Coefficient of friction of a Skateboard on a rampby Dielere Tags: coefficient, friction, ramp, skateboard 

#1
Mar612, 11:01 AM

P: 6

1. The problem statement, all variables and given/known data
The coefficient of friction is to be found from a skateboard and rider rolling down a ramp, which has an angle of 5 degrees. The skateboard accelerates at .5308 m/s^2. The mass of the rider and skateboard is 91.3 kg, or 895 newtons. The other data would be gravity. 2. Relevant equations Coefficient of friction = Fx/Fn 3. The attempt at a solution I found what I thought to be the acceleration the skateboard would have had if it was friction less. That was .8526, however now I am pretty sure I did something wrong. Then I found the force of gravity at 5 degrees, which was 9.39722. But I know there is something very wrong with what I'm doing. Then putting .5308/9.39722 to find a coefficient of about .05. I'm certain I'm doing something wrong, and the mass can be used to find it. But I am lost. Any help would be greatly appreciated! 



#2
Mar612, 01:59 PM

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P: 3,337

Hey Dielere, welcome to physics forums! You've got an interesting question. So the idea is that the skateboard is rolling down the slope on its wheels, but the wheels have some rolling resistance, with a dimensionless coefficient of rolling friction, which is the thing you're trying to find. Did I understand it all right?




#3
Mar612, 05:03 PM

P: 6

I was in Radians for that one part, but I used sin of 5 to multiply by gravity in order to find the new normal force that presses on the skateboard at that angle. Without radians its 9.76, so my thinking is that I would use that for normal force. However when I do it out, I feel like I'm missing something. I wouldn't need to multiply either by the mass, because the ratio of Fx/Fn would still be the same. I'm just kind of lost...




#4
Mar612, 06:12 PM

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P: 3,337

Coefficient of friction of a Skateboard on a ramp 



#5
Mar612, 06:58 PM

P: 6

Yeah, I've figured out the calculator stuff. It was just a mistake on my part. So I have no idea where to go from here....I feel like I've got to be missing some part, or is it as simple as just .5208 divided by 9.76?




#7
Mar612, 07:36 PM

P: 6

I didn't calculate it, that was the actual acceleration of the skateboard and rider moving down the ramp




#8
Mar712, 04:42 AM

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P: 3,337

Ok, its just that in the first post, you have it as .5308 I guess one of them you wrote down wrong?
About the problem, I don't think it is as simple as .5208/9.76 Because .5208 is the net force/mass (in the downhill direction), But you need just the friction force. So how would you get the friction force, considering you have the net force and the gravitational force? 



#9
Mar712, 09:38 AM

P: 6

Oh yes, I'm sorry I didn't even notice. But yes, I mistyped that. I'm not sure how I would do that...The way that I've thought of is just Fx/Fn. I could use the mass multiplied by 9.76 to find newtons, but I don't know what I would do with the downward acceleration without having a mass with it. If I divide .5308 by .8526 then that would tell me how much friction slowed the acceleration. Friction would slow it down by about 38 percent. I'm not sure how, or if, I would convert that into the coefficient of friction...




#10
Mar712, 12:13 PM

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P: 3,337

You shouldn't be dividing .5308 by .8526, but the answer is related to these two numbers. About not knowing the mass, you can just use the symbol m in your equations, and it will cancel out in the end. So for the problem, in the direction parallel to the slope, there is a friction force, a gravitational force, and a total force. So how are these related?




#11
Mar812, 09:47 AM

P: 6

Ah, that helps a bit! Friction slows down the skateboard, so the friction force would be .8526.5308, which is .3218.That force is opposite to the force of the skateboard sliding down. The total, net force is .5308 right? Is it the force of friction the number that needs to go on top for coefficient of friction? Perhaps .3218 over 9.76? I know it needs to be very low, because it rolls and doesn't take much to move it. Using the other numbers I have now it would be too high. Am I on the right track, or am I missing something?




#12
Mar812, 10:30 AM

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P: 3,337

You have got the right answer. But I get the feeling you don't really understand why. In your last post, you kept saying force, but then giving force divided by the mass of the object. For example, .3218 is the frictional force divided by the mass of the object. And 9.76 is the normal force divided by the mass of the object.
I think when you do these kinds of questions, it is best to leave in the symbol m, otherwise it gets more complicated to talk about. 


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