Question

A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-kg object be placed so as to experience a net force of zero from the other two objects?

Answer 1

1.045 m from 120 kg

Step-by-step explanation:

m1 = 120 kg

m2 = 420 kg

m = 51 kg

d = 3 m

Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

`(Gm_(1)m)/(y^(2))=(Gm_(2)m)/(\left ( d-y \right )^(2))`

`(m_(1))/(y^(2))=(m_(2))/(\left ( d-y \right )^(2))`

`(3-y)/(y)=\sqrt{(7)/(2)}`

3 - y = 1.87 y

3 = 2.87 y

y = 1.045 m

Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.