# Force acting on system

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#1
Hey. I need some help on this question (It's attached along with the diagram). So I first took to drawing free body diagrams of each of the masses but I don't think I've shown all the forces; I'm not particularly sure of my diagram for M2. Also, I'm not sure on the strategy to solve the question. I've assumed there is no friction (because of smooth surfaces). Am I meant to use Newton's second for each of the masses? I don't think I am because it talks of motion relative to the bodies. Any ideas please?
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6 years ago
#2
I think you will be using N2 - the condition referring to relative motion just means that all three bodies will have the same acceleration

There are a couple of issues with your force diagrams: F acts on the box, not the mass m2: you do however need horizontal normal contact forces in two of the diagrams.

I think you are better off considering the box and the pulley as a single object: however if you treat them separately you must not neglect the forces they exert on each other.
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#3
(Original post by Mr Gum)
I think you will be using N2 - the condition referring to relative motion just means that all three bodies will have the same acceleration

There are a couple of issues with your force diagrams: F acts on the box, not the mass m2: you do however need horizontal normal contact forces in two of the diagrams.

I think you are better off considering the box and the pulley as a single object: however if you treat them separately you must not neglect the forces they exert on each other.
Thanks for the reply. How're the diagrams now?
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6 years ago
#4
Normal reactions good - but you still need the forces on the box at the pulley, and the external force F on the box.
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6 years ago
#5
(Original post by samjohnny)
Hey. I need some help on this question (It's attached along with the diagram). So I first took to drawing free body diagrams of each of the masses but I don't think I've shown all the forces; I'm not particularly sure of my diagram for M2. Also, I'm not sure on the strategy to solve the question. I've assumed there is no friction (because of smooth surfaces). Am I meant to use Newton's second for each of the masses? I don't think I am because it talks of motion relative to the bodies. Any ideas please?
Only one tension in string.
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#6
(Original post by Mr Gum)
Normal reactions good - but you still need the forces on the box at the pulley, and the external force F on the box.
Here's what I've got for the free body diagram of the box M. I've assumed - as the question states - that the pulley is massless and frictionless but, because it's attached to the box, it is a part of it and hence is included in the box's diagram. Is that right?
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#7
(Original post by brianeverit)
Only one tension in string.
Ah yeah, thanks!
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6 years ago
#8
(Original post by samjohnny)
Here's what I've got for the free body diagram of the box M. I've assumed - as the question states - that the pulley is massless and frictionless but, because it's attached to the box, it is a part of it and hence is included in the box's diagram. Is that right?
Yes, that's how I'd do it. But don't forget the external force F.
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#9
(Original post by Mr Gum)
Yes, that's how I'd do it. But don't forget the external force F.
Thanks. Here's what I'm doing so far: for each of the 3 systems I'm adding together all their force vectors and equating them to m*a. How do I proceed on from there?
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6 years ago
#10
(Original post by samjohnny)
Thanks. Here's what I'm doing so far: for each of the 3 systems I'm adding together all their force vectors and equating them to m*a. How do I proceed on from there?
For one of the objects, both the horizontal and vertical directions are interesting: for the other two, only the horizontal equations are interesting. This gives you four equations which between them involve a tension, a normal reaction, the acceleration, M, m1, m2, g and F. Eliminate the first three of these and you can get F in terms of M, m1, m2 and g.
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#11
(Original post by Mr Gum)
For one of the objects, both the horizontal and vertical directions are interesting: for the other two, only the horizontal equations are interesting. This gives you four equations which between them involve a tension, a normal reaction, the acceleration, M, m1, m2, g and F. Eliminate the first three of these and you can get F in terms of M, m1, m2 and g.
I hope you don't mind, would it be possible for you to quickly check over my work. I think I might have missed a step or something because I've not quite got what you said I should.
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#12
^Anyone?
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6 years ago
#13
(Original post by samjohnny)
I hope you don't mind, would it be possible for you to quickly check over my work. I think I might have missed a step or something because I've not quite got what you said I should.
I think your equations 1 - 6 are right, except for a couple of sign errors in equation 1.

Your last line, that F= a(M-m2-m1) reflects these errors in that applying N2 to the whole system would give you F= a(M+m2+m1).

I think your answer is not expected to include a. From your equations 1,3,5,6, you should seek to eliminate a, t and N_2
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