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How do I do this 'Moments' question? I need help !

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Applying to Uni? Let Universities come to you. Click here to get your perfect place 20-10-2014
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    The picture needed for the question is attached.

    Oh and the weight of the trolley and its contents is 160N

    The question is :

    (a) Calculate the minimum force that needs to be applied vertically at A to life the front of the grounds.

    (b) State and explain, without calculation, how the minimum force that needs to be applied vertically at A to lift the rear wheels off the ground compares to the force you calculated in part (a)

    I just don't understand it all, I think I'm misunderstanding what the perpendicular distance from the pivot is. I'm trying to revise.. and this is really bugging me.

    Your help will be much appreciated.
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    First you need to find the force acting at P and Q (let them be P and Q).

    Do you agree that P+Q = W where W = weight of trolley ?

    Can you take moments about A and hence form an equation that will allow you to find the forces at P and Q in terms of W ?

    In case you get stuck
    Spoiler:
    Show
    W=P+Q
    10P + 100Q = 50W Taking moments about A

    W-P=Q

    10P + 100W - 100P = 50W
    50W = 90P

    5W/9 = P
    4W/9 = Q

    Take moments about the COG and now find the minimum force needed.
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    (Original post by Ari Ben Canaan)
    First you need to find the force acting at P and Q (let them be P and Q).

    Do you agree that P+Q = W where W = weight of trolley ?

    Can you take moments about A and hence form an equation that will allow you to find the forces at P and Q in terms of W ?

    In case you get stuck
    Spoiler:
    Show
    W=P+Q
    10P + 100Q = 50W

    W-P=Q

    10P + 100W - 100P = 50W
    50W = 90P

    5W/9 = P
    4W/9 = Q

    Take moments about A and now find the minimum force needed.
    Sorry, I missed some information out.

    The weight of the trolley is 160N.

    So the force at Q = 80N

    I'm confused after that...

    So the moment at Q = 80 x 50?

    So that's 1600.

    Is that the minimum force..I feel I've messed it up.
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    (Original post by Ari Ben Canaan)
    First you need to find the force acting at P and Q (let them be P and Q).

    Do you agree that P+Q = W where W = weight of trolley ?

    Can you take moments about A and hence form an equation that will allow you to find the forces at P and Q in terms of W ?

    In case you get stuck
    Spoiler:
    Show
    W=P+Q
    10P + 100Q = 50W

    W-P=Q

    10P + 100W - 100P = 50W
    50W = 90P

    5W/9 = P
    4W/9 = Q

    Take moments about A and now find the minimum force needed.
    1600 would be the moment wouldn't it?

    So wait is the answer just 80N for part (a)?

    And for part (b) the answer is ' they're both the same'?
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    (Original post by sunexotic)
    Sorry, I missed some information out.

    The weight of the trolley is 160N.

    So the force at Q = 80N

    I'm confused after that...

    So the moment at Q = 80 x 50?

    So that's 1600.

    Is that the minimum force..I feel I've messed it up.
    Hold your horses. How did you determine the force at Q was 80N ?
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    (Original post by Ari Ben Canaan)
    Hold your horses. How did you determine the force at Q was 80N ?
    Force x perpendicular distance from the pivot.

    160 x 0.5 ( It's 50cm ) = 80

    Or am I confusing the perpendicular distance again?
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    (Original post by sunexotic)
    Force x perpendicular distance from the pivot.

    160 x 0.5 ( It's 50cm ) = 80

    Or am I confusing the perpendicular distance again?
    You are wrong since you've forgotten about the force at P.

    I suggest you look at the spoiler in my first post
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    (Original post by Ari Ben Canaan)
    You are wrong since you've forgotten about the force at P.

    I suggest you look at the spoiler in my first post
    I'm sorry but where does 100W come from?

    I understand that :

    10P + 100Q = 50W

    So that means

    10P + 100Q = 8000

    But what do you do next?

    ( I realise.. I'm being infuriating )
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    (Original post by sunexotic)
    I'm sorry but where does 100W come from?

    I understand that :

    10P + 100Q = 50W

    So that means

    10P + 100Q = 8000

    But what do you do next?

    ( I realise.. I'm being infuriating )
    The important point to remember is that the sum of the UPWARD forces is equal to the sum of the DOWNWARD acting forces.

    If this was not the case then the trolley would fall through the ground.

    Thus, P+Q = W

    Using the above you can substitute for P or Q in your equation.

    That is, P = W - Q or Q=W-P
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    (Original post by Ari Ben Canaan)
    The important point to remember is that the sum of the UPWARD forces is equal to the sum of the DOWNWARD acting forces.

    If this was not the case then the trolley would fall through the ground.

    Thus, P+Q = W

    Using the above you can substitute for P or Q in your equation.

    That is, P = W - Q or Q=W-P
    Oh thank you !

    So p = 88.8 and q = 71.1

    The minimum force needed to lift the front wheels.. would be.. yeah I'm clueless here.

    Anti clockwise moment = clockwise moment ..

    So you would need to calculate the moment around the rear wheels?

    Yeah.. I'm clueless here.
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    Oh wait it's

    160 x 40 = 10 x Force !

    So it's 640N !
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    (Original post by sunexotic)
    Oh thank you !

    So p = 88.8 and q = 71.1

    The minimum force needed to lift the front wheels.. would be.. yeah I'm clueless here.

    Anti clockwise moment = clockwise moment ..

    So you would need to calculate the moment around the rear wheels?

    Yeah.. I'm clueless here.
    The right terminology is taking moments about the rear wheel. 'Around' is very ambiguous :P

    Yes, if you wanted to you could consider the rear wheel as your pivot and use your value for the force at Q and find the necessary force at A.
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    Do you have a markscheme for this paper i.e. the answer ?
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    It's a question I got wrong in an exam, so I've been redoing it, so unfortunately no, but I can ask someone from my class on facebook or something.

    Thanks for all your help. I'm really glad I've done it.

    Thanks again

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