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Two blocks on a plane with a pulley

  1. Oct 22, 2015 #1
    1. The problem statement, all variables and given/known data
    A block of mass m1 = 3.70 kg on a frictionless inclined plane of angle 30.0° is connected by a cord over a massless, frictionless pulley to a second block of mass m2= 2.30 kg hanging vertically. What are (a) the magnitude of the acceleration of each block and (b) the direction of the acceleration of m2? (c) What is the tension in the cord?

    upload_2015-10-21_22-40-44.png


    2. Relevant equations
    F=ma

    3. The attempt at a solution
    ?temp_hash=0b2545f925e8c27f7444b5711d612232.jpg
    I set the coordinate system according to the diagram above.

    System: m1
    Ft+Fg=m1a
    Ft-m1gsinθ=m1a

    System: m2
    Ft+Fg=m2a
    Ft-m2g=m2a
    Ft=m2g+m2a

    Then I plug in both of them.
    m2g+m2a+m1gsinθ=m1a
    m2a-m1a=-m1gsinθ-m2g
    Then I sub in the values
    a=3.15?

    I dont know what I did wrong. The correct answer is 0.735.
     

    Attached Files:

  2. jcsd
  3. Oct 22, 2015 #2
    You've mad a mistake with the signs. Consider the cord as your coordinate and define its positive direction from m1 to m2. Now redo your force balances with all forces in each direction with the same sign.
     
    Last edited: Oct 22, 2015
  4. Oct 22, 2015 #3
    Why are all forces positive? That doesnt sound right.
     
  5. Oct 22, 2015 #4
    Sorry, maybe I didn't express myself clearly. Of course you have to distinguish the direction of the forces by using the correct signs. After defining the positive direction, you know all forces pointing in this direction have positive signs and the rest has negative signs.
     
  6. Oct 22, 2015 #5
    So how would you define it? Which part of my signs is wrong?
     
  7. Oct 22, 2015 #6
    It's up to you, but as already proposed in my previous post: m1 → m2: positive
     
  8. Oct 22, 2015 #7
    So if to the right of M1, which is on the x axis to the right was positive then the M2 going up would be positive too?
     
  9. Oct 22, 2015 #8
    If your coordinate s along the chord is positive from m1 → m2, then for

    m1: sgn(Ft) = + 1 & sgn(Fg) = - 1 ... according to the x-axis you defined
    m2: sgn(Ft) = - 1 & sgn(Fg) = + 1 ... as s is positive along the chord in direction m1 → m2
     
  10. Oct 22, 2015 #9
    Why is Fg positive for m2 Fg? Can you draw it out on the whiteboard which sign should be where?
     
  11. Oct 22, 2015 #10
    Sorry, whiteboard doesn't work on my computer. To see it just draw an arrow parallel to the chord direction pointing from m1 to the pulley and a second one after the pulley pointing in direction of m2. These are the positive directions of your cordinate s. Now compare the direction of each force at m1 with the first arrow: If the force points in the same direction the sign is +. Same for m2 and the second arrow.
     
    Last edited: Oct 22, 2015
  12. Oct 22, 2015 #11

    haruspex

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    In your equation for m1, you have effectively defined the acceleration a as positive up the slope. That means it must be positive downwards for m2, but you wrote your m2 equation as though it is still positive up.
     
  13. Oct 22, 2015 #12
    Nice post
     
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