Finding friction without coefficient?

In summary: That's the calculation you did in post #10, just doing the divide the right way around this time. And yes, that's the force acting on the system.So, what's next?So, what's next?Next, you need to think about how to find the frictional force acting on the cart only. Any ideas?Next, you need to think about how to find the frictional force acting on the cart only. Any ideas?Maybe by subtracting the force acting on the cart from the force acting on the whole system?Maybe by subtracting the force acting on the cart from the force acting on the whole system?Yes, that's one way to do it. Can you think of another way
  • #1
kelseybrahe
44
0

Homework Statement


We did a Prac where we attached ticker tape to a cart (905g), and the cart to a mass (249g) through a pulley. We then had to assess the tape and answer questions. We are asked to calculate the force acting upon the whole system and then calculate the force acting only upon the cart. We are not given the coefficient of friction.
I do not know the applied force (can I work this out?)

Average acceleration = 1.05m/s (initial velocity was zero/at rest?)

There is no inclination. The cart was on a flat, smooth desk.

Homework Equations


F=ma a(av)= v-u/t
Maybe Ek=1/2mv2

The Attempt at a Solution


Read somewhere that Ek=1/2mv2 can give you the coefficient of friction, but that didn't seem right at all. Other than that, completely lost.
 
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  • #2
kelseybrahe said:
the force acting upon the whole system
Define the system.
kelseybrahe said:
the force acting only upon the cart
Net force, or just the force from the string tension?
 
  • #3
haruspex said:
Define the system.

Net force, or just the force from the string tension?

I'm not entirely sure about either of these. The questions are very vague. I'm assuming by 'whole system' it means the combined mass of the cart and the mass pulling it. Which is 1154g.

For the second, the question simply states "it could be possible to use the same ticker tape to determine the friction acting on the cart only. How?"
 
  • #4
kelseybrahe said:
I'm not entirely sure about either of these. The questions are very vague. I'm assuming by 'whole system' it means the combined mass of the cart and the mass pulling it. Which is 1154g.
Ok. This is awkward, and I rather suspect the question is phrased wrongly. But let's take it at face value for now. What are the forces acting on the system (cart+string+weight).
kelseybrahe said:
For the second, the question simply states "it could be possible to use the same ticker tape to determine the friction acting on the cart only. How?"
Can you calculate what the acceleration would have been without friction?
 
  • #5
haruspex said:
Ok. This is awkward, and I rather suspect the question is phrased wrongly. But let's take it at face value for now. What are the forces acting on the system (cart+string+weight).Can you calculate what the acceleration would have been without friction?

Okay, thank you.
If it helps this is the first question:
"Calculate the difference in theoretical and measured acceleration, and then calculate The force of friction acting on the system."

Cart, string, pulley, weight and also the ticker time itself?

Yes! If I've done it correctly it's 0.47m/s^2.
 
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  • #6
kelseybrahe said:
Yes! If I've done it correctly it's 0.47m/s.

Sorry, part of my response has gone to the wrong place in there.
 
  • #7
kelseybrahe said:
If it helps this is the first question:
"Calculate the difference in theoretical and measured acceleration, and then calculate The force of friction acting on the system."
do I have this right:
The first part asks for the frictional force acting on the system (as you have now defined it).
The second part does not ask you to calculate the frictional force acting on the cart only, but does ask how you could do that using the equipment.
?

You were advised to calculate the theoretical acceleration without friction. Do that.
 
  • #8
haruspex said:
do I have this right:
The first part asks for the frictional force acting on the system (as you have now defined it).
The second part does not ask you to calculate the frictional force acting on the cart only, but does ask how you could do that using the equipment.
?

You were advised to calculate the theoretical acceleration without friction. Do that.

Yes, I think so.
You've definitely got the first part right.
For the second (I have a cold and am a bit foggy - I'm sorry if I seem slow!) I think that it's asking for a formula or something. Do you know of one? If you don't, just go with equipment and see where that leads?

The theoretical acceleration is 0.47m/s^2
 
  • #9
kelseybrahe said:
The theoretical acceleration is 0.47m/s^2
That cannot be right. Please post your working.
 
  • #10
haruspex said:
That cannot be right. Please post your working.

m+M (mass total) divided by weight

(m= 249 (mass on string) M=905 (mass of cart) )

= 1.154/2.44
= 0.47

This was being told to assume the weight force was the only force acting on the system for that question.
Perhaps that can extend to the other questions as well.

Have I used the wrong formula?
I converted it to Kg, is that incorrect?
 
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  • #11
kelseybrahe said:
m+M (mass total) divided by weight
Small but important error. What is the relationship between force, mass and acceleration?
 
  • #12
haruspex said:
Small but important error. What is the relationship between force, mass and acceleration?

They are proportional to each other?
Increase force and mass, acceleration increases?
 
  • #13
kelseybrahe said:
They are proportional to each other?
Yes, but what is the equation?
 
  • #14
haruspex said:
Yes, but what is the equation?

