Find the acceleration of a pulley system

[AndrejN96]

Homework Statement

Find the acceleration of the system shown on the image, given m1=0.1kg and m2=0.3kg. Mass of pulleys is to be disregarded, and rope is non-stretchable.

Homework Equations

m1*a=m1*g-T

The Attempt at a Solution

I am unsure of how the forces of the second body would act and in what direction. Any help is appreciated.

 

[NTW]

Look at m₂. It obviously pull the pulley with a force m₂ * g. And that force divides in two parts...

Imagine that your mass is, for example, 70 kg, and hence your weight 70 * 9,8 = 686 N, and that you sit on the table of a swing that is fastened to an horizontal beam above with two chains. How much of your weight do you think is supported by each chain...? Well, the same thing happens with m₂

 

[AndrejN96]

Look at m₂. It obviously pull the pulley with a force m₂ * g. And that force divides in two parts...

Imagine that your mass is, for example, 70 kg, and hence your weight 70 * 9,8 = 686 N, and that you sit on the table of a swing that is fastened to an horizontal beam above with two chains. How much of your weight do you think is supported by each chain...? Well, the same thing happens with m₂

My first thought was that it spreads the force applied in half, but how would the equations look like in that case?

 

[NTW]

My first thought was that it spreads the force applied in half, but how would the equations look like in that case?

Leave the equations apart for the moment, and just think about the forces involved. ¿What force is caused by m₁? And what forces are caused by m₂?

 

[AndrejN96]

Leave the equations apart for the moment, and just think about the forces involved. ¿What force is caused by m₁? And what forces are caused by m₂?

Well m₁ would cause tension on the rope, and m₂ would cause tension on both parts of the rope but with lower intensity (?)

 

[NTW]

Well m₁ would cause tension on the rope, and m₂ would cause tension on both parts of the rope but with lower intensity (?)

Yes, but how much lower...? You have already written the right answer above...

Once you decide which are the forces involved, write them down and/or calculate them, and then you should deduce the net force.

 

[AndrejN96]

Yes, but how much lower...? You have already written the right answer above...

Once you decide which are the forces involved, write them down and/or calculate them, and then you should deduce the net force.

I just did some more research in literature I have on physics. It says the tension would be the same in all 3 places of the pulley system, because the rope and pulley are light enough to be disregarded.

 

[NTW]

Look at the right rope, the one that is anchored to the floor. Along that length of rope, there is a force that is one-half of m₂ * g. It is canceled by the same reaction from the roof.
Now, we look at the left rope rising from the pulley that holds m₂. Along that length of rope, you also have one-half of the force caused by m₂ Let's write it down: F1 = (m₂ * g)/2.

But that length of rope is also affected by another pulling force, that coming from m₁. We also write it down: F2 = m₁ * g.

It's easy to see that F1 and F2 are opposed. Evidently, the net force is their difference...

Go on now... There is little left to solve the problem...

 

[Orodruin]

Look at the right rope, the one that is anchored to the floor. Along that length of rope, there is a force that is one-half of m2 * g. It is canceled by the same reaction from the roof.

This is only true of the system is in equilibrium. In general the system will be accelerated and the only thing you can say about the tension in the rope is that it is the same in all parts. You will then have to relate the tension to the acceleration and geometry of the problem.

 

[AndrejN96]

This is only true of the system is in equilibrium. In general the system will be accelerated and the only thing you can say about the tension in the rope is that it is the same in all parts. You will then have to relate the tension to the acceleration and geometry of the problem.

Yes, thanks, this was the confirmation that I needed. Cheers!