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Dynamic FBD - Man and box sliding on an incline

  1. Aug 23, 2012 #1
    1. The problem statement, all variables and given/known data
    http://img59.imageshack.us/img59/493/capturemml.png [Broken]


    3. The attempt at a solution
    Here is my attempt. I assumed that both the man and the box would slide down the incline. (The box is the left diagram and the man is on the right)
    http://img35.imageshack.us/img35/7628/photorxq.jpg [Broken]
    The acceleration I get from this attempt tells me that they accelerate the other way. So do I have to change the directions of the friction forces to find the answer? I am not sure what to do next. Also how do I use the coefficient of static friction in this question?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 23, 2012 #2
    First find the maximum force can be applied(or pulled) by the man before he slips.
     
  4. Aug 23, 2012 #3
    Is that just 784sin(37)? Since it's limited only by the friction force.
     
  5. Aug 23, 2012 #4
    2 objects, man and box.

    3 forces acting on the man.
    The components of his weight, the force he pulls the box and static frictional force-Nμs.
    Pulling force is limited by static force.

    Check if his pulling force greater than the net downward force acting on the box.
     
    Last edited: Aug 23, 2012
  6. Aug 23, 2012 #5
    Okay so I was given a hint by TA to assume there is no rope and to find the accelerations of both the box and the man. Turns out, the acceleration of the box is greater than the man so they will accelerate down the slope as one system and there will be tension in the rope.

    As for what you said about the pulling force, do you mean the tension in the rope? I am confused :S

    Next I proceeded to solve the system of equations but I still get the wrong answer:
    http://img18.imageshack.us/img18/408/photo1feh.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  7. Aug 23, 2012 #6
    Just imagine there is a spring in the middle connected between the man and the box.
    If the spring is contracted, it will pull the man down and the box up.
    The spring is the muscles of the hands that being contracted to produce the force.

    From the question, the man is trying to pull.
    So he is applying a force and is transmitted to the box via the rope.
    This force must be equal and opposite, according to Newton's 3rd law.
     
  8. Aug 24, 2012 #7
    I'm not quite sure how to redraw the FBD with the muscles. Is my FBD correct first of all?
     
  9. Aug 24, 2012 #8
    a=g(Sinθ-Cosθμ)

    Since kinetic friction for the man less than the box, the rope will not be taut.

    Thus the box will accelerate independently.
     
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