Coefficient of kinetic friction?

Question

A 1100 N crate is being pushed across a level floor at a constant speed by a force of 300 N at an angle of 20.0° below the horizontal.
(a)What is the coefficient of kinetic friction between the crate and the floor?

(b)If the 300 N force is instead pulling the block at an angle of 20.0° above the horizontal, what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in (a).

I know the number is close to .2562 for (a). I was trying different things and that answer was within 10%, but I have no idea what I did, and no idea what I did wrong

Answer 1

> ...answer was within 10%...



My first guess is maybe you forgot to count the downward component of the push, when calculating the normal force?



> a constant speed...



That means zero acceleration, which means the net force on the crate is zero. In turn, that means that the force of friction exactly matches (and cancels out) the horizontal component of the push:



Force of friction = (Fn)(μ)

(where Fn = normal force, and μ = coefficient of friction)



Horizontal component of push = (300N)(cos(20))



Since they are equal:

(Fn)(μ) = (300N)(cos(20))



Next step: How much is Fn ? It equals the upward force on the crate (exerted by the floor), and it exactly matches (and cancels with) the downward force on the crate.



The downward force on the crate is the sum of its weight (1100N) plus the downward component of the push (300N)(sin(20)). So:



Fn = 1100N + (300N)(sin(20))



Putting it all together (substituting for Fn):

(1100N + (300N)(sin(20)))(μ) = (300N)(cos(20))



Now use algebra to solve for μ

μ = (300N)(cos(20)) ⁄ (1100N + (300N)(sin(20)))

Answer 2

Make a free-body diagram: I'll tell you that there is gravity, the floor, the frictional force and that other unnamed force acting on the crate.

Separate everything into X and Y axes, and since the crate is moving at a constant speed, you know the net force is 0 (I'm assuming this is non-relativistic...if you're in highschool then its not).

You should probably get F_unnamed cos (20) = F_friction.



Once you do a), you know the magnitude of all the forces. It should be pretty easy from there...aka F_net = F_1 + F_2

Answer 3

The coefficient of kinetic friction is the ratio between the kinetic friction rigidity and the traditional rigidity performing on the object. This ratio is greater often than not a relentless variety finding at the variety of fabrics in touch. Kinetic friction is that actual resisting rigidity which varies with the traditional rigidity performing on the object. they selection linearly, meaning that if the traditional rigidity doubles, kinetic friction doubles, however the ratio will proceed to be a relentless variety. *Edit to function: Kinetic friction will proceed to be a relentless fee of F = (coefficient of kinetic friction) * N, so as which will extremely be the optimal friction. With static friction on the different hand, it stages from 0 to (coefficient of static friction)*N.