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Stuck on Tension

  1. May 13, 2013 #1
    1. The problem statement, all variables and given/known data

    Two blocks m1 and m2 are at rest, each on a frictionless plane and are connected by a massless rope passing over a frictionless pulley. The pully is at the top of where the two planes connect to make an angle of 90 degrees. the plane with m1 is to the left and makes an angle of 30 degree.

    Find the acceleration and tension where m1 = 25kg, m2 = 50kg, gravity = 10 m/s2

    2. Relevant equations

    2||N F = MA

    3. The attempt at a solution

    for acceleration I found the y component of the force of each block

    block 1 F1 = m1*a*sinθ1
    F = (25)*(10)*sin(30)
    F = 125 N

    block 2 F2 = m2*a*sinθ2
    F = (50)*(10)*sin(60); angle 2 is 60 because it's a 30-60-90 triangle.
    F = 433 N

    then I solve for acceleration as the sum of force over the sum of masses

    a = (F1 + F2)/(m1 + m2) = 4.1 m/s2

    Here we can say that the acceleration of both blocks is 4.1 m/s2

    Also, tension is the same as well. (this is the part I don't get, how is tension the same)

    I solve for tension as

    T1 - m1*g*sinθ1 = m1*a
    -T2 + m2*g*cosθ2 = m2*a \\ I don't understand why they use cos, shouldn't it be sin for the y

    As 3||N every force has an equal and opposite reacting force.

    T1 = T2 // I don't understand how T1 = T2 like this.

    T1 = m1*a + m1*g*sinθ1
    T2 = -(m2*a - m2*g*cosθ2)

    T1 = (25)*(4.1) + (25)*(10)*(sin(30))
    T1 = 102.5 + 125 = 227.5 N
    Thus T2 = 227.5 N as well because T1 = T2
     
  2. jcsd
  3. May 14, 2013 #2

    haruspex

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    to what - the vertical or the horizontal?
    what are you choosing as the y direction? what is a?
    Don't these forces oppose in some way?
    To avoid confusing yourself, introduce an unknown for the tension. Consider the free body diagram and equations for each mass separately.
     
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