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## Homework Statement

Hello, How do i find coefficient of friction knowing only mass, velocity and incline for a inclined plane?

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- Thread starter amazondog
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- #1

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Hello, How do i find coefficient of friction knowing only mass, velocity and incline for a inclined plane?

- #2

Matterwave

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Mass of what? Velocity of what? Initial velocity or final velocity? Any information on acceleration? Is the velocity constant?

...

- #3

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A biker going up a hill

Mass of what? Velocity of what? Initial velocity or final velocity? Any information on acceleration? Is the velocity constant?

...

Mass = 98.7kg

Constant velocity = 5.95m/s

So there is no acceleration.

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- #5

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Ok.. How do i fund mu though?

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Is it the tangent of the incline?

- #7

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How do you find the force that is needed to go up the hill at the same velocity?

- #8

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[tex]\mu F_N-F_g=0[/tex]

But since he is on an incline you need to find the component of the gravitational force that is in the same direction as the normal force... and then plug that in to find [tex]\mu[/tex].

Edit: You need to know the inclination angle of the hill. I don't know if that was given to you or not.

- #9

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How do you find the force that is needed to go up the hill at the same velocity?

I see what you're getting at. I just tried doing it and I don't know how velocity comes into play.

- #10

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[tex]\mu F_N-F_g=0[/tex]

But since he is on an incline you need to find the component of the gravitational force that is in the same direction as the normal force... and then plug that in to find [tex]\mu[/tex].

Edit: You need to know the inclination angle of the hill. I don't know if that was given to you or not.

umgcos(theta) - gcos(theta) = 0

That would be a very small number.....

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What friction , static,rolling please clarify

- #12

nasu

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A biker going up a hill

Mass = 98.7kg

Constant velocity = 5.95m/s

So there is no acceleration.

It cannot go up the hill with constant velocity unless there is a force acting on it.

Do you know this force?

- #13

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All I have is: A biker rides up an 9.75degree hill at a constant speed of 5.95m/s. The biker and bicycle together have a mass of 98.7kg.

Find the coefficient..

What force does the system need to make the biker go up the hill at the same velocity..

What must the bikers power output be in order to go up the hill at the same speed.

Thanks guys I really appreciate it.

- #14

nasu

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All I have is: A biker rides up an 9.75degree hill at a constant speed of 5.95m/s. The biker and bicycle together have a mass of 98.7kg.

Find the coefficient..

What force does the system need to make the biker go up the hill at the same velocity..

What must the bikers power output be in order to go up the hill at the same speed.

Thanks guys I really appreciate it.

Then something is wrong. If in the first case it goes up, why do they ask then to find the force "to make the biker go up the hill at the same velocity.."????

You could solve the problem if in the first place it goes DOWN the hill with constant velocity. You don't need any other force besides weight and friction for this.

Then it makes sense to ask what extra force is required for it to go up.

Maybe there is some error in text with the problem. Or you misread it.

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- #16

PhanthomJay

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When the bike goes up the hill, the force that counteracts the weight component down the plane

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Question:

A bicyclist is traveling down a hill which has a 9.75 degree incline at a constant velocity of 5.95m/s. The mass of the cyclist and bike combined is 98.7kilograms.

- What is the coefficient of friction of the surface the biker is traveling on?

- What force is needed to go up the hill at the same speed, 5.95m/s.

- What power output must the cyclist generate to go up the hill at the same velocity.

- #18

nasu

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When the bike goes down hill with constant velocity , the friction (uphill, along the plane) must be equal to the component of the weight that acts downhill, along the plane.

This equation will provide the value of the friction coefficient.

Do you know how to write the expressions of these forces?

- #19

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umg = mgsin(theta ?

- #20

PhanthomJay

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Thanks for clarifying the question. Your equation is incorrect, but you already answered part a in your post #6, so you must have done something right at that time. For part b, nasu has already given you some great hints in posts 14 and 18. Then move onto part c.umg = mgsin(theta ?

- #21

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Ok, so the coefficient of velocity is tangent to the path. That makes sense.

- #22

nasu

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Ok, so the coefficient of velocity is tangent to the path. That makes sense.

Maybe if you read again and reformulate.

Do you mean the coefficient of friction is equal to the tangent of the angle of the path?

- #23

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Fa - Ff = FN

Facos(theta) - umgcos(theta) = mgsin(theta)

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Fa - Ff - F// = 0

Fa - umgsin(theta) - mgsin(theta) = 0

Solve for Fa.

- #25

PhanthomJay

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almost. The friction force is not umgsin theta...your geometry and trig is a bit off..,, it's umg (_?__)theta.

Fa - Ff - F// = 0

Fa - umgsin(theta) - mgsin(theta) = 0

Solve for Fa.

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