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Anyone know how to solve these problems (DYNAMICS)?

  1. May 5, 2009 #1
    1. The problem statement, all variables and given/known data
    http://img117.imageshack.us/img117/2513/55200915619am.jpg [Broken]

    http://img217.imageshack.us/img217/2817/55200915545am.jpg [Broken]

    For the first one I got 1100 N for F(A) and 981 N for F(G)... how do I do the rest?

    2. Relevant equations

    F(T)-F(K) = MA to get F(A)

    3. The attempt at a solution

    F(K) = 500 N
    F(A) = 600 + 500 = 1100 N
    F(N) = mg = 981 N?

    My attempt was for the first problem... have no idea about second =(.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 5, 2009 #2

    LowlyPion

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    Homework Helper

    Welcome to PF.

    Consider the force on the sledge.

    Frictional resistance is 500N. The mass is 100kg

    Since the sledge is accelerating at 1 m/s2 the force to accelerate that is only 100 N.

    For the rest of it you have action and reaction. The horse must be pushing forward against friction. The sledge must be pulling backward with the 500 N of friction. And of course the horse is accelerating its 500 kg.
     
  4. May 5, 2009 #3

    LowlyPion

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    In 2 consider the forces needed to overcome friction on the 2 surfaces. Those forces diminish the force available to accelerate the 10 kg box.
     
  5. May 5, 2009 #4
    so how i go about solving these =[
     
  6. May 5, 2009 #5
    For the second questions:
    there is friction between the 2 bodies and with the surface, thus the force 45N is overcoming 2 forces, the friction between the bodies and the surface.:
    what is the magnitude of each friction force?
    then use Newton's second law to determine acceleration.
    Good Luck

    oh, and a bout the tension in the string:
    1 of the action-reaction forces is friction, so on the upper body there is friction that is directed to the right((which the 10KG tries to take the KG to a walk,and he would've succeeded if there wasn't a string)) which is equal to the tension(equilibrium).
     
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