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    hi all, i have no idea how to start this question so any help would be appreciated! i have attached a copy of the question...thanks..
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    Gravity acts vertically. The reaction forces act vertically upwards.

    Resolve and take moments about an axle.
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    how would you draw the free body diagram? i figured out how to do parts B n C now thanks
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    you draw it with the wheel A in the middle.
    I'm no good with paint, but you could take wheel A as the middle and find out all the forces clockwise and anticlockwise to it
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    (Original post by TC88)
    how would you draw the free body diagram? i figured out how to do parts B n C now thanks
    Well your best bet is to model the combined fork lift truck and crate as a block of some shape (I'd recommend having a rectangle with base 1.3m), and then you just draw all of the forces acting on it... that is, the weight acting at the centre of gravity of the whole shape, and the normal contact forces with the ground.
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    (Original post by nuodai)
    Well your best bet is to model the combined fork lift truck and crate as a block of some shape (I'd recommend having a rectangle with base 1.3m), and then you just draw all of the forces acting on it... that is, the weight acting at the centre of gravity of the whole shape, and the normal contact forces with the ground.
    when it says 'fork lift truck and crate combined' does that mean the weight will just be 1500g downwards in the centre of the block?
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    (Original post by TC88)
    when it says 'fork lift truck and crate combined' does that mean the weight will just be 1500g downwards in the centre of the block?
    I think it should be 0.5m from the right hand side of the block. You could probably put the separate weights on separately, or you could take moments... the centre of mass of the combined object works 0.5m from the right-hand side of the block.
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    where would the 0.4m be? :confused: or can i just ignnore it since the truck and crate are now combined?
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    (Original post by nuodai)
    I think it should be 0.5m from the right hand side of the block. You could probably put the separate weights on separately, or you could take moments... the centre of mass of the combined object works 0.5m from the right-hand side of the block.
    I get reaction at A 10500N and reaction at B 4200N.

    Does anyone agree?
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    (Original post by nuodai)
    I think it should be 0.5m from the right hand side of the block. You could probably put the separate weights on separately, or you could take moments... the centre of mass of the combined object works 0.5m from the right-hand side of the block.
    can u draw the free body diagram for me if u dont mind? i understand how to do parts b and c... i just don't know what the free body diagram should look like to get the full 6 marks? thanks
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    (Original post by steve2005)
    I get reaction at A 10500N and reaction at B 4200N.

    Does anyone agree?
    reaction at A is 4.09 kN
    reaction at B is 3.27 kN
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    I forgot to divide by two because of two wheels at A and two wheels at B. BUT , of course my answer is still wrong.
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    i can't seem to get 3.27kN too... but i got 4.09kN! i have the numerical answers to the question
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    • Thread Starter
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    thanks for the detailed solution i got the same answers now.. im just wondering how to get the full 6 marks for the free body ?
 
 
 
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