Applying physics knowledge question

Question

A box that weighs 5.00x10^2 N is sliding down a ramp at a constant speed. The angle the ramp makes with the horizontal is 25 degrees. What is the coefficient of friction between the box and the ramp? 

Answer 1

GIVEN:

Weight = mg = 500N and the angle theta θ = 25^°

The frictional force is equal to the coefficient of friction multiplied by the normal force: F_{f =} μ F_N

The coefficient of friction (mu) is equal to the ratio of the frictional force to the normal force: μ = (F_f / F_N).

UNKNOWN:

So how do you calculate the magnitude of the frictional force and the normal force?

 

The answer is found in the free body diagram of the box on the inclined plane.

We can see 5 forces acting on the box. The weight of the box (mg), the perpendicular force (F_⊥), the parallel force (F_ιι), the frictional force (F_f) and the normal force (F_N).

The normal force is equal in size to the perpendicular force, F_N = F_⊥. 

Since the box is sliding at constant speed then the net force down the ramp is zero. This means that the frictional force is equal in size to the parallel force, F_f = F_ιι .

 

So our equation for the coefficient of friction is now: µ = (F_ιι / F_⊥)

The parallel force and the perpendicular force are components of the weight of the box.

Thus the F_ιι = mg(sinθ) and F_⊥ = mg(cosθ).

Substituting:

µ = [mg(sinθ) / mg(cosθ)]

µ = [500(sin25°) / 500(cos25°)]

µ = [211.3 / 452.1]

µ = 0.467 

 

Now for the shortcut:

µ = tanθ ⇒ because (sinθ/cosθ) = tanθ and 500/500 = 1

µ = tan25°

µ = 0.466