Coefficient of kinetic friction


by jillgotcher
Tags: coefficient, friction, kinetic
jillgotcher
jillgotcher is offline
#1
Sep17-08, 04:51 PM
P: 1
A 1000-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 degrees 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 degrees above the horizontal, what will be the acceleration of the crate? (Assume that the coefficient of friction is the same as in a)

I have not a clue how to do this problem.
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LowlyPion
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#2
Sep17-08, 06:08 PM
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P: 5,346
Quote Quote by jillgotcher View Post
A 1000-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 degrees 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 degrees above the horizontal, what will be the acceleration of the crate? (Assume that the coefficient of friction is the same as in a)

I have not a clue how to do this problem.
Draw a diagram of the crate on the floor. Now draw the forces on the crate.
One force vertically toward the floor is 1000N. This is part of the normal force. Now someone is pressing downward at an angle of 20 degrees below the horizontal. There are 2 components. There is the normal component into the floor. What is that? That plus the 1000N is the total into the floor.

Next you know that it says constant velocity. That means the net horizontal forces are 0, or it would be accelerating right? What are the net forces horizontally?

There is the horizontal component of force and it is offset by the frictional force. This means that the Horizontal component of force times = the Total normal force times the coefficient of friction they have asked you for.

This means that the coefficient u = Horizontal Force/Total normal force.

For the second part the force is pulling upward along the same line. Rather than adding to the total normal force that vertical component is subtracted. Multiplying that times the u that you calculated gives the resistance. Subtracting that from the horizontal gives you the net force. This means that it is in motion with a net force and you will need to calculate that.
Dschumanji
Dschumanji is offline
#3
Sep17-08, 06:15 PM
P: 138
Starting with part (a), what is the relation between what is given and the coefficient of kinetic friction?

fk = [tex]\mu[/tex][tex]_{k}[/tex]n

It is clear what you must find to solve for the value of [tex]\mu[/tex][tex]_{k}[/tex] from the above equation. You know that the crate is moving with constant velocity so it is in equilibrium according to Newton's first law. Remember, n is not always equal to the weight of the object on a surface.

carltouss619
carltouss619 is offline
#4
Oct5-08, 05:26 PM
P: 8

Coefficient of kinetic friction


I'm going to try to help because I need practice doing these problems so.
Draw out a Free Body Diagram.

g= 9.8 & m= F/g & ()=theta

the horizontal Forces I got F= mgsin()+300N-Fk= 0 ( no acceleration)

Solve for Fk (the friction force acting on the block)

Fk= -mgsin()-300N

the vertical forces I got F= n-mgcos()-1000N=0


Next solve for n=mgcos()+1000N

then since Fk= u n (normal force) solve for u

u= Fk/ n


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