# Determine Coefficient of Friction

1. Sep 12, 2014

### killaI9BI

1. The problem statement, all variables and given/known data

You are pushing a 53 kg crate at a constant velocity up a ramp onto a truck. The ramp makes an angle of 22 degrees with the horizontal. If your applied force is 373 N, what is the coefficient of friction between the crate and the ramp?

2. Relevant equations

μ = Ff/Fn
Fn = m X g X cos θ
Fnet = Fa – mg sin θ +or- μmg cos θ

3. The attempt at a solution

Fn = 53 X 9.81 X cos 22
Fnet = 373 – 53 X 9.81 sin 22 + μ X 53 X 9.81 cos 22

The book's answer is 0.37 but I just can't get the rest of the equation. Any suggestions are appreciated.

2. Sep 12, 2014

### BiGyElLoWhAt

Friction always opposes motion, not acceleration, if you're pushing up the ramp, which way should friction be?

You are applying 373N to move the crate at a CONSTANT VELOCITY, what does that tell you about the net force?

And to be honest, I'm not sure I know what you mean by the rest of the equation? Can you clarify please?

3. Sep 12, 2014

### Staff: Mentor

Is the frictional force in the same direction as the applied force or in the opposite direction? If the crate is moving at constant velocity, what is the net force?

Chet

4. Sep 12, 2014

### killaI9BI

Me either - just using equations from notes that I took during the lesson.

I'm not really sure what it tells me about the net force....

the formula that I have for net force is:
Fnet = Fa – mg sin θ +or- μmg cos θ

where Fa is applied force, m is mass, g is acceleration due to gravity and μ is the coefficient of friction

5. Sep 12, 2014

### killaI9BI

I'm thinking the frictional force would be in the opposite direction as the applied force.

I'm going to see if I can find another formula for net force that doesn't require the value of coefficient of friction.

6. Sep 12, 2014

### killaI9BI

ohhh I see now; because it's moving at a constant velocity, the net force must be zero so the applied force equals the force of friction

right?

7. Sep 12, 2014

### killaI9BI

I'm still in left field somewhere

8. Sep 12, 2014

### Staff: Mentor

Go back to your equation and set the net force equal to zero. Then solve for the coefficient of friction, since you know everything else in the equation. In your problem, the applied force does not equal the force of friction because there is also a component of gravitational force present along the slide. Did you draw a free body diagram, or haven't you learned about that yet?

Chet

9. Sep 12, 2014

### BiGyElLoWhAt

Not quite. Look at your sum of the forces. How many forces are there? Youre right the net force is 0, but if the applied force equalled the force of friction, what would you say about this
$F_{applied}-F_{friction}=???$
?

Look at your sum of forces equation. The one you posted recently is right. Im not sure if you changed the original one, but there was a sign error in it. Youre almost to the answer, you just have to step back and look at what you have.

10. Sep 13, 2014

### killaI9BI

that worked! I have learned about the FBDs but didn't think it'd really help. I will try them in the future.

Thank you.

11. Sep 13, 2014

### Staff: Mentor

The value of using FBDs cannot be overemphasized.

Chet

12. Sep 13, 2014

### killaI9BI

Fapplied - Ffriction = Fnet - FgII
where FgII is the gravitational force parallel to the incline

.... according to my notes

13. Sep 14, 2014

### BiGyElLoWhAt

Looks good man, good job.

I just wanna add one thing: don't focus on your notes, sum of the forces = ma. done and done.

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