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Tension Exercise on frictionless inclined plane

  1. Dec 19, 2016 #1
    1. The problem statement, all variables and given/known data
    A mass ##m_1## is attached to a second mass ##m_2 > m_1## by an Acme (massless, unstretchable) string. ##m_1## sits on a frictionless inclined plane at an angle ##\theta## with the horizontal; ##m_2## is hanging over the high end of the plane, suspended by the taut string from an Acme (frictionless, massless) pulley. At time ##t = 0## both masses are released from rest.
    a) Draw the force/free body diagram for this problem.
    b) Find the acceleration of the two masses.
    c) Find the tension T in the string.
    d) How fast are the two blocks moving when mass m2 has fallen a height H (assuming that m1
    hasn’t yet hit the pulley)?
    Immagine.png

    2. Relevant equations
    Tension
    Newton's Second Law

    3. The attempt at a solution
    First thing I drew the free body diagram this way:
    photo_2016-12-19_19-15-27.jpg
    Now I write the forces acting on both masses with Newton's Second Law, knowing that the accelerations of both masses are the same, so ##a = a_1 = a_2##.
    First mass:
    $$\begin{cases}
    F_{1 x} = m_1 a_x = T - m_1 g sin \theta \\
    F_{1 y} = m_1 a_y = N - m_1 g cos \theta = 0
    \end{cases}$$
    Second mass:
    $$\begin{cases}
    F_{2 x} = m_2 a_x = m_2 g sin \theta - T sin \theta \\
    F_{2 y} = m_2 a_y = m_2 g cos \theta - T cos \theta
    \end{cases}$$
    Now what I don't understand is how I find ##a_y##. Because the only way for ##m_1 a_y = 0## is that ##a_y = 0## but this would go in contrast with the fact that in the second mass ##a_y## is something, since the mass moves along the y-axis. Or it is ##0## because the mass moves at a constant speed? In that case I would have these:
    First mass:
    $$\begin{cases}
    a_x = \frac{T}{m_1} - g sin \theta \\
    N = m_1 g cos \theta
    \end{cases}$$
    Second mass:
    $$\begin{cases}
    a_x = g sin \theta - \frac{T}{m_2} sin \theta \\
    m_2 g cos \theta = T cos \theta
    \end{cases}$$
    With ##m_2 g cos \theta = T cos \theta## becoming ##T = m_2 g##.
    Is this way correct?
     
  2. jcsd
  3. Dec 19, 2016 #2
    I can't really see the details of your diagram too well, but why are you assuming that block 2 has an acceleration in the x-direction?
     
  4. Dec 19, 2016 #3

    nrqed

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    you do not have to use the same coordinate system for the two masses. You may use a vertical y axis for mass 2 in which case there are only forces along y for mass 2 and the equation is simple. The key point is to use indices to distinguish the accelerations of the two masses, you should use ##a_{x1}, a_{y1}, a_{x2},a_{y2}##. You should notice that then the values of ##a_{x1}## and ##a_{y2}## are not independent, they are related by a simple equation.
     
  5. Dec 19, 2016 #4
    It's because I used the same diagram for both masses, so I end up having the second mass moving on the x-axis too.

    But I've been taught that since the two masses are linked by a string, we should use one single coordinate system for both.
    Anyway, following your way, the first one remains the same and the second one becomes like this:
    $$\begin{cases}
    a_{x 2} = 0 \\
    a_{y 2} = g - \frac{T}{m_2}
    \end{cases}$$
    Being not independent because of the string, ##a_{x 1}## and ##a_{y 2}## are equal and so we have:
    $$\frac{T}{m_1} - g sin \theta = g - \frac{T}{m_2}$$
    With some passages I end up with:
    $$T = \frac{m_1 m_2}{m_1 + m_2} g (1 + sin \theta)$$
    Substituting this into one of the two I get the acceleration and substituting it into this equation ##v_f = \sqrt{2 a H}## I get the last point of the exercise.
    Am I right?
     
  6. Dec 19, 2016 #5

    nrqed

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    Science Advisor
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    Gold Member

    Looks good!
     
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