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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

Respuesta :

As we know that friction force on box is given by

[tex]F_s = \mu_s N[/tex]

here we know that

[tex]N = mg[/tex]

here we have

m = 12 kg

[tex]\mu_s = 0.42[/tex]

so now we have

[tex]N = 12(9.8) = 117.6 N[/tex]

now we will have

[tex]F_s = 0.42(12)(9.8)[/tex]

[tex]F_s = 49.4 N[/tex]

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

Histrionicus

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

                                       


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