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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?

Respuesta :

Answer:

h = 0.638 m

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

[tex]F h - W\dfrac{a}{2} = 0[/tex]

[tex]320\times h = 530\times \dfrac{0.77}{2}[/tex]

[tex]h = \dfrac{204.05}{320}[/tex]

h = 0.638 m

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

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