Question

A box of mass 12 kg is at rest on a flat floor. The coefficient of static friction between the box and floor is 0.42. What is the minimum force needed to set the box in motion across the floor?

31 N

62 N

49 N

119 N

Answer 1

Given data:

mass of the box (m) = 12 kg,

W = m.g

= 12 kg × 9.8 m/s²

= 117.6 N

coefficient of static friction (μ) = 0.42,

determine the minimum force needed to set the box in motion(F) = ?

Important Point: The minimum force required to move a body is proportional to the weight of the body. Please find the attached figure.

F ∝ W

F = μ. W Newtons

μ = It is the constant proportionality also known as coefficient of static friction.

The normal reaction of the body R = W

Therefore,

F = μ . R Newtons

Where,

F = Minimum force required to move a body, N

F = 0.42 × 117.6

= 49.39 N

The minimum force is needed to move a body on flat floor is 49 N

Answer 2

As we know that friction force on box is given by

`F_s = \mu_s N`

here we know that

`N = mg`

here we have

m = 12 kg

`\mu_s = 0.42`

so now we have

`N = 12(9.8) = 117.6 N`

now we will have

`F_s = 0.42(12)(9.8)`

`F_s = 49.4 N`

so it required minimum 49 N(approx) force to move the block