Acceleration of rising mass and tension in cord

[Ithilrandir]

Homework Statement

In the system shown in Fig. 5-5, M1 slides without friction on the inclined plane. Θ=30°, M1 = 400g, M2 = 200g. Find the acceleration a of M2 and the tension T in the cords.

Relevant Equations

...

First I determined the direction the masses will be traveling by comparing the change in potential energy, so when M2 rises by Δy, M1 drops by Δy. Potential energy decreases when M1 drops and M2 goes up.

Tension force for string pulling against M1= 400 x gsin30°, and since there are two strings pulling against the pulley, force caused by the lower pulley = 400g.

Tension force due to M2= 200g.

Net force=400g-200g=200g=0.200 x 9.81= 1.962N

Acceleration = 1.962/(0.400+0.200)=3.27 m s^-1

T=1.962N.

My answer feels off, so I checked the answer, and the answer's acceleration is g/9= 1.09, T=2.17782N

Could anyone help point my mistakes?

 

[PeroK]

Tension force due to M2= 200g.

Net force=400g-200g=200g=0.200 x 9.81= 1.962N

Acceleration = 1.962/(0.400+0.200)=3.27 m s^-1

This makes no sense to me Where is $\theta$? Where is effect of the second pulley?

 

[Ithilrandir]

Does θ and the second pulley have any effect? I couldn't see it so it's my bad.

 

[PeroK]

Does θ and the second pulley have any effect? I couldn't see it so it's my bad.

Yes, they do!

 

[Ithilrandir]

Hmm... I'll give it a good look again after work tmr. Maybe I'll see something extra.

 

[PeroK]

Hmm... I'll give it a good look again after work tmr. Maybe I'll see something extra.

You can do it by forces or think about energy - at least to see the effect of the slope and pulley.

 

[Steve4Physics]

Could anyone help point my mistakes?

There are many mistakes. Here are just a few points to think about.

When using the formula ‘weight = mg’ with SI units, what units are used for m?

Why is it wrong to equate the tension acting on M₁ to the downhill component of M₁’s weight (which I think is what you have done) in this problem?

In the diagram there are 2 separate ropes. Are the tensions in the 2 ropes equal? If not, how are they related?

Would it be a good idea to draw a fre-body diagram of each mass?

This problem is based on what is called an ‘Atwood machine’. If you are not familiar with Atwood machine calculations, I would highly recommend reading about them (and understanding) before attempting this problem.

 

[Lnewqban]

Hmm... I'll give it a good look again after work tmr. Maybe I'll see something extra.

Because they are linked together, M1 and M2 act like a system of masses.
Gravity is the only external agent acting on that system (there are no frictional forces).

The geometry of the slope and arrangement of pulleys and cords will determine the value and direction of the acceleration (if any) of the whole system and of the requested acceleration of M2.

Tip: In any direction along the surface that supports each particle, M1 will move twice as fast as M2 will (because the mobile pulley acts like a class 2 lever).
Please, see:
https://en.m.wikipedia.org/wiki/Lever#Classes_of_levers

 

[Ithilrandir]

When using the formula ‘weight = mg’ with SI units, what units are used for m?

Please forgive my omission of units in my calculations.

Why is it wrong to equate the tension acting on M₁ to the downhill component of M₁’s weight (which I think is what you have done) in this problem?

I think it's because M1 is accelerating, so you can't equate it. However, one doubt I have in my mind is whether the tension on m1 should be greater or smaller than the downhill component of M1. My guess is that it should be smaller so that M1 can accelerate downwards.

In the diagram there are 2 separate ropes. Are the tensions in the 2 ropes equal? If not, how are they related?

After looking at it again, I think they are probably not equal. My best guess is that the string pulling against M2 has a higher tension than the one pulling against M1, but the next effect of string pulling M1 is higher than the tension on M2.

 

[haruspex]

one doubt I have in my mind is whether the tension on m1 should be greater or smaller than the downhill component of M1

It is not necessary to guess. Just choose which way you are taking as positive for its acceleration and write the equations accordingly.

they are probably not equal.

Quite so, so assign two different variables.
What forces act parallel to the slope on m₁?

 

[Ithilrandir]

What forces act parallel to the slope on m₁?

Letting the tension of the ropes correspond to the masses, i.e T1, T2.

Force on M1 is M1gsin30-T1. Force on M2 is T-M2g. Force on pulley on the slope is 2T1 - T2. Is that right so far?

 

[Steve4Physics]

Please forgive my omission of units in my calculations.

Omission of units was not my point.

Is the weight of a 400 gram mass given by W = mg = 400grams *9.81 m/s² = 3924N?
(Hint: no it is not!)

I think it's because M1 is accelerating, so you can't equate it. However, one doubt I have in my mind is whether the tension on m1 should be greater or smaller than the downhill component of M1. My guess is that it should be smaller so that M1 can accelerate downwards.

Correct. The downhill component of M₁'s weight is only equal to the tension on it when M₁ is in equilibrium, i.e. not accelerating.

Don't guess the direction of acceleration. Assume it is in a certain direction and, when you calculate it, if it turns out negative then you know your assumption was wrong.

After looking at it again, I think they [tensions] are probably not equal. My best guess is that the string pulling against M2 has a higher tension than the one pulling against M1

Correct. Are the 2 tensions related in some way?

Did you read about Atwood machine calculations? You will even find YouTube videos about them. Unless you understand how to deal with a simple Atwood machine calculation, you won't be able to do this problem.

It's past my bedtime here in the UK, so I'll wish you a goodnight!

