Physics Forces?

Question

A 25 kg block is at rest on a 18 degree incline...What is the force of static friction between the ramp and the block?...what is the coefficient of static friction between the ramp and the block?

All help is appreciated

Answer 1

Hope you have your answer.

Answer 2

To answer the last question, we need to assume the 18 deg angle is the maximum angle just before the block begins to slide.

The frictional force is F = kN = k mg cos(theta); where k is the coefficient of friction, m = 25 kg mass, g = 9.81 m/sec^2, and theta = 18 deg angle of incline.

To find k, note that f = ma = W sin(theta) - k mg cos(theta) = 0; where W = mg the weight of the block and f is the net force on the block, which is not yet sliding; so a = 0 the acceleration. Thus, mg sin(theta) = k mg cos(theta) and k = tan(theta) = tan(18 deg). You can look this up.

So F = kN = [sin(theta)/cos(theta)] mg cos(theta) = mg sin(theta) = 25*9.81*sin(18). You can do the math.

NB: k does not equal tan(18) if theta = 18 deg is not the maximum incline just before the block slides. If you testbook did not specify it as the maximum, then the question is not a good question.

Answer 3

According to Newton's second law: Unbalanced forces cause acceleration, or no acceleration when the net force is 0N. Because the block is resting on the incline, it is not moving, thus, the net force the acts on the block is 0N

First, assume there is no friction, the block will acclereate downward. The force the cause it to accelerete is the weight that is taken down parellel to the incline (called weight(x)). In this problem, the block is not moving, the static friction force must equal to the weight(x) so that the net force is 0N

draw a digram, and you'll see weight(x) = sin(theta)mg

Weight(x) = sin(18)(25 x 10)
Weight(x) = 77.25N

Friction force must be at least equal to 77.25N

we know that Friction = μ(Fnormal)

77.25 = μ(Fnormal)

The normal force in this case is equal to the weight of the block that is taken perpendicular to the incline.

Fnormal = cos(theta)mg

77.25 = μ(cos(theta)mg)

77.25 = μ (cos(18)25 x 10)

μ = 77.25 / cos(18)(25x10)

μ = .325

Answer 4

Nothing more to add. Thanks for the previous answerers