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Math Help - Mechanics 1, string and tension Problem, Home work due

  1. #1
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    Mechanics 1, string and tension Problem, Home work due

    Hi, This is silver doing 12th year, I have clearly anyidea regarding inextensible string, could you please explain and help me in this problem. Thank you well in Advance.

    A mass of 2kg lies on a rough plane whichis inclined at 30 degree to the horizontal. One end of a light inextensible string is attached to this mass and the string passes up the line of greatest slope and over a smooth pulley fixed at the top of the slope; a freely suspended mass of 5 kg is attached to its other end. The system is released from rest; as the 2kg mass accelerates up the slope, it experiences a constant resistance of motion of 14 Ndown the slope due to the rough nature of the surface. Find the tension in the string?


    MY solution: force = mg= 2x9.8= 19.6N at 30 degree

    Force parallel to the plane= 196x sin 30 = 98N
    A system of mass 7kg is acclerated. Force applied is 14N down
    Object B weigh 9.8x5= 49N
    Net force = 98N-49N-14N
    = 35 N


    I have no idea. I would appreciate if you take time to explain this.
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  2. #2
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    Re: Mechanics 1, string and tension Problem, Home work due

    forces acting on the hanging mass ...

    Mg - T = Ma

    forces acting on the incline's mass ...

    T - mg\sin{\theta} - f = ma

    combining the equations ...

    Mg - mg\sin{\theta} - f = (M+m)a

    solving for the acceleration ...

    \frac{Mg - mg\sin{\theta} - f}{M+m} = a

    calculate the acceleration, then use the first equation to determine the tension in the string, T

    ... next time, please post physics questions in the "Math Topics" forum.
    Last edited by skeeter; October 13th 2012 at 11:47 AM.
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  3. #3
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    Re: Mechanics 1, string and tension Problem, Home work due

    Thank you Skeeter. Thank you once agin for the speedy response. Sorry, wil post on Math tics
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    Re: Mechanics 1, string and tension Problem, Home work due

    Hanging mass
    T-Mg=Ma
    5x 9.8 = 49N
    T- 49N= Ma-------------> (1)
    Force acting on incline mass
    T-mg sin theta -f= Ma
    -2x9.8x sin 30 -f = Ma---------> (2)
    Mg- mgsin theta -f= (M+m)a
    19.6- 9-8/7=a
    1.4= a

    Mg-T= Ma
    49-T = 14
    -T= -35
    T= 35N


    Please verify and correct me . Thank you very much for your timely help.
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  5. #5
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    Re: Mechanics 1, string and tension Problem, Home work due

    incorrect ...

    Mg = 5g

    mg\sin(30^\circ) = 2g \cdot \frac{1}{2} = g

    f = 14


    \frac{5g - g - 14}{5+2} = a
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  6. #6
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    Re: Mechanics 1, string and tension Problem, Home work due

    So it is 4g- 14= 7a
    4x 9.8-14= 7a
    39.2-14= 7a
    3.6= a

    T-Mg= Ma
    T- 5g = Ma
    T- 5x9.8= 3.6 x5
    T- 49= 18
    T= 18+49
    = 67 N

    thank you very much for explaining. Still not sure about the answer. Once again thank you very much for your time and help.
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  7. #7
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    Re: Mechanics 1, string and tension Problem, Home work due

    no again ...

    the net force equation was ...

    Mg - T = Ma

    not

    T - Mg = Ma
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  8. #8
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    Re: Mechanics 1, string and tension Problem, Home work due

    So 5g -T= Ma
    5x9.8-T= 3.6x5
    49-T = 18
    -T= -31
    T= 31

    Thank you very much for your immense help.
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