Question

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 2.9 N weight is placed 0.25 m to the right of the support.

Answer 1

`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,

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