Finding friction without coefficient?

[haruspex]

Oh! I messed up the equation.
Okay.

Friction + weight = mass x acceleration?

Yes. In the context of this question, what exactly is the expression for weight there, and what exactly the expression for mass?

 

[kelseybrahe]

Yes. In the context of this question, what exactly is the expression for weight there, and what exactly the expression for mass?

What do you mean by expression?
Weight = 2.44N
Mass = 1.154kg ?

 

[kelseybrahe]

What do you mean by expression?
Weight = 2.44N
Mass = 1.154kg ?

Or weight is 'mg' and mass is 'm'?

 

[David Lewis]

To re-cap:
You have correctly calculated the hypothetical acceleration without friction.
You have experimentally measured the actual acceleration with friction.
You have found the difference between the two accelerations (1.10 m/s²)
You know how much gravitational force is tending to accelerate the system
Now you want to find the frictional force

 

[haruspex]

What do you mean by expression?
Weight = 2.44N
Mass = 1.154kg ?

Yes.

 

[kelseybrahe]

Yes.

So, therefore,

Friction + 2.44N = 1.154kg x 1.05m/s^2

Friction = 1.154 x 1.05 - 2.44

And to find only the friction of the cart, I would replace the total mass by only the mass of the cart?

 

[haruspex]

Friction = 1.154 x 1.05 - 2.44

Yes. Note that this gives a negative result because you effectively chose the positive direction as being in the same direction as the acceleration.

And to find only the friction of the cart, I would replace the total mass by only the mass of the cart

By what logic? It worries me that you persist in making wild guesses instead of getting to grips with the laws of mechanics and the equations that come from them.

 

[kelseybrahe]

Yes. Note that this gives a negative result because you effectively chose the positive direction as being in the same direction as the acceleration.

By what logic? It worries me that you persist in making wild guesses instead of getting to grips with the laws of mechanics and the equations that come from them.

Is that a problem? Do I just ignore the negative sign?

Oh. I assumed that I could use the same equation but narrow it down to find only the friction acting on the cart by swapping the numbers we used for the whole system to be only the numbers for the cart.

 

[haruspex]

Is that a problem? Do I just ignore the negative sign?

It is not a problem. In general, if you are clear in how you define the positive directions for forces and accelerations, and write your equations accordingly, then the sign in your answer can tell you something useful. In the present case, you wrote the sum of forces as friction+mg. in reality, it is clear that they will oppose each other, so it is to be expected that you will get a negative sign for the friction. It is not an error.

There are two aspects of the whole question I'm not happy with.

It is not good practice to add forces that are acting in different directions as though they are just numbers (that is, scalars rather than vectors). Forces have direction. In this problem, gravity acts vertically on the suspended mass and friction acts horizontally on the other mass. The right approach is to let the tension in the string be T. So in the horizontal direction for the mass on the table we have an equation m_table a = T- friction; in the vertical direction for the suspended mass we have m_sus a = m_sus g - T. Eliminating T between these two gives the equation you found. Yes, it is a longer route, but it helps you understand what is going on, and can be applied in more general circumstances.

Secondly, it asks you to find the total frictional force when some of the friction may be in the axle of the pulley. That would not be a force, but a torque. However, you can equate it to a force by dividing by the radius of the pulley. A frictional torque τ at the axle will have the same effect as a frictional force τ/r at the periphery of the pulley.

I assumed that I could use the same equation but narrow it down to find only the friction acting on the cart by swapping the numbers we used for the whole system to be only the numbers for the cart.

Friction slows the acceleration no matter where it occurs. Knowing what the total effective frictional force is gives you no clue as to how it is distributed in the system. It could be entirely axial torque in the pulley.
To answer this part of the question you need to think up a variant of the experiment, or some extra observation you could make.

 

[kelseybrahe]

It is not a problem. In general, if you are clear in how you define the positive directions for forces and accelerations, and write your equations accordingly, then the sign in your answer can tell you something useful. In the present case, you wrote the sum of forces as friction+mg. in reality, it is clear that they will oppose each other, so it is to be expected that you will get a negative sign for the friction. It is not an error.

There are two aspects of the whole question I'm not happy with.

It is not good practice to add forces that are acting in different directions as though they are just numbers (that is, scalars rather than vectors). Forces have direction. In this problem, gravity acts vertically on the suspended mass and friction acts horizontally on the other mass. The right approach is to let the tension in the string be T. So in the horizontal direction for the mass on the table we have an equation m_table a = T- friction; in the vertical direction for the suspended mass we have m_sus a = m_sus g - T. Eliminating T between these two gives the equation you found. Yes, it is a longer route, but it helps you understand what is going on, and can be applied in more general circumstances.

