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Tention of boxes ona hill

  1. Oct 5, 2005 #1
    [​IMG] (for image of boxes)
    Two masses of 2.00 kg each, connected by a string, slide down a ramp making an angle of 41° with the horizontal. The coefficient of kinetic friction between m1 and the ramp is 0.26. The coefficient of kinetic friction between m2 and the ramp is 0.15. Find the magnitude of the acceleration of the masses.
    I found the acceleration
    4.91 m/s^2


    What is the tension in the string?
    is what im having trouble with.
     
  2. jcsd
  3. Oct 5, 2005 #2

    Pyrrhus

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    You need to show some work. What have you done?
     
  4. Oct 5, 2005 #3

    i calculated the accel by using
    fk=uk*m*g*cos(theta) on each of the two blocks. adding them then using m*g*sin(theta)-fk=m*a
    fk values:
    m1 = 3.84
    m2 = 2.218

    as for tension, i tried to do
    f=ma
    f=(2)*accel which doesnt work.

    m*g (since its kinda hanging)
    2*9.8
    2*9.8cos(41) (since its at angle and not straight down)
    2*9.8sin(41) (same as above but dif angle)
     
  5. Oct 5, 2005 #4

    Pyrrhus

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    On m1 block you got the tension force, the gravitational force, and the friction force acting!

    Applying Newton's 2nd Law

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = m \vec{a} [/tex]

    On m1

    [tex] \vec{T} + m_{1} \vec{g} + \vec{F}_{friction}_{1} = m_{1} \vec{a} [/tex]

    On m2

    [tex] \vec{T} + m_{2} \vec{g} + \vec{F}_{friction}_{2} = m_{2} \vec{a} [/tex]
     
  6. Oct 5, 2005 #5
    [tex] \vec{T} [/tex] = Tension?
     
  7. Oct 5, 2005 #6

    Pyrrhus

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    Basicly, you have the scalar equations for each body

    For the m1

    [tex] \sum F_{x} = T - \mu m_{1}g \cos \theta + m_{1} g \sin \theta = m_{1} a [/tex]

    For the m2

    [tex] \sum F_{x} = m_{2} g \sin \theta - T - \mu m_{2}g \cos \theta = m_{2} a [/tex]
     
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