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Inclined pully table

  1. Nov 5, 2012 #1
    1. The problem statement, all variables and given/known data
    block1 2kg
    block2 2.5kg

    coefficient of static friction =.5
    coefficient of kinetic friction= .3


    A)how can you find the acceleration and tension?
    which coefficient do you use?
     

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    Last edited: Nov 5, 2012
  2. jcsd
  3. Nov 5, 2012 #2
    For the Tension, you calculate the amount of tension in each section of string and add them together. The tension in T1 is calculated by the force of friction with is
    [itex]F_{friction} = F_N\mu_k[/itex]

    For this, you use the coefficient of kinetic friction, as I am assuming that the force is being calculated when the blocks are already moving.

    So,

    [itex]T = T_1 + T_2[/itex]

    [itex]T_1 = F_N\mu_k[/itex]

    [itex]T_1 = 5.886 N[/itex]

    [itex]T_2 = mg[/itex]

    [itex]T_2 = 24.525 N[/itex]

    [itex]T = 30.411 N[/itex]

    For the acceleration you use Newton's Second:
    [itex]\Sigma\vec{F} = m\vec{a}[/itex]

    The sum of the forces in the system that we are looking at is contained within the tension of the wires, so

    [itex]30.411 N = (4.5 kg)a[/itex]

    [itex]a = 6.758 \frac{m}{s^2}[/itex]
     
  4. Nov 5, 2012 #3
    are you sure those are the right answers? did you look at the picture?
     
  5. Nov 6, 2012 #4

    haruspex

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    I disagree entirely with danielu3's analysis and answers.
    Since the (presumed) pulley is massless, the tension, T, is the same throughout the string.
    The first thing, strictly speaking, is to check that the masses will move. Will the tension overcome the static friction?
    Assuming so, the two masses each undergo the same acceleration, a.
    Write down the forces on each mass. What net force is required in each case to produce the acceleration a? What two equations does that give you?
     
  6. Nov 6, 2012 #5
    Yes, the tension in the string is constant. However, there are two components to the tension in the string; one is caused by the force of gravity acting on mass 2 and the other is the force of friction acting on mass 1. I may have issues where I used a positive instead of negative. Tension is one of the things that has always caused me issues though.
     
  7. Nov 6, 2012 #6

    haruspex

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    You can't add tensions like that. Consider two equal masses, m, hung over a light pulley by a string. The tension in the string is mg, not 2mg. Now make the masses unequal. What is the tension now?
     
  8. Nov 6, 2012 #7
    These pulley problems have always confused me for some reason. If there were two unequal masses on a pulley would the tension be

    [itex]T = g\frac{m_1+m_2}{2}[/itex]

    ?
     
  9. Nov 6, 2012 #8

    haruspex

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    Don't just guess, try to reason it out. Think about the acceleration.
     
  10. Nov 6, 2012 #9
    Well, based on what you said, I was thinking that the tension is based on the even distribution of the masses, which is why I'm thinking the average of the masses. But I have some feeling that it should be g(m2-m1) since the pulley makes the gravitational forces counteract each other. Am I at least thinking in the right direction right now?
     
  11. Nov 6, 2012 #10

    haruspex

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    Let the acceleration be a (down for the heavier, up for the lighter) and the tension be T.
    Draw a free body diagram for each mass and write out the F=ma equation for each.
     
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