Secondly, it asks you to find the total frictional force when some of the friction may be in the axle of the pulley. That would not be a force, but a torque. However, you can equate it to a force by dividing by the radius of the pulley. A frictional torque τ at the axle will have the same effect as a frictional force τ/r at the periphery of the pulley.

Friction slows the acceleration no matter where it occurs. Knowing what the total effective frictional force is gives you no clue as to how it is distributed in the system. It could be entirely axial torque in the pulley.
To answer this part of the question you need to think up a variant of the experiment, or some extra observation you could make.

Okay, thank you. This all makes sense.

Okay, so the friction could be spread across the system in any way and the way I tried to isolate that was thus incorrect and wouldn't work at all. Alright, that seems really obvious now. Thank you.

So contributing to the friction acting on the cart we've still got the vertical mass of gravity and the mass of the cart itself, plus the frictional torque's acting upon the string and the pulley etc. (but these are still insignificant and basically cancel out).
So, is there something else we have to consider? Or am I not going about this the right way?

 

[haruspex]

but these are still insignificant and basically cancel out).

Friction doesn't cancel friction.

With essentially the same set-up, what could you easily vary that would give a different acceleration?
If the friction consists of friction on the cart and friction at the pulley, which of those changes would alter the contribution of one friction source but not the other?

 

[kelseybrahe]

Friction doesn't cancel friction.

With essentially the same set-up, what could you easily vary that would give a different acceleration?
If the friction consists of friction on the cart and friction at the pulley, which of those changes would alter the contribution of one friction source but not the other?

Well the pulley is controlling the speed that the accelerating mass falls, so that would affect the cart, right?

So it must be friction on the cart?

 

[haruspex]

Well the pulley is controlling the speed that the accelerating mass falls

No, why do you say so? If you were to increase the mass of the cart that would also reduce the acceleration of the falling mass.

With the same arrangement, a cart, a pulley etc., what could you easily vary that would affect the acceleration?

 

[kelseybrahe]

No, why do you say so? If you were to increase the mass of the cart that would also reduce the acceleration of the falling mass.

With the same arrangement, a cart, a pulley etc., what could you easily vary that would affect the acceleration?

Affect the acceleration of the cart?
Well if you vary the mass of the weight pulling the cart, it will accelerate faster. If you reduce it, the cart will travel slower.
The weight of the cart itself. Increase the weight and it will travel slower, reduce the weight and it will travel faster.

 

[haruspex]

Affect the acceleration of the cart?
Well if you vary the mass of the weight pulling the cart, it will accelerate faster. If you reduce it, the cart will travel slower.
The weight of the cart itself. Increase the weight and it will travel slower, reduce the weight and it will travel faster.

Right. Now, suppose the friction just comes from the cart and the pulley axle. How would those two frictions vary if you change the suspended weight, say? Would they change in exactly the same way, i.e. stay in the same ratio?

 

[kelseybrahe]

Right. Now, suppose the friction just comes from the cart and the pulley axle. How would those two frictions vary if you change the suspended weight, say? Would they change in exactly the same way, i.e. stay in the same ratio?

Yes, I think so. If the weight falls faster it will pull the cart faster too, but there will be greater friction on the pulley too, won't there?
And the same if it falls slower. So the ratio would stay the same.

But I'm not sure. it seems strange to say that things have an increase in encountered friction if they move faster.

 

[haruspex]

there will be greater friction on the pulley too, won't there?

There will, but why? Does that reason apply also to the friction on the cart, which has the same mass as before?

And the same if it falls slower. So the ratio would stay the same.

I did not say the rationbetween the friction and the acceleration. I said the ratio between the two frictions.

it seems strange to say that things have an increase in encountered friction if they move faster.

They don't. It is not the increased speed that would cause the increase in friction. What does kinetic friction between two surfaces depend on?

 

[kelseybrahe]

There will, but why? Does that reason apply also to the friction on the cart, which has the same mass as before?

I did not say the rationbetween the friction and the acceleration. I said the ratio between the two frictions.

They don't. It is not the increased speed that would cause the increase in friction. What does kinetic friction between two surfaces depend on?

It depends on the type of materials the surfaces are made of?
And their mass too?

Oh, right. So, the friction acting on the cart will stay the same because it's mass doesn't change, but the friction acting on the pulley will vary as the mass of the weight varies.
Is this because of the tension from the string and mass?

 

[kelseybrahe]

To re-cap:
You have correctly calculated the hypothetical acceleration without friction.
You have experimentally measured the actual acceleration with friction.
You have found the difference between the two accelerations (1.10 m/s²)
You know how much gravitational force is tending to accelerate the system
Now you want to find the frictional force

Sorry, missed this. Yes, that's right.

