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Contact Force

  1. Feb 8, 2008 #1
    Two blocks are in contact on a frictionless table. A horizontal force F is applied to M2, as shown. If M1 = 1.51kg, M2 = 3.96kg, and F = 4.55N, find the size of the contact force between the two blocks.

    So first I found the acceleration (F=ma)

    4.55/(1.51+3.96) = 0.83 = acceleration

    Im suck from here
  2. jcsd
  3. Feb 8, 2008 #2
    Try drawing a free-body diagram of M1, to make this easier. What forces are acting on it? You also know its acceleration, so what forces are causing this acceleration?
  4. Feb 8, 2008 #3
    the horizontal force that was applied to M2...
  5. Feb 8, 2008 #4
    If the force on M2 was also applied to M1 like you said, M1's acceleration would not be the 0.83 m/s^2 you calculated. So, a different force must be causing this acceleration...
  6. Feb 8, 2008 #5
    would gravity play a factor?
    Last edited: Feb 8, 2008
  7. Feb 8, 2008 #6
    No, let me explain this differently. Now, draw the free body diagram of M2. There's the 4.55 N force acting on it, but there is another horizontal force acting in the opposite direction, the contact force. Newton's Third Law states there are equal and opposite forces, so this applies to the contact force. So, the contact force is acting on M1, giving it its acceleration, if I'm thinking of this problem right (force acting on M2, and M2 is pushing on M1). Does this help? I deeply apologize if I overcomplicated the problem.
    Last edited: Feb 8, 2008
  8. Feb 8, 2008 #7
    So the contact force is the force acting in the opposite direction?

    Which would be .83*1.51 =1.25 N is that the contact force?

    or do I do 4.55 N - 1.51 N = 3.3 N

    and ya that's the problem.. m2 is pushing m1
  9. Feb 8, 2008 #8
    The contact force is merely the mass of M1 times its acceleration. Remember, the contact force is the force giving M1 its acceleration. Do you see why?
  10. Feb 8, 2008 #9
    im sorry terbum, this is just so confusing for me.. probably b/c it's getting late..

    so how do i find the acceleration of M1 if it isn't 0.83m/s^2
  11. Feb 8, 2008 #10
    The acceleration you solved for was the acceleration for the system. So, BOTH M1 and M2 have an acceleration of 0.83 m/s^2.

    Yeah, I had a tough time with this topic when I learned it, too. It does get a little confusing. I guess if you want to learn it, keep doing problems. Practice never hurts.
  12. Feb 8, 2008 #11

    Shooting Star

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    The accn of M1 is 0.83 m/s^2, as you have calculated. A force of 4.55 N is pushing on M2, which in turn is pushing on M1, and both masses are moving together, and so their accn must be same.

    The last step remaining is to find out what force acting on M1 produces an accn of 0.83 m/s^2? You know mass of M1, and accn 'a'. Apply Newton's 2nd law.
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