What is the coefficient of kinetic friction

[ilkjester]

Homework Statement

If you use a horizontal force of 30.0 N to slide a 12.0-kg wooden crate across a floor at a constant velocity, what is the coefficient of kinetic friction between the crate and the floor.

Homework Equations

Fkf=μFn

The Attempt at a Solution

Im not sure how to find the μ

 

[PhanthomJay]

Homework Statement

If you use a horizontal force of 30.0 N to slide a 12.0-kg wooden crate across a floor at a constant velocity, what is the coefficient of kinetic friction between the crate and the floor.

Homework Equations

Fkf=μFn

The Attempt at a Solution

Im not sure how to find the μ

You need to identify all the forces acting on the crate, and then apply Newton's laws in both the x and y direction.

 

[ilkjester]

Humm I don't think i know how to do it still but i don't know ill try my attempt at the problem and post what i get.

 

[ilkjester]

Ok well i got 117.6 when i did the problem not sure if its right or not.

 

[PhanthomJay]

Ok well i got 117.6 when i did the problem not sure if its right or not.

You got 117.6 for which force? Please show your work.

 

[ilkjester]

Well i have no idea how i did it or if i even did it right. But this is what i did. I kinda just copied a similar problem they did in the book and filled in the differences.
Fp= μmg
fp= μ(12kg(9.8)
fp-117.6

 

[pooface]

no 117.72 N is the Normal force on the crate.

 

[ilkjester]

no 117.72 N is the Normal force on the crate.

aww man and i thought i had something. Well I guess ill have to keep trying something I don't know why I am not getting this problem the rest of them were easy i think once i get it I am going to be like duhh.

 

[sdstewar]

Think about Newtons second law. The object is in constant velocity, so there is no NET force on the object. It's in equilibrium.

 

[ilkjester]

Homework Statement

If you use a horizontal force of 30.0 N to slide a 12.0-kg wooden crate across a floor at a constant velocity, what is the coefficient of kinetic friction between the crate and the floor.

Homework Equations

Fkf=μFn

The Attempt at a Solution

Im not sure how to find the μ

ok i think I've got this now. But its assuming that the floor is wooden.
Kinetic friction force Ff kinetic =μkFn
Now μk wood on wood equals 0.20
so Ff kinetic=0.20(30)
Ff kinetic=6
does this sound right.

 

[sdstewar]

Well.. the question says your trying to find the coefficient of kinetic friction, not the force of kinetic friction.

 

[ilkjester]

Man looks like its back to the drawing boards again. Thank you all for the help and tips so far though.

 

[Sourabh N]

Kinetic friction force Ff kinetic =μkFn

You can find Fn by balancing forces in vertical direction. Now frictional force is balanced by external force (30 N). Thus you can find $$\mu$$$$_{k}$$

Cheers

 

[ilkjester]

Ok now i think i might have something or its just my bizarre attempt at something.
The coefficient of kinetic friction is defined as the ratio of the kinetic friction force (F) between the surfaces in contact to the normal force: Ff/N.
So Ff/N which i found out earlier the friction force is 6 and the normal force is 117.72 so 6/117.72=.0509

 

[ilkjester]

You can find Fn by balancing forces in vertical direction. Now frictional force is balanced by external force (30 N). Thus you can find $$\mu$$$$_{k}$$

Cheers

so would it be 6=μk(30N)
then divide by 30 on both sides so 6/30=.2μk

 

[D H]

ok i think I've got this now. But its assuming that the floor is wooden.
Kinetic friction force Ff kinetic =?kFn
Now ?k wood on wood equals 0.20
so Ff kinetic=0.20(30)
Ff kinetic=6
does this sound right.

You are supposed to compute $u_k$, not look it up. And no, this is not correct. The crate will accelerate at 24N/12kg=2 m/s² if a person applies a 30N horizontal force in one direction and friction only applies a 6N force in the opposite direction.

so would it be 6=?k(30N)
then divide by 30 on both sides so 6/30=.2?k

You are making this too hard. The crate is moving at a constant velocity. This means that the net horizontal force acting on the crate is zero. The person is applying a 30N force in one direction. To make the net horizontal force zero, friction must therefore be applying a 30N force in the opposite direction. The coefficient of kinetic friction is just the ratio of the frictional force (which you now know) to the normal force (which you calculated in post #4).

 

[ilkjester]

You are supposed to compute $u_k$, not look it up. And no, this is not correct. The crate will accelerate at 24N/12kg=2 m/s² if a person applies a 30N horizontal force in one direction and friction only applies a 6N force in the opposite direction.



You are making this too hard. The crate is moving at a constant velocity. This means that the net horizontal force acting on the crate is zero. The person is applying a 30N force in one direction. To make the net horizontal force zero, friction must therefore be applying a 30N force in the opposite direction. The coefficient of kinetic friction is just the ratio of the frictional force (which you now know) to the normal force (which you calculated in post #4).

so would it be 6=μk(30N)
then divide by 30 on both sides so 6/30=.2μk

so was i right in my one post then

Ok now i think i might have something or its just my bizarre attempt at something.
The coefficient of kinetic friction is defined as the ratio of the kinetic friction force (F) between the surfaces in contact to the normal force: Ff/N.
So Ff/N which i found out earlier the friction force is 6 and the normal force is 117.72 so 6/117.72=.0509

 

[Sourabh N]

sorry to tell you but frictional force is 30 N, exactly equal to the horizontal force applied (since the object is moving with constant velocity) and normal force is exactly equal to gravitational force (since object is not moving in vertical direction). Their ratio is what you need.

 

[D H]

so was i right in my one post then

How did you get that out of what I wrote?? The frictional force is 30N.

 

[ilkjester]

I have no idea but thank you for all the help.

 

[pooface]

the frictional force is 30N, the Normal force is 117.72N. These are the two components you need to figure out mu.

30N/117.72N = 0.2548 (this is a ratio).

This works so easily only when the acceleration is 0. (constant velocity)
If there was acceleration applied, you have to do it like this:
sigma F = (mass of object + pulling/pushing mass)*a(acceleration given)
Ffriciton force= (pulling/pushing weight) – sigma F