## On friction

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Darkstorm
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### On friction

So I'm doing this starter course in physics and I can't figure out how to do this one assignment
I don't need to have people tell me how to do it, just point me in the general direction, because I'm stumped

So here goes

A rock with the mass of M is sliding at v=6.0 m/s onto a floor with the friction coefficient of 0.25

a) How long does it take before the rock stops
b) How far did it go?
c) What would happen if the rock hadd the mass of 2M

Is it just me, or am I missing a piece of information? Shouldn't I at least know something more about the forces or mass involved to be able to solve this one?

SDK
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### Re: On friction

You have the mass. That's all you need to find the force of gravity. Once you have that you can find the normal force. Once you have that you can find the force of friction. Once you have that you can calculate acceleration (or deceleration). Once you have that, everything else is easy!
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Darkstorm
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### Re: On friction

SDK wrote:You have the mass. That's all you need to find the force of gravity. Once you have that you can find the normal force. Once you have that you can find the force of friction. Once you have that you can calculate acceleration (or deceleration). Once you have that, everything else is easy!

Nope, I have the mass as an unknow, it's just listed as M
If I did have a number, then I'd be able to do it

jaap
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### Re: On friction

Darkstorm wrote:
SDK wrote:You have the mass. That's all you need to find the force of gravity. Once you have that you can find the normal force. Once you have that you can find the force of friction. Once you have that you can calculate acceleration (or deceleration). Once you have that, everything else is easy!

Nope, I have the mass as an unknow, it's just listed as M
If I did have a number, then I'd be able to do it

Then just do it with M instead of an explicit number. Who knows, maybe it will disappear further on in the calculations...

Darkstorm
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### Re: On friction

jaap wrote:
Darkstorm wrote:
SDK wrote:You have the mass. That's all you need to find the force of gravity. Once you have that you can find the normal force. Once you have that you can find the force of friction. Once you have that you can calculate acceleration (or deceleration). Once you have that, everything else is easy!

Nope, I have the mass as an unknow, it's just listed as M
If I did have a number, then I'd be able to do it

Then just do it with M instead of an explicit number. Who knows, maybe it will disappear further on in the calculations...

Huh, didn't think of that. I'll give it a shot

SDK
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### Re: On friction

That's just convention. Based on how the question is worded, it's known. Questions like this sometimes don't result in a solid number answer, just an answer in terms of M (or whatever other variable). It's a way for teachers to get their students to recognize relationships between different variables (in this case, the relationship between mass and deceleration due to friction) since the M won't get lost in a meaningless number.

Find your answer in terms of M. You might be pleasantly surprised.
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Darkstorm
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### Re: On friction

Hold my horses

so µ=0.25
m is unknown

My calculations give me 0.25=F/m*9.81 => F= m*9.81/4 => 0.25=(m*9.81/4)/m*9.81

a=(((m*9.81)/4)-0.25*m*9.81)/m => a=0/m or ma=0

Therefore, mass and acceleration does not exist

douglasm
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### Re: On friction

The equations you set up seem rather confused and don't mean much of anything. What, exactly, is the 9.81m/39.24m supposed to represent? And what does the 39.24m number in the second equation represent, and why are you subtracting a quarter of it from gravity?

Try breaking it into smaller steps:
What is the force due to gravity?
Based on that, what is the force due to friction?
How much does gravity affect horizontal acceleration?
How much does friction affect horizontal acceleration?
Of the forces that are relevant to horizontal acceleration, what is the sum of their horizontal components?
How long does that sum total force take to stop the rock?

Answers in terms of m are acceptable and expected.

eSOANEM
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### Re: On friction

In general, it's best to avoid putting numbers into your equations until the very end. If you leave your equations in terms of μ, g, and u, (I'm using u for initial velocity so we can use x, v and a for displacement, velocity and acceleration at a time t) it should be easier for us to read your equations.

I'm not sure you're doing what you think you're doing though, it looks like the thing you've posted is kind of circular and not very meaningful. Try breaking it up into douglasm's steps and try not to put any numbers in until right at the last moment.
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Dopefish
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### Re: On friction

I would also be wary of using m to represent both meters (as in meters per second) and mass at the same time. Use uppercase M for the mass to limit the potential of confusing yourself (although as mentioned, just using g and similar with the dimensions built in until the end can avoid that problem).

SDK
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### Re: On friction

Your very first equation doesn't line up with anything I recognize. Friction force is easy, just

Ff=µFN

where FN is the normal force. Work from there, solving for the friction force, and you should get this.
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Sizik
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### Re: On friction

SDK wrote:Your very first equation doesn't line up with anything I recognize. Friction force is easy, just

Ff=µFN

where FN is the normal force. Work from there, solving for the friction force, and you should get this.

