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?

the maximum static force that can be applied is equal to the normal forcethe frictional force. the normal force on the box is equal to mg since the floor is flat using 9.81m/s^2 for gravity 12kg9.81m/s^2 = 118N multiplying the normal force by the frictional force you get a 118*.42= 49.6N so overcome the force of static friction on the box a minimum of 49.6N would need to be applied.

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abidemiokin abidemiokin · Answer 2 · 2019-12-16T10:14:11-05:00

Answer: 50.4N

Explanation:

Frictional force (Ff) is the force that opposes the moving force acting on a body.

This frictional force (Ff) is the product of the coefficient of friction(n) and normal reaction (R)

Mathematically, Fm = Ff = nR (since both forces are equal and opposite forces along the horizontal axis.)

NB: Fm = Ff (true for static bodies only)

Also note that the weight (W) of the body will be equal to the normal reaction (R) (forces acting along the y axis on the body) i.e W = R

R = Weight = mg = 12×10 = 120N

n = 0.42

Substituting the datas given into the formula of friction force

Ff = nR = Fm

Fm = 0.42×120

Fm = 50.4N

Therefore the minimum force needed to set the box in motion across the floor is 50.4N