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Math Help - Dynamics Problem

  1. #1
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    Question Dynamics Problem


    A mass of 25kg is being pulled to the right by an applied force of 85 N. A mass of 8 kg is located between the applied force and the 25 kg mass. The coefficient of friction is 0.2
    What is the friction force of each mass?
    (25*9.8)*0.2= 49 N

    (8*9.8)*0.2=15.7 N

    So for the 25 kg mass the friction force is 49 N. The friction force for the 8 kg mass is 15.7 N


    What is the acceleration of each system?
    F=ma<br />

    Fa-Ff=ma


    Fa=85

    Ff=49+15.7=64.7

    m=25+8=33<br />

    85-64.7=20.3<br />

    20.3=33a<br />

    \frac{20.3}{33}=a=0.615<br />

    So the acceleration is 0.615 m/s^2


    What is the tension in the connecting rope?

    The answer is supposed to be 64.4N however I do not know how to get this answer. Help!
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  2. #2
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    Quote Originally Posted by iamanoobatmath View Post





    (25*9.8)*0.2= 49 N

    (8*9.8)*0.2=15.7 N

    So for the 25 kg mass the friction force is 49 N. The friction force for the 8 kg mass is 15.7 N




    F=ma<br />

    Fa-Ff=ma


    Fa=85

    Ff=49+15.7=64.7

    m=25+8=33<br />

    85-64.7=20.3<br />

    20.3=33a<br />

    \frac{20.3}{33}=a=0.615<br />

    So the acceleration is 0.615 m/s^2




    The answer is supposed to be 64.4N however I do not know how to get this answer. Help!
    Well done showing your work (which I haven't actually checked but it gives the correct answer for tension - see below - so is probably OK).

    Block B:

    F_{net} = T - \mu R = T - (0.20)(25g) = T - 49 .... (1)

    F_{net} = ma = (25)(0.615) = 15.375 .... (2)

    Equate the above equations and easily solve for T.
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