 

[Ithilrandir]

Did you read about Atwood machine calculations? You will even find YouTube videos about them. Unless you understand how to deal with a simple Atwood machine calculation, you won't be able to do this problem.

I did read up a little, however in the context of this question, I don't know how to change the pulley pulling M1 into a mass I can use.

 

[haruspex]

Letting the tension of the ropes correspond to the masses, i.e T1, T2.

Force on M1 is M1gsin30-T1. Force on M2 is T-M2g. Force on pulley on the slope is 2T1 - T2. Is that right so far?

Yes. Since you are not given a mass for the pulley, assume it has none. What equation does that give you?

 

[Ithilrandir]

Yes. Since you are not given a mass for the pulley, assume it has none. What equation does that give you?

Is that not just 2T1 - T2? I didn't include the mass in this.

 

[haruspex]

Is that not just 2T1 - T2? I didn't include the mass in this.

That's an expression, not an equation. Include the mass and acceleration to make an equation.

 

[Ithilrandir]

That's an expression, not an equation. Include the mass and acceleration to make an equation.

Well at the right end we have F = M₂a = T₂ - M₂g.

Since M₁ will be twice as fast as M₂, acceleration of M₁ = 2a.

M₁g - T₁ = 2M₁a.

M₁(g-2a) = T₁

Subbing this back into the previous few equations,

2M₁(g-2a) - M₂(a+g) = M₂a

Putting in the values of the masses, a= 0.3g, which is not the correct answer of g/9 or 1.09.

 

[haruspex]

M1g - T1 = 2M1a.

Which way do those two forces act on the mass? Are they in the same straight line?

Subbing this back into the previous few equations,

To do that you need the relationship between the two tensions. Have you figured that out?

 

[Ithilrandir]

Which way do those two forces act on the mass? Are they in the same straight line?

My bad I had forgotten about the sin30θ. That said, I don't see a different relation the two tension should have.

 

[haruspex]

My bad I had forgotten about the sin30θ. That said, I don't see a different relation the two tension should have.

I can't see anywhere where you have posted what that relationship is.
You wrote

2T1 - T2

Did you think you wrote 2T1=T2?

 

[Ithilrandir]

I had meant 2T₁-T₂= net force at that point.

 

[haruspex]

I had meant 2T₁-T₂= net force at that point.

Yes, that's what I thought you meant at the time, that it's the net force on the pulley. So how did you go from there to finding the relationship between the two tensions?
There is a way, and it's quite easy.

 

[Ithilrandir]

Yes, that's what I thought you meant at the time, that it's the net force on the pulley. So how did you go from there to finding the relationship between the two tensions?
There is a way, and it's quite easy.

I had equated the net force to be M₂a, with a being the acceleration of M₂ I didn't think there was a direct relation between the two tensions and solved it and got the wrong answer.

 

[haruspex]

I had equated the net force to be M₂a, with a being the acceleration of M₂ I didn't think there was a direct relation between the two tensions and solved it and got the wrong answer.

I do not understand how you could get a solution without finding the relationship between the two tensions. I told you how to do that in post #16.

It is very hard having to guess what you have done. Please post all your working.

 

[Ithilrandir]

I do not understand how you could get a solution without finding the relationship between the two tensions. I told you how to do that in post #16.

It is very hard having to guess what you have done. Please post all your working.

F = M₂a = T₂ - M₂g

2T₁ - T₂ = F

(M₁gsin30 - T₁)/M₁ = 2a

These are the equations I used to get my wrong answers.

 

[haruspex]

2T₁ - T₂ = F

How did you get that equation?

 

[Ithilrandir]

How did you get that equation?

I had assumed the net force to the be same across the string. Judging by your reply it seems to be wrong.

 

[Steve4Physics]

I'm chipping in.
1. Draw a free body diagram of the pulley (the one near M₁).
2. What are the forces directly acting on the pulley?
3. What is the resultant force on the pulley?
4. Now apply 'F =ma' to the pulley (remembering the pulley has negligible mass).
5. You should now have an equation relating the tensions. If you don't, post your working for each of the above steps.

 

[haruspex]

I had assumed the net force to the be same across the string. Judging by your reply it seems to be wrong.

As I wrote in post #14

Since you are not given a mass for the pulley, assume it has none. What equation does that give you?

Or to put it in more detail, what @Steve4Physics wrote in post #28.

 

[Ithilrandir]

I'm chipping in.
1. Draw a free body diagram of the pulley (the one near M₁).
2. What are the forces directly acting on the pulley?
3. What is the resultant force on the pulley?
4. Now apply 'F =ma' to the pulley (remembering the pulley has negligible mass).
5. You should now have an equation relating the tensions. If you don't, post your working for each of the above steps.

1. I've uploaded a free body diagram of the pulley near M₁.
2. Forces directly acting on the pulley are the two T₁ acting down the slope and the T₂ acting up the slope.
3. Resultant force = 2T₁ - T₂
4. If the pulley has negligible mass, F = ma shouldn't be calculable?

 

[haruspex]

3. Resultant force = 2T₁ - T₂
4. If the pulley has negligible mass, F = ma shouldn't be calculable?

If 2T₁ - T₂=ma and m=0 then...?

 

[Ithilrandir]

2T₁ = T₂?

 

[haruspex]

2T₁ = T₂?

Yes!

 

[Ithilrandir]

Yes!

Thanks, I solved the answer. I was under the impression that the pulley accelerates with force = ma and trying to solve it abstractly that way.