 

[haruspex]

Oh, right. So, the friction acting on the cart will stay the same because it's mass doesn't change, but the friction acting on the pulley will vary as the mass of the weight varies.

Yes, but to be accurate it's the normal force that matters. Because the cart's mass stays the same, the weight stays the same, so the normal force stays the same.

Is this because of the tension from the string

Yes. The tension in the string leads to a normal force between the pulley's axle and its support, and that affects the frictional torque.

Can you see how this will allow you to distinguish the two sources of friction?

 

[kelseybrahe]

Yes, but to be accurate it's the normal force that matters. Because the cart's mass stays the same, the weight stays the same, so the normal force stays the same.

Yes. The tension in the string leads to a normal force between the pulley's axle and its support, and that affects the frictional torque.

Can you see how this will allow you to distinguish the two sources of friction?

It must have something to do with distinguishing the normal force of the cart from some total? Because if the normal force stays the same, you can always extract that from a total result.
But what total?
The total friction?

 

[haruspex]

It must have something to do with distinguishing the normal force of the cart from some total? Because if the normal force stays the same, you can always extract that from a total result.
But what total?
The total friction?

Yes. With the increased mass of the suspended weight you run the experiment again and calculate the new total friction. You can try several different masses and plot the relationship between mass and total friction. There are reasons to expect it to be a straight line. You could then extrapolate to the case of no suspended mass, which would mean no pulley friction. The friction that remains must be at the cart.

 

[kelseybrahe]

Yes. With the increased mass of the suspended weight you run the experiment again and calculate the new total friction. You can try several different masses and plot the relationship between mass and total friction. There are reasons to expect it to be a straight line. You could then extrapolate to the case of no suspended mass, which would mean no pulley friction. The friction that remains must be at the cart.

Alright, cool. That makes sense.
Thank you so much for all your help.
You have saved me from so much stress. And I feel like I understand these things a bit better now.

Thank you very much.

So for the second (terribly worded) question I just talk about what I would do to find the friction acting on the cart with the same equipment (by repeating the experiment but removing the cart and then finding the difference in the two totals).

 

[haruspex]

Alright, cool. That makes sense.
Thank you so much for all your help.
You have saved me from so much stress. And I feel like I understand these things a bit better now.

Thank you very much.

So for the second (terribly worded) question I just talk about what I would do to find the friction acting on the cart with the same equipment (by repeating the experiment but removing the cart and then finding the difference in the two totals).

I've had second thoughts on what I just posted about the second problem. It sort of works, but the justification is not obvious.

We need a reasonable model for what determines each of the two frictional forces. For the cart, it is clearly the weight of the cart and the coefficient of kinetic friction there. For the pulley there will be another such coefficient, but what will play the role of the weight? Any ideas?

 

[kelseybrahe]

I've had second thoughts on what I just posted about the second problem. It sort of works, but the justification is not obvious.

We need a reasonable model for what determines each of the two frictional forces. For the cart, it is clearly the weight of the cart and the coefficient of kinetic friction there. For the pulley there will be another such coefficient, but what will play the role of the weight? Any ideas?

I'm not entirely sure. Perhaps the weight of the mass that pulls the cart and goes through the pulley?

My physics teacher was away when I handed it in, but I talked to the other science faculty and they seemed to think that we were supposed to look at the distance between the dots on the ticker tape and talk about that. So something along the lines of friction causing deceleration, displayed as the distance between dots slowly decreasing.
Where to go from there I am completely lost. Do you have any ideas?

 

[haruspex]

Perhaps the weight of the mass that pulls the cart and goes through the pulley?

When thinking about the force acting on some part of a system, a common mistake is to look too far away from it. The pulley knows nothing about the mass or the cart. What does it experience directly?

they seemed to think that we were supposed to look at the distance between the dots on the ticker tape and talk about that. So something along the lines of friction causing deceleration, displayed as the distance between dots slowly decreasing

You had to do that for the first part, finding the total friction. As I wrote, it will not help you distinguish the different sources.

 

[David Lewis]

Looks like the exercise asks for force acting on the whole system (which you've already calculated) and force acting on the cart. Net force on the cart, however, is force acting on the whole system minus friction of the cart. There will be frictional losses from the pulley and the acceleration measuring apparatus but they might be small compared to the friction of the cart.

I noticed an arithmetic error in one of my earlier posts:
2.44 N/1.154 kg = 2.15 m/s²
Should be 2.11 m/s² = acceleration if there were no friction

The measured acceleration (1.05 m/s²) is about half of 2.11 m/s²
So you might be able to solve the problem in your head. I'm assuming acceleration is measured imprecisely enough in this experiment to swamp whatever effect the pulley and ticker tape have.

 

[kelseybrahe]

When thinking about the force acting on some part of a system, a common mistake is to look too far away from it. The pulley knows nothing about the mass or the cart. What does it experience directly?

