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Tenison/Force Problem

  1. Sep 13, 2008 #1
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    Two 100 kg boxes are dragged along a frictionless surface with a constant acceleration of 0.6 m/s2, as shown below. Each rope has a mass of 1 kg. Find the force F and the tension in the ropes at points A, B, and C.
    F = N
    TA = N
    TB = N
    TC = N


    What I have done:
    Equations:
    Weight = M * G
    Force = M * A

    The weight for each box is 981N.
    The force of gravity is offset by the normal force because it is on a surface, and there is no friction force.

    Can I say that the force of Tension A is = the weight of the first box which is 981N?

    And since Tension A and B are attached, Tension B = Tension A?

    Since there is a force pulling at C to accelerate the 2 boxes can I say
    Tension C = F?

    F = M * A
    Force of Tension A, B, C = 202kg * 0.6m/s^2
    Force of Tension C + 1962N = 121 kg m/s^2
    Then I would get a negative force for Tension C which is the Force...

    Can someone point out what I did wrong?
    Thanks so much!
     
  2. jcsd
  3. Sep 13, 2008 #2
    Ah! No magic massless ropes in this problem!

    Draw free body diagrams for each object in the problem showing the forces acting on that object, and apply F = mA, and Newton's 3rd law to move from object to object. Vertical forces in this problem are irrevelant (they are in balance).

    The FBD for the box on the left has only one horizontal force acting to the right. F = mA and so the force is 100 kg (0.6 m/s^2) = 60 N. Now the important part: the force acting on the left box is equal and opposite the force acting on the LEFT END of the rope between the left box and the right box, i.e., TA = 60N. Since the rope has mass and is also accelerating, the forces acting on the rope (draw its FBD) cannot be in balance, and so TB cannot be = TA.

    Can you see that TA > TB > TC > F ?
     
    Last edited: Sep 13, 2008
  4. Sep 13, 2008 #3
    Would Tension B be Force of Tension A + F = 101kg * 0.6(m/s^2) = 1.01N?
     
  5. Sep 13, 2008 #4
    Draw a FBD of the rope attached between points A and B.

    There's a 60 N force at A acting to the left, TB acting to the right, it has a mass of 1 kg, and its accelerating right. Sum of all F = mA, if to the right is taken to be the (+) direction:

    - 60 N + TB = (1 kg) (0.6 m/s^2) = 0.6 N

    TB = 60.6 N

    By the way, TA > TB > TC > F is incorrect; TA < TB < TC < F.
     
  6. Sep 13, 2008 #5
    ooh ok, I think I got the concept.

    I drew the FBD.

    I have 60N force at A and 60.6N force at B and Force C
    -60 N - 60.6 N + TC = 201kg * 0.6 m/s^2
    =-120.6 + TC = 120.6
    TC = 241.2N

    Is this correct?

    And F is 362.4N?
     
  7. Sep 13, 2008 #6
    Which object does this last FBD represent?

    If you take the mass on the right (only),

    - 60.6 + TC = ma = 100 (0.6) = 60

    TC = 120.6 N

    When you draw a FBD for an object or collection of objects, you must include only those forces that act externally on object/objects.
     
  8. Sep 14, 2008 #7
    So the last one for the Force is
    -120.6N + F = 101kg * 0.6
    F = 181.2N

    How come this is wrong?
     
  9. Sep 14, 2008 #8
    I figured it out, thanks for helping.

    -120.6N + F = 1kg * 0.6 m/s^2
    F = 121.2N
     
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