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Size:  255.6 KB 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.
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    I understand this method but I don't don't know why a is only considered when the q only asks for b Name:  1465119147336-1421123930.jpg
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    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.
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    (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
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    (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.
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    (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
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    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)
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    (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
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    (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}=6.18N.
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    (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.
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    (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 R_{A}-0.6g=0.6\times0.5 so R_{A}=6.18N.
    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.
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    (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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