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Calculating mass given maximum tension of a rope

  1. Nov 3, 2015 #1
    If a string can handle a max tension of 85 N what is the largest mass that could be lifted with this string if it is lifted with an acceleration of 2.0 m/s^2 up on the moon?

    Let up be positive in value.

    Ft = - 85 N, a = 2.0 m/s^2, Fg = -1.6 m/s^2

    2. Relevant equations

    F = ma -> Fg = mg -> Ft = ma + mg -> Ft = m ( a + g)



    3. The attempt at a solution

    When I do the problem, I get a negative mass ...

    Ft = m (a + g) -> -85 N = m ( 2 m/s^2 - 1.6 m/s^2 ) -> m = - 212.5 kg

    Do you see the mistake?
     
    Last edited: Nov 3, 2015
  2. jcsd
  3. Nov 3, 2015 #2

    haruspex

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    By what logic do you add these accelerations? Is it accelerating at rate a faster than gravity?
    Draw a free body diagram. What are the forces acting on the body? What is the resulting acceleration?
     
  4. Nov 3, 2015 #3
    Oh sorry. In the downward direction there is Fg and Ft.

    By what logic do you add these accelerations? Is it accelerating at rate a faster than gravity?

    Hm, I guess I added them because they were in two different directions (the lifting of the string was upward and the gravity as downward, but I realize that this is probably a mistake right?)

    Oh, and by the way the orignal equation I did was Ft = mg + ma so that lead to Ft = m ( a + g)
     
  5. Nov 3, 2015 #4

    haruspex

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    I understand that seems intuitively right, but it fails when you plug in the signed values.
    Remember, the basic equation is ##\Sigma F_i = ma##. Plug Ft and mg into that.
     
  6. Nov 3, 2015 #5
    I will try to do what you said

    ∑F = Fg + Ft = ma

    And Fg = mg

    so (mg) + Ft = ma

    m (-1.6 m/s^2) - 85 N = m (2 m/s^2)

    -1.6 m - 85 = 2 m

    m = 23.611111111 kg -> 24 kg

    how does that look?
     
  7. Nov 3, 2015 #6

    haruspex

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    You changed 1.6 to 1.5, but otherwise that looks right.
     
  8. Nov 3, 2015 #7
    oooooh sorry :( but anyway thanks a million!
     
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