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# M1 a very confusing lift question Watch

1. hi, can I get some help with part c of this question pls? I don't understand why the weight of a is also consider although the question says 'on B' only . Also by the force exterted on mass B , don't they want the reaction force and just consider B only ? In the other lift questions I have done , when they ask for the force exerted on the person by the lift, only the weight of that person is considered though. I am getting confused . Thanks.
2. I understand this method but I don't don't know why a is only considered when the q only asks for b
3. Because the question requires all of the forces acting on B. As A lies on top of B, the force exerted by the support must include the weight of both A and B. The answer doesn't say that A should be considered, it just says that the force that A exerts should be included.
4. (Original post by RKM21)
Because the question requires all of the forces acting on B. As A lies on top of B, the force exerted by the support must include the weight of both A and B. The answer doesn't say that A should be considered, it just says that the force that A exerts should be included.
Okay thanks. So if b instead is on top of A does it mean the weight of A is not considered then for part c.? Thanks
5. (Original post by coconut64)
Okay thanks. So if b instead is on top of A does it mean the weight of A is not considered then for part c.? Thanks
Yeah i think you're right. For part C, just use B and nothing else.
Support-(Mass of B x9.8) = Mass of B x Acceleration.

We use that book at my college too.
6. (Original post by RKM21)
Yeah i think you're right. For part C, just use B and nothing else.
Support-(Mass of B x9.8) = Mass of B x Acceleration.

We use that book at my college too.
Are you sure because the method shows that both weights are considered though ... It's p58 by the way thanks
7. Yeah, sorry about that. I thought you were referring to part B.
For part C, both weights would be considered so you would pretty much get the same answer as you got in Part A. I would just use the scale pan tbh and resolve the forces from there. For part B, if you used B and resolved the forces you would get the same answer as Part C - which is obviously wrong. If you look at the wording of the question:

Find the force exerted on mass B by mass A

It is kind of opposite to answering the impulse questions where we get told

Find the impulse exerted on X by Y. But here we look at The impulse for X only.
(Not to confuse you more but in the impulse example above, the impulse for Y on X would also be exactly the same)
8. (Original post by RKM21)
Yeah, sorry about that. I thought you were referring to part B.
For part C, both weights would be considered so you would pretty much get the same answer as you got in Part A. I would just use the scale pan tbh and resolve the forces from there. For part B, if you used B and resolved the forces you would get the same answer as Part C - which is obviously wrong. If you look at the wording of the question:

Find the force exerted on mass B by mass A

It is kind of opposite to answering the impulse questions where we get told

Find the impulse exerted on X by Y. But here we look at The impulse for X only.
(Not to confuse you more but in the impulse example above, the impulse for Y on X would also be exactly the same)
Okay.... Can i just ask if b is on top of A does it mean the weight or A is not considered. Or A will also be considered since A is also in the pan? Yeah when I add the two R together it gives me 10.3. Cheers
9. (Original post by coconut64)
Okay.... Can i just ask if b is on top of A does it mean the weight or A is not considered. Or A will also be considered since A is also in the pan? Yeah when I add the two R together it gives me 10.3. Cheers
You are correct! The force exerted by the scale will still be 10.3N for an acceleration upwards of 0.5ms^-2 and this force will be exerted on A but the force exerted by A on B will be so N.
10. (Original post by coconut64)
Okay.... Can i just ask if b is on top of A does it mean the weight or A is not considered. Or A will also be considered since A is also in the pan? Yeah when I add the two R together it gives me 10.3. Cheers
For which part of the question. For A it would make no difference which is on top of the other, for B the force exerted on mass B by A would be equal to its weight as every reaction has an equal and opposite reaction plus the extra force required to produce the acceleration of 0.5ms-2 upwards. For C, the scale pan technically only supplies a force to A(as it is touching the pan) - which includes the weight of B on top, so IMO it would be 0N applied to B as the scale pan applies a force only to mass A.
11. (Original post by Cryptokyo)
You are correct! The force exerted by the scale will still be 10.3N for an acceleration upwards of 0.5ms^-2 and this force will be exerted on A but the force exerted by A on B will be so N.
You mean the 10.3 N force is exerted on B right? that's the part c. So the pan does not exert any force on A since it's on top of B? Thanks.
12. (Original post by coconut64)
You mean the 10.3 N force is exerted on B right? that's the part c. So the pan does not exert any force on A since it's on top of B? Thanks.
Yes you are correct. I was doing it for the case where B is on top of A.

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