Abox 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

As we know that friction force on box is given by

here we know that

here we have

m = 12 kg

so now we have

now we will have

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

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