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Average Force using Newton's Laws of Motion

  1. Mar 8, 2012 #1
    1. The problem statement, all variables and given/known data

    A 89.6 kg man steps off a platform 4.52 m above the ground. He keeps his legs straight as he falls, but at the moment his feet touch the ground his knees begin to bend, and, treated as a particle, he moves an additional 0.88 m before coming to rest. Treating our rigid legged friend as a particle, what is the average force his feet exert on the ground while he slows down?
    Note: Assume the acceleration while he is slowing down is constant.

    Give your answer in Newtons to the nearest whole number.

    Hint: Draw a free body diagram to aid in seeing the forces.

    2. Relevant equations

    4 equations of constant acceleration and F=ma

    3. The attempt at a solution

    I was rather confused at the question's requirements to begin with, specifically with the terms 'average force' (I looked up average force to be only related to Δmomentum/Δtime, and this question is only with respect to "Newton's laws of motion") and "while he slows down". Any help would be greatly appreciated.
     
  2. jcsd
  3. Mar 8, 2012 #2

    Doc Al

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    Staff: Mentor

    What is the man's acceleration as he comes to rest? To find the 'average force', just pretend that the force exerted by the ground on him is constant. What other forces act on him?
     
  4. Mar 8, 2012 #3
    Well, since the question hinted to assume acceleration is constant while slowing down, I assumed a=g, but at rest a=0, would it not? From what I can understand, there are two forces on the man - gravity (excuse the poor terminology) and the normal force.
     
  5. Mar 8, 2012 #4

    Doc Al

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    Staff: Mentor

    You want his acceleration while his feet are touching the ground and he's coming to rest. Sure, before he hits the ground the acceleration = g and after he stops his acceleration is zero. But that's not relevant here. Hint: Use kinematics to solve for the acceleration. What's his speed just as he first touches the ground?
    Good!
     
  6. Mar 8, 2012 #5
    Alright, so his speed just before he touches the ground is 9.41 m/s.
    Then, his acceleration (or deceleration?) is 50.3 m/s^2.
    Now, given that I have - okay, I think I got it. The answer was supposed to be something in the order of 5388 N and I got 5386 N, rounding it up.
    Thank you so very much.
    The final equation stood as
    F-mg=ma where a=50.3 m/s^2 and solving for F gave me the aforementioned answer.
     
  7. Mar 8, 2012 #6

    Doc Al

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    Staff: Mentor

    Good!
     
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