, the second child must be 130 m away from the pivot point.

b) Force exerted by pivot = m1g + m2g

= (26 + 32) 10

= 580 N

Therefore the supporting force exerted by the pivot will be 580 N.