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Thread: pulley system on block

  1. #1
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    pulley system on block

    pulley system on block-win_20170511_18_39_22_pro.jpgClick image for larger version. 

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    Im stuck on question 22 the first part
    i resolved the forces paralel and horizontal to the slope.
    i got the 2 equations T-√3 mg/2=ma using F=ma and kgm/2-T=kma
    substituting these into each other I get a=kg+
    √3 g /2k+2

    From here i can not see how to prove k>
    √3 if the weight of mass m is to accelerate towards P
    Ive done most of the work but i am missing something small. maybe some one can tell me?
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  2. #2
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    Re: pulley system on block

    Parentheses would help. Do you mean kg+(√3 g/2k)+ 2 or (kg+ √3g)/(2k+ 2)?
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  3. #3
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    Re: pulley system on block

    (kg+ √3g)/(2k+ 2) is what I mean. Sorry I should of used brackets
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  4. #4
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    Re: pulley system on block

    You calculated $a$ wrong.

    $T-\dfrac{mg\sqrt{3}}{2}=ma$
    $\dfrac{kmg}{2}-T = kma$

    So,
    $\dfrac{kmg}{2}-T = k\left(T-\dfrac{mg\sqrt{3}}{2}\right)$
    implies
    $T = \dfrac{kmg(1+\sqrt{3})}{2(k+1)}$

    Plugging that back in, you have:

    $\dfrac{k\cancel{m}g}{2}-\dfrac{k\cancel{m}g(1+\sqrt{3})}{2(k+1)} = \cancel{m}a$

    Thus, after simplification, you get:

    $a = \dfrac{kg(k-\sqrt{3})}{2(k+1)}$

    Now, the particle is moving towards P if $a>0$, so:

    $0 < \dfrac{kg(k-\sqrt{3})}{2(k+1)} \Longrightarrow \sqrt{3}<k$
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  5. #5
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    Re: pulley system on block

    Thanks for the answer.
    I'm not sure how you go from
    Kmg/2 -kmg(1+✓3)/2(k+1)=ma
    To a=kg(k-✓3)/2(k+1)
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  6. #6
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    Re: pulley system on block

    $T-\dfrac{\sqrt{3}}{2}mg = ma$

    $\dfrac{1}{2}kmg - T = kma$

    summing the two equations to eliminate $T$ yields ...

    $\dfrac{mg}{2} \left(k-\sqrt{3}\right) = ma(k+1)$

    solving for $a$ ...

    $a=\dfrac{g(k-\sqrt{3})}{2(k+1)}$

    Since the initial net force equations were formed under the assumption that $a > 0$, then $k > \sqrt{3}$
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