Oh! F = ma
 
  • #15
kelseybrahe said:
Oh! F = ma
Right. So is that what you wrote here?
kelseybrahe said:
m+M (mass total) divided by weight
 
  • #16
haruspex said:
Right. So is that what you wrote here?

No... Oh, okay.
So I would re-arrange it to be a=f/m?
 
  • #17
kelseybrahe said:
No... Oh, okay.
So I would re-arrange it to be a=f/m?
Yes.
 
  • #18
haruspex said:
Yes.

Right. But how do I find f?
I have an answer for it but that is using the calculated acceleration of 1.05m/s^2. And that wouldn't be accurate, would it?

(Thank you for taking the time to go through this with me - I really appreciate it!)
 
  • #19
kelseybrahe said:
Right. But how do I find f?
Let's get this step done first:
kelseybrahe said:
"Calculate the difference in theoretical and measured acceleration..."
 
  • #20
haruspex said:
Let's get this step done first:

Oh sorry, I mean the f in the equation a=f/m
Because that's force, right? But the only way I can find force is by doing f=ma. I think?
 
  • #21
kelseybrahe said:
Oh sorry, I mean the f in the equation a=f/m
Because that's force, right? But the only way I can find force is by doing f=ma. I think?
That's the calculation you did in post #10, just doing the divide the right way around this time.
 
  • #22
haruspex said:
That's the calculation you did in post #10, just doing the divide the right way around this time.

Oh sorry! I'm such a ditz.
Okay.
2.44/1.154

= 2.15m/s^2
 
  • #23
kelseybrahe said:
Oh sorry! I'm such a ditz.
Okay.
2.44/1.154

= 2.15m/s^2
Right. So how much less is the observed acceleration?
 
  • #24
haruspex said:
Right. So how much less is the observed acceleration?

1.10m/s^2
 
  • #25
kelseybrahe said:
1.10m/s^2
Right. Can you figure out the total frictional force from that?
 
  • #26
haruspex said:
Right. Can you figure out the total frictional force from that?

I'm not sure how to do that... Do you use the difference in acceleration and multiply that by the mass?
 
  • #27
kelseybrahe said:
I'm not sure how to do that... Do you use the difference in acceleration and multiply that by the mass?
That will work, but it would be better to understand why.
If you do not, I urge you to use the more standard approach of solving ΣF=ma. You know one force, you know the total mass, and you observd the accelration. There is one force, friction, that you do not know.
 
  • #28
haruspex said:
That will work, but it would be better to understand why.
If you do not, I urge you to use the more standard approach of solving ΣF=ma. You know one force, you know the total mass, and you observd the accelration. There is one force, friction, that you do not know.

I'm really sorry, I don't understand what you mean.
So I would substitute the total mass and observed acceleration into that equation in order to find (sum)F. So, friction is only included in that total, it is not THE total, right?
Is there something I have to do to isolate friction, or does that answer cover it?

(Sorry for my ignorance here)
 
  • #29
kelseybrahe said:
So I would substitute the total mass and observed acceleration into that equation in order to find (sum)F. So, friction is only included in that total, it is not THE total, right?
Yes.
kelseybrahe said:
Is there something I have to do to isolate friction,
Yes. What other force(s) are there affecting the motion?
 
  • #30
haruspex said:
Yes.

Yes. What other force(s) are there affecting the motion?

String, pulley, ticker timer, Gravity?
 
  • #31
kelseybrahe said:
String, pulley, ticker timer, Gravity?
The string is internal to the system. We are only concerned with external forces here.
Yes, the pulley exerts an external force, but it does no work, so does not contribute to the system's acceleration. It also may have some friction at its axle. Strictly speaking that is a frictional torque, but for the purpose of the question we can just consider that part of the frictional force.
The ticker timer may slow things a bit, but that is just part of the total frictional force, so we do not need to separate that out either out.
That leaves gravity, and you know that force. So what equation do you get?
 
  • #32
kelseybrahe said:
2.44/1.154 = 2.15m/s^2

The way I would write my scratch pad calculation is 2.44 N/1.154 kg = 2.15 m/s2
just to keep it all straight in my mind. Less chance of error.
And makes it easier for others to follow along.
 
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  • #33
[QUOTE="haruspex, post: 5539034, member: 334404]
That leaves gravity, and you know that force. So what equation do you get?[/QUOTE]

Will it be the sumF multiplied by gravity (9.8)?
 
  • #34
David Lewis said:
The way I would write my scratch pad calculation is 2.44 N/1.154 kg = 2.15 m/s2
just to keep it all straight in my mind. Less chance of error.
And makes it easier for others to follow what you're doing.

Thank you, that makes sense!
 
  • #35
kelseybrahe said:
[QUOTE="haruspex, post: 5539034, member: 334404]
That leaves gravity, and you know that force. So what equation do you get?

Will it be the sumF multiplied by gravity (9.8)?[/QUOTE]

No. I'll divide it by gravity?
 

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