Find the magnitude of the force which when applied to a point (0.40m) to the right of the support will keep the meter stick in equilibrium?

A uniform rigid (1.00m) long bar is supported horizontally at the center. There is no physical connection between the triangular support and the bar. A 4.9 N force is applied to the extreme left-hand end of the stick and a enter image source here 2.9 N weight is placed 0.25 m to the right of the support.

1 Answer
Mar 1, 2018

4.30N

Explanation:

Clearly,the system tends to rotate down,along its left,so let,we need to apply force F at a distance 0.40 m from the mid point of the bar along its left,so that it can maintain rotational equilibrium by acting along with the weight of 2.9N to nullify the excess torque due to the 4.9N force,

enter image source here

so we can write,

4.9 * 0.50 = 2.9*0.25 + F*0.40

so, F=4.30 N

Note here we consider the bar is very very light,so it has almost negligible mass