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Forces (Newton's Laws)

  1. Oct 26, 2004 #1
    Hi.

    I have a physics question. I cannot get the right naswer.

    The question:
    Three boxes rest side-by-side on a smooth horizontal floor. Their masses are 1.10 kg, 3.45 kg, and 5.55 kg, with the 3.45 kg one in the center. A horizontal force of 20.0 N pushes on the 1.10 kg mass which pushes against the other two masses. What is thte contact force between the 3.45 kg and 5.55 kg boxes?

    My solution:
    I know that the force applied (20.0 N) is not constant throughout. However, acceleration of the full system is. Therefore I thought I could use Newton's 2nd Law. Fnet = ma.

    Fnet of mass 1.10 kg = ma
    rearranging gives a = 18.18 m/s^2.

    Fnet of mass 3.45 = ma = 62.7 N

    I thought this would be my contact force as this is the force that the 3.45 kg block pushes against the 5.55 kg box?/ Isn't this the answer? Can someone point me in the right direction or help me out.?? Thanks.
     
  2. jcsd
  3. Oct 26, 2004 #2
    Have you drawn the free body diagram of the boxes individually ?
    Have you applied newton's third law as well ?For example, two forces acts on the 1.10 kg block which are the 20.0 N force and the contact force exerted on it by the middle block.
    The 20.0 N force applied is constant of course or the question will get quite complicated.
     
  4. Oct 27, 2004 #3
    i assume friction is negligible.
    i would like to use newton's second law like this
    first, the 20N force accelerates the whole system:
    20N = (1.10 + 3.45 + 5.55)(kg)*a; => a_total = 1.99m/^s^2
    so, the acceleration of the whole system of boxes is 1.99m/s^2.
    Next, for the last box of weight 5.55kg to get an acceleration of 1.99m/s^2 you will need the contact force
    F = m*a = (5.55)(kg)*(1.99)(m/s^2) = 10.95N.
    This should be the force excerted on the last box of 5.55kg. Haven't had my morning coffee yet though, so i might be wrong :)

    /edit
    although the easiest way to solve this problem would be to examine the quota F/m. since the acceleration is the same for all parts of the system you could solve it like
    F_tot/m_tot = F_3/m_3 and thus get the force on the third box (F_3)
     
    Last edited: Oct 27, 2004
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