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Simple exercise with tension

  1. Dec 18, 2016 #1
    1. The problem statement, all variables and given/known data
    A mass ##m_1## is attached to a second mass ##m_2## by an Acme (massless, unstretchable) string. ##m_1## sits on a frictionless table; ##m_2## is hanging over the ends of a table, suspended by the taut string from an Acme (frictionless, massless) pulley. At time ##t = 0## both masses are released.
    Find:
    a) The acceleration of the two masses.
    b) The tension T in the string.
    How fast are the two blocks moving when mass ##m_2## has fallen a height ##H## (assuming that ##m_1## hasn’t yet hit the pulley)?
    Immagine.png

    2. Relevant equations
    Newton's Second Law
    Tension

    3. The attempt at a solution
    So, the forces in the first mass are null on the y-axis while it's ##T## on the x-axis. (Putting right as the positive direction of the x-axis and up as the positive direction of the y-axis)
    So, we have:
    $$F_1 = m_1 a_1 = T$$
    $$a_1 = \frac{T}{m_1}$$
    And this is the first acceleration.
    The second mass has instead two forces acting on it on the y-axis. (nothing on the x-axis since it is just going down) These two forces are ##T## and ##P = m_2 g##. Since ##P## goes down, it is a negative force.
    So we will have:
    $$F_2 = m_2 a_2 = T - m_2 g$$
    $$a_2 = \frac{T}{m_2} - g$$
    And this is the second acceleration.
    Now the problem asks for the tension ##T##. Isn't this one just ##m_1 a_1##? So I basically ended up answering both at the same time, right?

    For the last question, I have to find the final velocity from the initial point until ##H##. This should be easy. Finding the final time ##t_f## and then substituting in the motion equation with ##\Delta x = H## we end up with this equation:
    $$v_f = \sqrt{2 g H}$$
    Is this way of doing correct?
     
  2. jcsd
  3. Dec 18, 2016 #2

    mfb

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    You can calculate a1, it should not appear in the final formula for the tension. In the same way, the tension should not appear in the formulas for the acceleration.
    There is one equation that you missed, it is related to the fixed string length.

    Your answer for the velocity is wrong, and I don't understand how you got it.
     
  4. Dec 18, 2016 #3
    You mean the fact that I should take the two masses as one? Like ##m = (m_1 + m_2)## since the string just attaches them together.

    Ops. I got a sign wrong and thought I did it. Because I wanted to arrive at this equation:
    $$v_f^2 - v_0^2 = 2 a \Delta x$$
    With ##v_0 = 0, \Delta x = H, a = -g##.
    But, obviously, I end up with ##v_f^2 = - 2 g H##, which is wrong.
     
  5. Dec 18, 2016 #4

    mfb

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    The total mass can be useful at some point, but that's not the point where you need to introduce it.

    Are the two accelerations independent?

    The acceleration is not -g (and not g either). You didn't calculate the acceleration yet.
    a and delta x should have the same sign, either both positive or both negative.
     
  6. Dec 18, 2016 #5
    If one of the two accelerates, the other one will too because they are attached by an Acme string. So no, they are not independent.

    Oh true. And ##\Delta x## would be negative since it's going down.
     
  7. Dec 18, 2016 #6

    mfb

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    Express this as formula and it will allow you to calculate everything.
     
  8. Dec 18, 2016 #7
    This means that the two accelerations are the same? So ##a = a_1 = a_2##?
     
  9. Dec 18, 2016 #8

    mfb

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    Right.
     
  10. Dec 18, 2016 #9
    So ##a = \frac{m_2}{m_1 + m_2} g##, ##T = \frac{m_1 m_2}{m_1 + m_2} g##, and ##v_f = \sqrt{\frac{m_2}{m_1 + m_2} 2 g H}##?
     
    Last edited: Dec 18, 2016
  11. Dec 18, 2016 #10
    Maybe double-check your result for T.
     
  12. Dec 18, 2016 #11
    Yeah, sorry, I forgot to write the ##g##. Now it's fixed.
     
  13. Dec 18, 2016 #12

    mfb

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    Correct.
     
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