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Math Help - Mechanics String and PUlley problems

  1. #1
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    Mechanics String and PUlley problems

    A mass of 10 kg lies on a smooth plane which is inclined at THETA to the horizontal. The mass is 5m from the top measured along the plane. One end ofa light inextensible string is attached to this mass; the string passes up the line of greatest slope and over a smooth pulley fixed at the top of the slope. The other end is attached to a freely suspended mass of 15 kg. This 15 kg mass is 4m above the floor. The system is released from rest and the string first goes slack 1 3/7 s later. Find the value of THETA?

    Please Help me
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    Re: Mechanics String and PUlley problems

    Quote Originally Posted by mysur View Post
    A mass of 10 kg lies on a smooth plane which is inclined at THETA to the horizontal. The mass is 5m from the top measured along the plane. One end ofa light inextensible string is attached to this mass; the string passes up the line of greatest slope and over a smooth pulley fixed at the top of the slope. The other end is attached to a freely suspended mass of 15 kg. This 15 kg mass is 4m above the floor. The system is released from rest and the string first goes slack 1 3/7 s later. Find the value of THETA?

    Please Help me
    we've done this before in another problem, no?

    start by making a sketch, then by writing individual scalar net force equations for the hanging mass and the mass on the incline.
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    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Mechanics String and PUlley problems

    Let me know if you get:

    \theta =30.067{}^{\circ}
    Last edited by MaxJasper; October 15th 2012 at 11:49 AM.
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    Re: Mechanics String and PUlley problems

    Thank you I did used the mg- mgsin theta = ma ( two masses)
    15g - 10g sintheta= 25a
    s= 4m a=? t= 1 3/7
    s= 1/2 at ^2
    4= 1/2 x a x (10/7)^2
    a= 3.92 m/s^2

    using 15g- 10gsin theta = 25a
    where a = 3.92
    15g- 10x9.8 sin theta= 25x3.92
    we get sin theta= 15g-98/10g = 49/98= .5
    theta= 30 degrees

    can you please verify, Thank you very much
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    Re: Mechanics String and PUlley problems

    since it is theta to the horizontal it is cos theta is it
    I m confused
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    Re: Mechanics String and PUlley problems

    Could you please verify and let me know where I went wrong?
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    Senior Member MaxJasper's Avatar
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    Re: Mechanics String and PUlley problems

    30 deg looks OK
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    MHF Contributor ebaines's Avatar
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    Re: Mechanics String and PUlley problems

    Quote Originally Posted by mysur View Post
    Thank you I did used the mg- mgsin theta = ma ( two masses)
    ... can you please verify, Thank you very much
    Yes, this is correct.

    Quote Originally Posted by mysur View Post
    since it is theta to the horizontal it is cos theta is it
    I m confused
    The force due to the weight of the 10Kg block that is opposing the falling of the 15 Kg weight is mg \sin \theta just as you have it. I sometimes find it easiest to think through whether it's sine or cosine by imaging what happens if theta changes - for example if the slope was 0 degrees to the horizontal (perfectly level) then the weight of the 10Kg block doesn't present any force at all, and if theta = 90 then the full weight of the 10Kg mass would come into play. So it's pretty clear that you want to use sine theta in this problem.
    Last edited by ebaines; October 15th 2012 at 12:24 PM.
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  9. #9
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    Re: Mechanics String and PUlley problems

    Thank you for clairfying that
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