M1 a very confusing lift question

OP

I understand this method but I don't don't know why a is only considered when the q only asks for b

Reply 2

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.

Reply 3

OP

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

Reply 4

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.

Reply 5

OP

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

Reply 6

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)

Reply 7

OP

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

Reply 8

Cryptokyo

14

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

R_{A}-0.6g=0.6\times0.5

so

R_{A}=6.18

N.

(edited 7 years ago)

Reply 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

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.

Reply 10

OP