What Is the Impact of Friction on Acceleration in a Dual-Mass Pulley System?

[Strawer]

Homework Statement

Two blocks with equal masses m1=m2=5kg are connected via a pulley, as shown. Take $\mu$ = 0.25. Find the acceleration given that a) m1 is moving down, b)m1 is moving up. If m2=6 kg for what values of m1 will the pair move att constant speed?

Homework Equations

F=ma, f=$\mu$N

The Attempt at a Solution

I've managed to solve a) and c) but not b).
I don't understand how m1 can be moving up, since m1=m2 and the force m2 affects m1 with is sin(37)*m2g < m1g.
If m1 is moving up doesn't that mean that it will not affect the other mass? I mean shouldn't the tension on m2 be zero? If so then we have $\Sigma$F=5*9.8*sin(37)-0.25*9.8*cos(37)*5=19.7
Which gives a=19.7/5=3.94 m/s^2 However it should be 2.94 m/s^2 according to the answer sheet.

 

[cepheid]

I don't understand how m1 can be moving up, since m1=m2 and the force m2 affects m1 with is sin(37)*m2g < m1g.

No. I highly recommend that you draw a free body diagram for m₂, and then another free body diagram for m₁. Apply Newton's 2nd law, namely that ƩF = ma. You should be able to solve for a. It doesn't matter that it is impossible for m₂ to pull m₁ up by gravity alone. All that matters is that you're told that right now m₂ happens to be moving down the incline, while m₁ is moving upward. These are the initial conditions you've been given. (Maybe somebody applied a force to make this happen. Who cares? It doesn't matter, just accept this as the initial condition).

Indeed, if you choose the "down the incline" direction to be positive in your force balance equations for m₂, you'll find that you get a negative value for a, meaning that although m₂ is presently moving down the incline, its acceleration points up the incline, meaning that it is slowing down and will eventually reverse direction.

If m1 is moving up doesn't that mean that it will not affect the other mass? I mean shouldn't the tension on m2 be zero?

No, the fact that m₁ is moving up doesn't mean that the rope is going slack. As I said before, it means that m₂ is moving down the incline while m1 moves upward (and the rope remains taut).

 

[Strawer]

No. I highly recommend that you draw a free body diagram for m₂, and then another free body diagram for m₁. Apply Newton's 2nd law, namely that ƩF = ma. You should be able to solve for a. It doesn't matter that it is impossible for m₂ to pull m₁ up by gravity alone. All that matters is that you're told that right now m₂ happens to be moving down the incline, while m₁ is moving upward. These are the initial conditions you've been given. (Maybe somebody applied a force to make this happen. Who cares? It doesn't matter, just accept this as the initial condition).

Indeed, if you choose the "down the incline" direction to be positive in your force balance equations for m₂, you'll find that you get a negative value for a, meaning that although m₂ is presently moving down the incline, its acceleration points up the incline, meaning that it is slowing down and will eventually reverse direction.



No, the fact that m₁ is moving up doesn't mean that the rope is going slack. As I said before, it means that m₂ is moving down the incline while m1 moves upward (and the rope remains taut).

Experience tells me that if I would give m₁ a push the rope would go slack. Or is that because i accelerate it? Also wouldn't the rope go slack if v₁ > v₂ because s=vt and if the rope is taut then s₁=s₂?

Anyways, new try
ƩF=ma

The relevant forces acting on m₂ is

friction, f=0.25*cos(37)*5*9.8
Tension T=5*9.8
Gravitational component, F_x=sin(37)*5*9.8

And so
f-F_x+T=ma
a=(0.25*cos(37)*5*9.8-sin(37)*5*9.8+5*9.8)/10=2.93 m/s^2

 

[cepheid]

Experience tells me that if I would give m₁ a push the rope would go slack. Or is that because i accelerate it?

That's true. An example of the adage, "you can't push on a rope." However, nobody is saying that you (or anyone else) is pushing on m₁. All that is being said is that, presently, m₁ is moving upwards, whilst m₂ is simultaneously moving down the ramp. That's it. Do you understand?

Yeah, that's right. I busted out a "whilst."

Also wouldn't the rope go slack if v₁ > v₂ because s=vt and if the rope is taut then s₁=s₂?

Once again, you're not wrong. Any upward advancement of m₁ must be accompanied by an advancement of m₂ down the ramp by an equal distance. That, in fact, is precisely what is happening (i.e. this is the situation you are asked to consider). It's just the exact reverse of what happens when m₁ falls down, and m₂ consequently gets pulled up the ramp. The difference is that the latter situation can happen spontaneously, while the former cannot.

Anyways, new try
ƩF=ma

The relevant forces acting on m₂ is

friction, f=0.25*cos(37)*5*9.8
Tension T=5*9.8

This part in red is not correct. If the tension were just equal to the weight, then the net force on m₁ would be zero, and it would have zero acceleration. However, if m₁ had 0 acceleration, then the whole system would have zero acceleration. It doesn't.

To get T, you need to do the force balance for m₁, which tells you that T - mg = ma.

Gravitational component, F_x=sin(37)*5*9.8

And so
f-F_x+T=ma
a=(0.25*cos(37)*5*9.8-sin(37)*5*9.8+5*9.8)/10=2.93 m/s^2

It's a fluke that you got the right answer here. You made a second mistake that negated the first one that I mentioned above (in red). Your second mistake was to assume that the 'm' in the ma term was the total system mass. However, since the force balance equation you wrote above was for m₂ only, it should be that m = 5 kg, not 10 kg.

Without this second mistake, you see that your answer is too large by a factor of 2.

Also, I'd really recommend keeping everything algebraic, until the last possible step, and not plug in numbers until you've simplified things as much as possible. If you'd done that, you'd have seen that 'm' canceled from both sides of the equation, making it unnecessary to plug in the 5 kg. Furthermore, you'd see that g factors out of every term on the left-hand side.

Anyway, if you correct your first mistake (the one in red) then you'll have this problem nailed.

 

[asteriatic]

I would like to clarify a few things about the problem and your solution attempt. First, it is important to note that in this system, m1 and m2 are not directly connected to each other. They are connected via a pulley, which means that they are indirectly connected through the tension in the string. This tension is what allows the blocks to move together.

Now, in part b) of the problem, m1 is indeed moving up. This means that the tension in the string will be pulling m2 upwards, causing it to accelerate in the same direction as m1. This also means that m2 will be exerting a downward force on the pulley, which will in turn affect m1.

Your attempt at solving the problem is on the right track, but there are a few errors. First, the force of friction should be calculated using the normal force on m1, not m2. This normal force is equal to m1g, not m2g. Second, the tension in the string should not be considered as a separate force acting on m1. Instead, it should be included in the net force acting on m1, along with the force of friction and the weight of m1.

With these corrections, the correct solution for part b) should be:

\sumF = m1a = T - \mu(m1g) - m1g = 0

T = (1 + \mu)m1g = (1 + 0.25)(5kg)(9.8m/s^2) = 61.25 N

Therefore, the acceleration of the system in part b) is:

a = T/m1 = 61.25N/5kg = 12.25 m/s^2

As for part c) of the problem, the blocks will move at constant speed when the net force on the system is equal to zero. In other words, when the tension in the string is equal to the weight of m1. This means that:

T = m1g

So, when m2 = 6kg, the value of m1 that will result in the system moving at constant speed is:

m1 = T/g = (6kg)(9.8m/s^2) = 58.8 kg

In conclusion, as a scientist, I would recommend double-checking your calculations and taking into account all the relevant forces acting on the system