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Newton's Second Law Equations

  1. Oct 12, 2007 #1
    1. The problem statement, all variables and given/known data

    After falling from rest at a height of 29.8 m, a 0.596 kg ball rebounds upward, reaching a height of 23.9 m. If the contact between ball and ground lasted 1.68 ms, what average force was exerted on the ball?

    2. Relevant equations

    3. The attempt at a solution

    so I figured out that the final velocity before the ball first hits the ground is 24.17 m/s, and the velocity when the ball begins to bounce back up its inital velocity is 21.64 m/s. If the a= vf-vi / t, that gives me 21.64 - 24.17 which is -2.53 / 0.00168 which equals -1505.95, which I think is totally unreasonable....where am I going wrong?
  2. jcsd
  3. Oct 12, 2007 #2

    Doc Al

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    Staff: Mentor

    Careful with signs. Downward velocity should be negative. That will give the average acceleration; you still have to calculate the force of ground on ball.
  4. Oct 12, 2007 #3
    ok thanks got it!
  5. Oct 12, 2007 #4
    All right new question :D

    The leg and cast in the figure below weigh 232 N (w1). (Assume w3 = 100 N and q = 39.2o). (See pic)

    So I know that Ft3 = Fg3 = 100N

    [tex]\Sigma[/tex]Fy2 = m2ay2
    Ftyx + Fty3 - Fg1 = 0
    Ft2 sin [tex]\alpha[/tex] + Ft3sin[tex]\vartheta[/tex] - Fg1 = 0
    Ft2 sin [tex]\alpha[/tex] +100sin39.2 - 232 N = 0
    Ft2 sin [tex]\alpha[/tex] = 168.797

    But now i'm stuck...I would like to solve for Fty2, because I have found that Ftx2 is 77.49, but I'm not sure how to solve fot Fty2 with the information I have

    Attached Files:

  6. Oct 12, 2007 #5

    Doc Al

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    Staff: Mentor

    It looks like you analyzed the vertical components, so now analyze the horizontal components.
  7. Oct 12, 2007 #6

    I would start a new thread for this, it will get much better traffic. Also, if you have an image host like Photobucket, I would upload though that. The PF image hosting has to be approved by a moderator first...which could take all day as they have a lot of them to deal with.


    Edit: But it looks like Doc Al was just waiting to prove me wrong:rolleyes:
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