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Finding the Tension in terms of weight and the angle of incline

  1. Feb 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Two blocks, each with weight ω, are held in place by a frictionless incline.
    In terms of ω and the angle θ of the incline, calculate the tension of the rope connecting the two blocks.

    2. Relevant equations

    ƩF=ma

    3. The attempt at a solution

    Since it is on a tilted incline, with angle θ, I made the x and y axis tilted for the free body diagram. This means that ω, the weight, can be broken down into components ωx, which is parallel to the tension, and ωy, perpendicular to tension. In this case, I assumed that all forces acting in the x-direction (parallel to the tension) were ωx and T. I then used the equation ƩF=max and plugged in values and solved for T. First, I put "ωsinθ" in place of ωx, making the equation ωsinθ+T=ma. Solving for T, I get T=ma-ωsinθ. Is this correct? It looks a little odd to me.
     
  2. jcsd
  3. Feb 16, 2013 #2

    tms

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    I think something is missing in the statement of the problem. How can a frictionless incline hold anything in place?
     
  4. Feb 17, 2013 #3
    No, they are held in place by tension.
     
  5. Feb 17, 2013 #4
    Since the blocks are held in place there is no acceleration so ma=0⇔ƩFx=0. And remember, there are two blocks.
     
    Last edited: Feb 17, 2013
  6. Feb 17, 2013 #5

    tms

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    I can't picture the setup.
     
  7. Feb 17, 2013 #6
    Sorry guys, this is the setup. I have looked at it a few more times and still get the same thing. Am I doing something wrong?
     

    Attached Files:

  8. Feb 17, 2013 #7

    tms

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    Since the blocks are not moving, what is the acceleration?
     
  9. Feb 17, 2013 #8
    Your answer above is wrong.

    It may help to tell you that the tension in that rope would only depend on the mass of B (and not A). Remember the tension in string is equal to the force with which gravity is pulling the mass down the slope.
     
    Last edited: Feb 17, 2013
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