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Fiesta28 [93]
3 years ago
9

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 horizonta

l 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.
Physics
2 answers:
Tems11 [23]3 years ago
7 0

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 .

Natasha_Volkova [10]3 years ago
7 0

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.

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