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20 May, 09:57

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?

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  1. 20 May, 10:27
    0
    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
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