It looks like they started with µ = Ff/FN and went on to rederive the value of µ.
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SDK
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### Re: On friction

Ah, yes. Strange approach. I'm sure they'll figure it out.
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Darkstorm
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### Re: On friction

Thanks to all contributing

eSOANEM wrote:In general, it's best to avoid putting numbers into your equations until the very end. If you leave your equations in terms of μ, g, and u, (I'm using u for initial velocity so we can use x, v and a for displacement, velocity and acceleration at a time t) it should be easier for us to read your equations.

I'm not sure you're doing what you think you're doing though, it looks like the thing you've posted is kind of circular and not very meaningful. Try breaking it up into douglasm's steps and try not to put any numbers in until right at the last moment.

This one got me thinking, and I found a formula that enabled me to do the problem without further information

µ(Friction)=Force/(mass*gravity) => (mass(velocity - starting velocity)/time)/(mass*gravity) => (velocity - starting velocity)/(time*gravity)

slinches
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### Re: On friction

In most of these types of problems I find it quite helpful to draw a free body diagram. For your problem it should look something like this, except with velocity and displacement included. From that, you can more easily see which forces are opposing and which pieces of information you have been given. Then it's just a matter of trying to figure out the remaining values. It also helps to do sanity checks. If you get something impossible happening, then you know that you've got something mixed up.

Hope that helps.

Eebster the Great
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### Re: On friction

Darkstorm wrote:µ(Friction)=Force/(mass*gravity) => (mass(velocity - starting velocity)/time)/(mass*gravity) => (velocity - starting velocity)/(time*gravity)

It looks like you actually found the answer on your own, only you solved for the coefficient of friction (which is not an unknown) instead of time (which is). Also, when you set up equations like this in class, you will want to represent most quantities with letters, not words, and you can define your variables first. When you keep things neat, it's very easy to see how you're getting the answer which means you will tend to get more answers right, and when you don't you will tend to get more partial credit. So I'll show you how I would write that answer, up to where you got:

Spoiler:
g = acceleration of gravity = 9.8 m/s2
m = mass
µ = coefficient of friction = 0.25
F = frictional force
a = acceleration
t = time
v = speed at time t
v0 = initial speed = 6.0 m/s

F = -µmg (definition of coefficient of friction; the negative sign is there because I'm saying the block is moving in the positive direction)
F = ma (Newton's second law)
a = -µg
a = (v-v0)/t (since acceleration is constant . . . at least until it stops moving)
(v-v0)/t = -µg

At that point, you just have to solve that equation for t (very straightforward), then plug in the values you were given in the problem, along with v = 0 (because you want to know when the block will stop). Part b will take slightly more time, but it's just more algebra. Set up the kinematics equations in terms of t, v0, x, and a, solve for x, then "plug and chug."

As you can see, the acceleration doesn't depend on mass! And it's easy to see why. When an object is in freefall, it will fall at the same acceleration g regardless of its mass, because the force of gravity is just mg. In this case the force is friction instead, but that's just proportional to the normal force, which in this case is still gravity. Whenever the net force is directly proportional to the mass, the acceleration will be independent of mass, because F = ma, so the m will cancel.

drachefly
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### Re: On friction

Eebster the Great wrote:Also, when you set up equations like this in class, you will want to represent most quantities with letters, not words, and you can define your variables first.

I find it helpful to use words while I'm figuring out what I'm doing. I can switch over to letters once I'm familiar with the meaning of every symbol in the problem. I think the early move to letters is one thing that confuses newbies.

Tchebu
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### Re: On friction

Darkstorm wrote:Hold my horses

so µ=0.25
m is unknown

My calculations give me 0.25=F/m*9.81 => F= m*9.81/4 => 0.25=(m*9.81/4)/m*9.81

a=(((m*9.81)/4)-0.25*m*9.81)/m => a=0/m or ma=0

Therefore, mass and acceleration does not exist

Your mistake in the second equation is that the first term doesn't belong there. Gravity (or the normal force, whichever one you meant when you wrote "9.81m") and friction don't act along the same axis so you can't just add them like that, forces and accelerations are vectors not scalars! This is probably the number one mistake that people make at this level.

Decompose the problem along the vertical and horizontal directions.

Along the vertical direction gravity just gets cancelled by the normal force of the floor. This is precisely why you can say that your friction force has magnitude mu*Mg, because Mg happens to also be the magnitude of the normal force. This is generally not the case, but only true on horizontal surfaces!!

Along the horizontal direction your only force is the friction force that we just figured out! This is what's causing your acceleration. Divide it by the mass to find the acceleration, you should find that the "unknown" mass of the object conveniently cancels out, whatever it is.
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