You had to do that for the first part, finding the total friction. As I wrote, it will not help you distinguish the different sources.

It experiences friction from the string?

 

[kelseybrahe]

Looks like the exercise asks for force acting on the whole system (which you've already calculated) and force acting on the cart. Net force on the cart, however, is force acting on the whole system minus friction of the cart. There will be frictional losses from the pulley and the acceleration measuring apparatus but they might be small compared to the friction of the cart.

Yeah, that makes sense and that's what we thought too.
But the teacher said we had to use the same ticker tape to determine the friction on the cart. Or more, determine how it could be possible to find the friction acting on the cart.

 

[haruspex]

It experiences friction from the string?

There should be no slipping between the two, so yes, there will be static friction. But for there to be any friction, what other force must there be between them?
It just occurred to me that we have not mentioned the rotational inertia of the pulley, but no matter. By using two different weights with the same mass (or vice versa) we can find the total resistance at the pulley. We do not care how much of that is from rotational inertia and how much from axial friction.

 

[haruspex]

the same ticker tape

Just noticed that reply... Do you mean the same ticker tape or the same apparatus generating a new ticker tape?
I see no way to discriminate the cart's friction from other frictions and the rotational inertia of the pulley from that.

 

[David Lewis]

Correction:
Net force on cart = force acting on the whole system minus cart friction, minus pulley friction, minus ticker tape friction.
Acceleration = force acting on the whole system divided by (cart mass plus ticker tape mass plus pulley moment of inertia).

Acceleration if there were no friction = 2.11 m/s²
Measured acceleration = 1.05 m/[sups]2[/sup]

 

[haruspex]

Net force on cart = force acting on the whole system minus cart friction, minus pulley friction, minus ticker tape friction.
Acceleration = force acting on the whole system divided by (cart mass plus ticker tape mass plus pulley moment of inertia).

I disagree with both. The net force on the cart is also affected by the acceleration of the suspended mass and of the pulley.
And you can't add a mass to a moment of inertia.

I'm not sure exactly where the ticker tape attaches, so I'll assume it is directly attached to the cart.
Net force on cart = Tension in horizontal part of string - friction from cart axles (if it's on whels) - friction from ticker tape
Acceleration of system = net force on cart / mass of cart = radius x net torque on pulley / moment of inertia of pulley = net force on suspended mass / mass suspended
Net torque on pulley = radius x (tension in vertical string - tension in horizontal string) - axial friction torque
Net force on suspended mass = weight - tension in vertical string

If we run experiments with different masses:
Friction from cart axles is proportional to mass of cart
Axial friction at pulley is proportional to the net force on the pulley from the two tensions.

 

[David Lewis]

If I understand the OP correctly, only the force acting on the whole system and the force acting on the cart is asked for.

 

[haruspex]

If I understand the OP correctly, only the force acting on the whole system and the force acting on the cart is asked for.

Yes, but your statement was not correct. If we take it back to the simplest analysis, no friction, massless pulley, cart mass m, suspended mass M, the acceleration would be Mg/(M+m), so the force on the cart would be Mmg(M+m). According to what you wrote it would be Mg.

 

[kelseybrahe]

Just noticed that reply... Do you mean the same ticker tape or the same apparatus generating a new ticker tape?
I see no way to discriminate the cart's friction from other frictions and the rotational inertia of the pulley from that.

The same ticker tape. Which is why it's so confusing, because I didn't think you could do that?

 

[kelseybrahe]

There should be no slipping between the two, so yes, there will be static friction. But for there to be any friction, what other force must there be between them?
It just occurred to me that we have not mentioned the rotational inertia of the pulley, but no matter. By using two different weights with the same mass (or vice versa) we can find the total resistance at the pulley. We do not care how much of that is from rotational inertia and how much from axial friction.

Gravity and/or some resistance force?
Like the equal/opposite force?
Or does acceleration or some kinetic force apply as well?

 

[haruspex]

The same ticker tape. Which is why it's so confusing, because I didn't think you could do that?

Neither do I. The ticker tape will tell you the acceleration. Maybe you can also observe some non-constant acceleration, but it doesn't help. The friction on the cart, the friction on the pulley, and the moment of inertia of the pulley will all affect the acceleration, but not in a way to make it non-constant. So I see no way to disentangle the effects.

 

[kelseybrahe]

Neither do I. The ticker tape will tell you the acceleration. Maybe you can also observe some non-constant acceleration, but it doesn't help. The friction on the cart, the friction on the pulley, and the moment of inertia of the pulley will all affect the acceleration, but not in a way to make it non-constant. So I see no way to disentangle the effects.

I thought the same thing. I'll let you know what he says when I get the assignment back.