Friction doesn't cancel friction.kelseybrahe said:but these are still insignificant and basically cancel out).
haruspex said: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?
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.kelseybrahe said:Well the pulley is controlling the speed that the accelerating mass falls
haruspex said: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?
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 said: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 said: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?
There will, but why? Does that reason apply also to the friction on the cart, which has the same mass as before?kelseybrahe said:there will be greater friction on the pulley too, won't there?
I did not say the rationbetween the friction and the acceleration. I said the ratio between the two frictions.kelseybrahe said:And the same if it falls slower. So the ratio would stay the same.
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 said:it seems strange to say that things have an increase in encountered friction if they move faster.
haruspex said: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?
David Lewis said: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
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.kelseybrahe said: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. The tension in the string leads to a normal force between the pulley's axle and its support, and that affects the frictional torque.kelseybrahe said:Is this because of the tension from the string
haruspex said: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?
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 said: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 said: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.
I've had second thoughts on what I just posted about the second problem. It sort of works, but the justification is not obvious.kelseybrahe said: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 said: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?
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?kelseybrahe said:Perhaps the weight of the mass that pulls the cart and goes through the pulley?
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.kelseybrahe said: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
haruspex said: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.
David Lewis said: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.
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?kelseybrahe said:It experiences friction from the string?
Just noticed that reply... Do you mean the same ticker tape or the same apparatus generating a new ticker tape?kelseybrahe said:the same ticker tape
I disagree with both. The net force on the cart is also affected by the acceleration of the suspended mass and of the pulley.David Lewis said: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).
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.David Lewis said: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 said: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.
haruspex said: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.
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 said:The same ticker tape. Which is why it's so confusing, because I didn't think you could do that?
haruspex said: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.