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    • PS Reviewer
    • Thread Starter

    PS Reviewer
    How do i do this????
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    a) Well the box is moving at constant speed so there is no acceleration --> no resultant force. So the horizontal component of P is exactly balanced by the frictional force caused by the block. So you resolve P through the 20 degrees to find the horizontal component of P,  P_{horizontal} , then set this equal to  \mu mg , the frictional resistance.

    b) The box will now accelerate because the tension P has increased and will overcome the frictional resistance, so use  ma = \underbrace{P_{Horizontal} - \mu mg}_{Resultant Force} with the new value of P = 150.

    Hope this helps
    • PS Reviewer
    • Thread Starter

    PS Reviewer
    It kind of does but I still cannot do part a). The answer is meant to be 110N or so.

    I've got it now - Thanks a lot
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Updated: January 10, 2008

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