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Homework Help: What is the force exerted by one crate on the other?

  1. Feb 23, 2010 #1
    1. The problem statement, all variables and given/known data

    Two crates rests on the floor, one next to the other. A man pushes with a horizontal force of 400 N on the larger (50.0 kg) crate, which, in turn, pushes the other (25.0 kg) so that both slide as he walks along. Each crate experiences a friction force opposing the motion equal to 40% of its weight.

    1.Determine the acceleration of the crates.
    Answer in m/s^2

    2.What is the force exerted by one crate on the other?
    Answer in Newtons


    2. Relevant equations and 3. The attempt at a solution

    1) total friction force Ff = 40 x (50+25) x 9.8/100 = 294 N
    =>F(net) = F(applied) - Ff
    =>F(net) = 400 - 294 = 106 N
    =>By F = ma
    =>106 = 75 x a
    =>a = 106/75 = 1.41 m/s^2 => CORRECT

    2)F(net) = 106 N => INCORRECT. Why? Please help!!!
     
  2. jcsd
  3. Feb 24, 2010 #2
    the answer is 200N. i did i have take one crate and i calculated the force.according to the force exerted by one is exactly opposite to athor
     
  4. Feb 24, 2010 #3
    the answer is 200N. i did i have take one crate and i calculated the force.according to the force exerted by one is exactly opposite to anthor
     
  5. Feb 24, 2010 #4
    the answer is 200N. i did i have take one crate and i calculated the force.according to the force exerted by one is exactly opposite to anothor
     
  6. Feb 24, 2010 #5
    The force exerted on one crate to the other is:

    A = 50 kg; B = 25 kg

    F = ma
    400 = 75.0 kg x a
    a = 5.3 m/s^2

    F + F(BA) = m1 x a
    F + F(BA) = 50 kg x 5.3 m/s^2
    F + F(BA) = 265 N

    F + F(BA) = 265 - F
    F(BA) = 265 - 400
    F + F(BA) = -135 N (But put in 135 because that's the correct answer)
     
  7. Feb 24, 2010 #6
    Thanks! I got it!
     
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