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Mechanics help?

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    (Original post by petrus123)
    OK, thanks! I'll give it a try.
    Awesome, good luck, see what you get and compare it to that of the bear
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    (Original post by CheeseIsVeg)
    Awesome, good luck, see what you get and compare it to that of thebear
    OK, I'm just slightly confused about why @thebear has his contact force arrow pointing upwards (it says down on the lift), although I did that at first too, so maybe I missed something.
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    (Original post by petrus123)
    OK, I'm just slightly confused about why @thebear has his contact force arrow pointing upwards (it says down on the lift), although I did that at first too, so maybe I missed something.
    Be summoned the bear
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    I know this should be ridiculously simple, but I'm not quite sure how to solve it. Could anyone help with the last bit?
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    (Original post by petrus123)
    OK, I'm just slightly confused about why @thebear has his contact force arrow pointing upwards (it says down on the lift), although I did that at first too, so maybe I missed something.
    the lift is pressing up against the passenger's feet. *
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    (Original post by the bear)
    the lift is pressing up against the passenger's feet. *
    Yes, but how do we know that that force is 294N?
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    (Original post by petrus123)
    Yes, but how do we know that that force is 294N?
    she is not sinking into the floor or lifting up from the floor. with respect to the floor she is in equilibrium. *
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    (Original post by the bear)
    she is not sinking into the floor or lifting up from the floor. with respect to the floor she is in equilibrium. *
    OK, thanks. But what about the 294N downward contact force? Sorry if I'm missing something.
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    (Original post by petrus123)
    OK, thanks. But what about the 294N downward contact force? Sorry if I'm missing something.
    since she is in equilibrium with respect to the floor the upwards and downwards forces must be equal. *
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    (Original post by the bear)
    since she is in equilibrium with respect to the floor the upwards and downwards forces must be equal. *
    So you could then use 9.8(40+m)-T-294=3(40+m)?
    If so, how do you find T?
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    (Original post by petrus123)
    So you could then use 9.8(40+m)-T-294=3(40+m)?
    If so, how do you find T?
    Anyone?
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    (Original post by petrus123)
    So you could then use 9.8(40+m)-T-294=3(40+m)?
    If so, how do you find T?
    if you mean T to be the force in the string then it is not relevant as there are two equal and opposite tensions in the string which cancel out. *
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    (Original post by the bear)
    if you mean T to be the force in the string then it is not relevant as there are two equal and opposite tensions in the string which cancel out. *
    OK, thanks!
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    (Original post by petrus123)
    So you could then use 9.8(40+m)-T-294=3(40+m)?
    If so, how do you find T?
    I solved the problem a while ago, but I came back to this because I'm curious as to why the 294N force is only included in the upwards direction in the diagram if it is actually in both directions (so that the girl is in equilibrium relative to the lift floor). Sorry if this is a stupid question and/or obvious.
 
 
 
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