Frictional force down a ramp

[arc5005]

The frictional force acting on a block when it begins to slide down an inclined surface is best described by:

A. µk mg cos
theta

B. µk mg sin
theta

C. µs mg cos
theta

D. µs mg sin
theta




A) µk mg costheta


The force of friction acting on a sliding object is: Fkineticfriction = µk*N. Because the object is sliding (as opposed to being stationary or rolling), we need µk, and not µs. C & D are eliminated. The question is reduced to determining whether the normal force is mg costheta or mg sinθ. When the surface is perfectly flat, the normal force is equal to mg. That angle is defined as 0°, so N must be equal to mg cosθ for the normal force to equal the weight, mg.
Fkineticfriction = µk*N
Fkineticfriction = µk*mgcostheta
Fkineticfriction = µk*mg*1


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Why is it the normal force we are looking for here? Usually when I have a problem like this, I would have used the limiting factor approach, and would have thought we were looking for when angle is 0°, then the F = 0, not mg, Or if angle was 90°, then the F = mg, so then it would be µk mg sinθ.

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[ZukosRedemption]

F (friction) = u x F(normal), the normal force is the force from the ramp on the object. Through Newton's third law, this force is mgcos(theta). This video explains why its cos and not sine. There's a KA video explaining why its cos and not sine, basically how the triangle is oriented.

 

[kris400158]

If the angle is 0 then the object is stationary. If you picture a block on a flat surface, such as on a table not moving. It still has a normal force going up, which is equivalent of the weight of the object (w=mg) going down. Therefore, even at an angle of 0, there is a normal force equivalent of mg exerting itself on the object.

 

[JustinM88]

Ultimately the q-stem wants you to find the Ffriction.

In short, it is the normal Force you're looking for because of:
Ffriction = u x Fnormal

(a way to remember this formula: Friction is FUN ^.^)

^I found this picture and I think it illustrates why its mgcostheta and not mgsingtheta without going overboard on details.

 

[BerkReviewTeach]

Why is it the normal force we are looking for here? Usually when I have a problem like this, I would have used the limiting factor approach, and would have thought we were looking for when angle is 0°, then the F = 0, not mg, Or if angle was 90°, then the F = mg, so then it would be µk mg sinθ.

The key word in that question is "sliding," which makes it a question of kinetic friction. You would be on the right path if the question asked for a block "resting" on the incline. For static friction, the magnitude of the static friction up the incline would be equal to the net force down the incline, which is mg[sin(theta)]. However, this question is asking for kinetic friction, which is defined as µk N, which equals µk[mg cos(theta)].