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Connected masses, friction enabled

  1. Oct 16, 2005 #1
    Two blocks connected by a rope of negligible mass are being dragged by a horizontal force F. F = 68.0N, m1 = 12.0kg, m2 = 18.0kg, and coefficient of kinetic friction between each block and the surface is 0.100. I need the tension of the rope and the acceleration of the system.

    [m1]---[m2]--->F

    I found the acceleration by subtracting the force of friction for both blocks from the force being applied, then dividing that by the total mass, getting 1.29m/s^2. I'm not sure how to find the tension. I tried a few things, and I'm not getting the correct answer of 27.2N. Any advice on how to go in the right direction?
     
  2. jcsd
  3. Oct 16, 2005 #2
    Start with Newton's second law. F=ma

    List all the forces acting upon the object. (A free body diagram will help with this)
     
  4. Oct 16, 2005 #3
    Thanks for the response. For m1, I have fk1 with a horizontal negative force, T with a horizontal positive force, n1 and m1g opposite and equal. For m2 I have fk2 and T horizontal negative, F horizontal positive, and n2 and m2g opposite and equal. The tension for both of those is equal... and I'm still not sure what to do next. Another hint, please?
     
  5. Oct 16, 2005 #4

    Doc Al

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    Staff: Mentor

    Apply Newton's 2nd law to each mass separately. You'll get two equations and two unknowns. Solve!
     
  6. Oct 16, 2005 #5
    For m1, T - fk1 = SigmaF1. SigmaF1 = m1 * a.
    T - 11.8N = 12.0kg * 1.29m/s^2, T = 27.28 or 27.3N which is almost right... did I do that correctly and I just rounded too early, or is that just a coincidence?
     
  7. Oct 16, 2005 #6

    Doc Al

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    Staff: Mentor

    You did it correctly.
     
  8. Sep 20, 2008 #7
    How do you find those forces of friction for each block? I know we have the coefficient, and the total force....
     
  9. Nov 21, 2008 #8
    He found it by using Fk1 = Uk(mu sub 'k')N with N = mg (but in the opposite direction)
    Force of Friction of Block 1 = (.1)(12.0 kg)(9.8 m/s²) = 11.76 N
    Force of Friction of Block 2 = (.1)(18.0 kg)(9.8 m/s²) = 17.64 N
     
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