Question

A uniform cubical crate is 0.770 m on each side and weighs 530 N. It rests on the floor with one edge against a very small, fixed obstruction.

At what least height above the floor must a horizontal force of magnitude 320 N be applied to the crate to tip it?

Answer 1

h = 0.638 m

Step-by-step explanation:

given,

side of the crate, a = 0.77 m

Weight of the crate,W = 530 N

Horizontal force of magnitude,F = 320 N

let 'h' be the position of force so, that crate is in equilibrium.

Weight of the crate will pass through center of gravity.

Let O be the position where the crate can tip

for a body to be in equilibrium moment about o be equal to zero.

Taking moment about o

`F h - W(a)/(2) = 0`

`320* h = 530* (0.77)/(2)`

`h = (204.05)/(320)`

h = 0.638 m

Hence, For crate to be in equilibrium force should be applied at 0.638 m from bottom.