Question

A 2 ft x 2 ft x 2 ft box weighs 100 pounds, and the weight is evenly distributed. What is the magnitude of the minimum horizontal force, T, required at the top edge of the box totip the box over? Assume that the box will not slide when the force is applied.

Answer 1

Answer:

Explanation:

Let the force required be F . It is applied at the top of the box . The box is likely to turn about a corner . Torque of this force about this corner

= F x 2

This torque will try to turn the box . On the other hand the weight which is acting at CM will create a torque about the same corner . This torque will try to prevent the box to turn around the corner.

This torque of weight

= 100 x 1

= 100 pound ft.

For equilibrium

Torque of F = torque of weight.

F x 2  = 100

F = 50 pounds .

Answer 2

Answer:

50 lb

Explanation:

Given:

Edge of the cubical box = 2 ft

weight of the box, F = 100 pounds

Horizontal force = T

As the box tip at the right bottom corner so take the moments of force about this point which are in equilibrium.

T x 2 = 100 x 1

T = 50 lb

Thus, the force T is 50 lb.