Question

A 123 kg box is resting on the ground. The coefficient of static friction is 0.34. What force must be applied to the box to start it moving?

Answer 1

Answer:

410.254 N

Explanation:

Force: This can be defined as the product of the mass of a body. The Unit of force is Newton (N)

Deduced from the question,

Force applied to the box to start it moving = Force of friction.

Fₐ = F

Where Fₐ = Force applied to the box to start it moving, F = Force of friction.

But,

F = μR.......................... Equation 1

Where R = normal reaction, μ = coefficient of static friction.

R = W =  mg ( on a level surface)

Where m = mass of the box = 123 kg, g = acceleration due to gravity = 9.81 m/s²

R = 123×9.81

R = 1206.63 N.

Also, μ = 0.34

Substituting into equation 1

F = 1206.63×0.34

F = 410.254 N.

Thus the force applied to the box to start it moving = 410.254 N