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Is there a net force as someone jumps?

  1. Apr 7, 2013 #1
    1. The problem statement, all variables and given/known data

    While someone is accelerating upwards but before the body leaves its feet, is there a net force? The person will eventually leave the ground.

    2. Relevant equations

    ƩFy = 0; or does ƩFy = ma?

    3. The attempt at a solution

    I pretty sure that the latter is true, this is so frustrating because I can see how either is true. ƩFy = 0 makes sense because the feet relative to the ground are not accelerating or even moving. But, I can also see how it should me ƩFy = ma because there is a "stationary" acceleration that the feet are exerting on the ground while the body accelerates up.
     
  2. jcsd
  3. Apr 7, 2013 #2
    While an object is accelerating, what is its net force equal to? (In variable terms)
    Also, for your problem is the human accelerating at a constant rate or not? This should help guide you to your answer.
     
  4. Apr 7, 2013 #3
    The answer is in the question

     
  5. Apr 7, 2013 #4
    If there's an acceleration then a net force is present, but sense the person hasn't left the ground , then it would make sense F=0 because the only forces present are normal and gravitational.

    Fy=N-mg=May. V(initial)=0m/s

    N=mg

    Fy=0


    Question: When you say body leaves feet do you mean feet leaving the ground?
     
    Last edited: Apr 7, 2013
  6. Apr 7, 2013 #5

    Doc Al

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    Staff: Mentor

    Do not assume that N = mg. That's only true when there's equilibrium.

    Even though the feet remain in contact with the ground, there is still a net force and thus the center of mass of the person is accelerating. That's what matters when applying ƩF = ma to a flexible body.
     
  7. Apr 7, 2013 #6
    Is that because the person is about to leave the ground in oppose to not not moving at all?
     
  8. Apr 7, 2013 #7

    Doc Al

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    Staff: Mentor

    Yes. The person is actively using their leg muscles to push against the ground. That's how you jump. The normal force is greater than their weight, thus there's a net force.

    If they just stood there, then ƩF = 0.
     
  9. Apr 7, 2013 #8
    But what troubles me is that the feet are not moving while they are still on the ground, so why do we need a net force. Won't the reacting forces of the feet and the ground equilize until the feet leave the ground? The fact that the weight is accelerating but not the feet seem to be messing me up.

    The feet seem to be in equilibrium with the ground the whole time, and I can understand that the body isn't because of its acceleration. So this is frustrating me because there seems to be two entirely different things happening, but, yet, I can't define them seperately.
     
  10. Apr 8, 2013 #9

    Doc Al

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    Staff: Mentor

    You need a net force on the body, not the feet.

    Until the feet begin to accelerate, the net force on them is zero. The forces on the feet include: the foot's weight, the upward force from the ground, the downward force from the rest of the body above the feet. As you begin to jump the upward force from the ground is not simply equal to the weight of the body.

    You can divide the body into any pieces you like and apply Newton's 2nd law to each piece. The body as a whole is accelerating--meaning: the body's center of mass is accelerating--so there must be a net external force acting on the body. And there is: You have to push down on the ground with a force greater than your weight in order to jump up. Which of course means that the ground is pushing on you with that force.
     
  11. Apr 8, 2013 #10
    Thanks, it makes sense now.
     
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