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Mechanics Question Regarding Newton's Third Law and Acceleration watch

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    Hey, I am quite stuck on this question. I've worked on it for ages but can't find out how they got their answer. Here is the question:

    A tape is wound round two smooth cylinders (in an tipped over S shape, 1 cylinder above the other; lower cylinder first then higher cylinder). The higher cylinder is fixed, but the lower cylinder sits in a loop formed by the tape. There is a 5kg block (attached to the bottom of the tape hanging from the higher cylinder) and the lower cylinder is 12 kg.

    a/ Explain why the upward acceleration of the 5 kg block is twice the downward acceleration of the lower cylinder

    b/ Find the acceleration of the 5kg block

    | |o| <--higher cylinder
    |o| |
    ......D <---5 Kg weight.

    The answer is 1.225m/s²

    I have no idea how they got that answer though. =S

    Any suggestions?

    I've let the acceleration of the weight be 2a and the acceleration of the cylinder be a. I also let T equal the tension in the tape.

    So I got these two equations after resolving vertically on the two components.

    5*9.8 - T = 5*2a
    2T - 12*9.8 = 12a

    However, solving this simultaneously gives me 0.3769m/s² which isn't the right answer.... What have I done wrong? =S

    Thank you in advance.
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    (Original post by Henry Lee)
    So I got these two equations after resolving vertically on the two components.

    5*9.8 - T = 5*2a
    2T - 12*9.8 = 12a
    Hummm....

    For the 5kg mass. Acceleration is UPWARDS at 2a.

    So, T - 5g = 5*2a


    For the 12 kg mass. Acceleration is DOWNWARDS at a.

    So, 12g - 2T = 12a.

    Now solve them.

    Edit: Actually, as both your equations were back to front, so to speak, you should have got the right answer, except with a negative sign in front, so you've made a slip in solving your simultaneous equations.
 
 
 
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