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Physics Problem! (coefficient of friction) Please help

Drake22

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Please, I need some help on part b:

On a level road with its brakes on, the shortest distance in which a car traveling with 93 km/hr can stop is 89 m. This shortest distance occurs when the driver uses anti-lock brakes which means that that the car brakes without skidding.

a) What is the magnitude of the acceleration of the car?

3.75 m/s^2 is the right answer

b) What is the coefficient of friction between the tires and the pavement?



How do I do this? Don't I need the mass of the car? Thanks.
 

Drake22

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Well, none of those worked, but I believe we are supposed to calculate it. I think they want you to assume that the force stopping the tires is just on the threshold of static friction, but in order to find that force, dont you need the mass of the car and simply multiply that by the deceleration? And they don't supply you with mass.....grrrr
 

xenten

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If it gave you the mass I believe you use the formula:

coefficient = Force/ma

We just got done with that.
 

Drake22

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Right, if I had a mass it would be simple, is it possible without a mass?

I have a creeping suspicion that this problem is simply wrong. On our mid-term, they threw out one problem because all the possible answers were wrong and the prof had to correct two of the other problems during the test, pretty poorly planned if u ask me.
 

WORKEROUTER

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Man go toa physics and science board forum. Noone wants to hear about your struggles with basic mechanics.
 

WORKEROUTER

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I'm going to help you though.

SO you know the acceleration, right?

Well draw a force diagram on the car. What's the force acting on the car? Friction and gravity. Ignore gravity.

F=ma, right?

So Force of friction=ma=m (v/t), right?

But you know that FF=CF(W)=CF(mg)

But if that's true and W=mg, then FF(m)(g)=(mv)/t, right?

Well, what do you konw, m cancels out and CF=v/(gt).

And it makes sense that the coefficient of friction, which is a property that is part of the surface, should have no direct correlation with an arbritary object located on the surface itself.
 

WORKEROUTER

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Re: Re: Physics Problem! (coefficient of friction) Please help

Originally posted by WORKEROUTER
I'm going to help you though.

SO you know the acceleration, right?

Well draw a force diagram on the car. What's the force acting on the car? Friction and gravity. Ignore gravity.

F=ma, right?

So Force of friction=ma=m (v/t), right?

But you know that FF=CF(W)=CF(mg)

But if that's true and W=mg, then FF(m)(g)=(mv)/t, right?

Well, what do you konw, m cancels out and CF=v/(gt).

And it makes sense that the coefficient of friction, which is a property that is part of the surface, should have no direct correlation with an arbritary object located on the surface itself.

But seriously, go to a physics forum or something next time.
 

insomniac

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This is a horribly worded question.

If the car stops without skidding, then all the friction is between the brake pads and the discs. The brakes slowing the rotation of the wheels is what causes the car to stop.

The wheels are simply rolling over the pavement. There's no sliding of the tires on the pavement, and hence no force of friction between the tires and pavement.

There is no answer to b. Tell whoever wrote this to think a little more next time.
 

insomniac

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Originally posted by insomniac
This is a horribly worded question.

If the car stops without skidding, then all the friction is between the brake pads and the discs. The brakes slowing the rotation of the wheels is what causes the car to stop.

The wheels are simply rolling over the pavement. There's no sliding of the tires on the pavement, and hence no force of friction between the tires and pavement.

There is no answer to b. Tell whoever wrote this to think a little more next time.
Correction, I'm wrong...there has to be an external force acting on the car, so there is some force of the pavement against the tires. Must be looking for the coefficient of rolling friction.
 

snintel

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Originally posted by Vincent


Also in the future these forums shouldn't be used as your homework help. I just felt like being nice this time.
Umm.... why not? It's called "Anything Else" and I think this falls under "anything". Would you rather him post another thread about some show or something stupid like that instead of one that can possibly induce some thinking? :rolleyes:
 

B-Lemond

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Fnet = Ffk + Fg


Fg = mg - - - - - Fg = m(-9.8m/s^2)


Ffk = uk + Fg

uk is what you are trying to solve for. You do need the mass of the car, or the Net force.
 

BrotherAP

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The friction force on the car is given by the equation

F = N * μ

where N is the normal force of the ground pushing against the car activing perpendicular to the surface at the point of contact - in this case equivalent to weight of the car since the car is on a flat surface. So, since the normal force is equivalent to the weight, you can substitute that out and you get

F = W * μ

Which means that the friction is directly porportional to the weight of the car, with a porpotionality constant of μ (aka the coefficient of friction)

So, to solve this problem, assume any weight for the car - I recommend 10kg, since the resulting math is easy, but you can choose whatever you want.

If you still have some doubt, try it with two different weights. You should get the same answer.
 
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