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Blocks on incline

  1. Jul 14, 2005 #1
    The problem is:

    two identical blocks tied together with a string which passes over a pulley at the crest of the inclined planes, one of which makes an angle q1 = 18° to the horizontal, the other makes the complementary angle q2 = 72°.

    If there is no friction anywhere, with what acceleration do the blocks move?

    -----------------------------------------------------------------
    I think my problem is that I don't know what to do with the angles. I tried drawing the free body diagram for each block, and I end up with

    some of the forces for x component for block 1: T

    (same as above but for y component for block 1): N - mg = 0

    same goes for block 2.

    But what should I go from here? Or how should I approach this?
     
  2. jcsd
  3. Jul 14, 2005 #2

    siddharth

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    First of all you have to choose a convenient co-ordinate axis to make this problem simple.

    Which way does the block accelarate? Along the incline. So take the x-axis as along the inclined plane and the y-axis as perpendicular to the incline for each block.

    Next, draw the Free body force diagram for the blocks and write the components of the forces along the incline (ie x-axis) and perpendicular to the incline (ie, y-axis).This is an important step. You should be able to write the components of the all the forces acting on the block along the axis parallel to the inclined plane and perpendicular to the inclined plane before you proceed to the next step.

    Do you understand till now? Are you able to write the components of the forces acting along the incline and perpendicular to it? If so, can you post and show what equations you get?
     
  4. Jul 14, 2005 #3
    I really have no clue on this. But I guess something like this, with W = mg

    On block 1: x comp: Tension + W sin (q1) = ma

    y comp: N - W cos (q1) = 0
    On block 2: x comp: -Tension - W sin (q2) = ma

    y comp: N - W cos (q2) = 0
     
  5. Jul 14, 2005 #4
    Ok I revise:

    On block 1: x comp: Tension - W sin (q1) = ma

    y comp: N - W cos (q1) = 0
    On block 2: x comp: -Tension + W sin (q2) = ma

    y comp: N - W cos (q2) = 0

    Then I solve for a when I plug tension in for another, and I got acceleration correct as 3.15! Thanks siddharth. But now I have another question:

    Now suppose the coefficient of sliding friction between the blocks and planes is µ = 0.02. With what acceleration do the blocks move in this case?
     
  6. Jul 14, 2005 #5

    siddharth

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    The force of friction acting in the direction opposing the motion of each block will be
    [tex] \mu N [/tex].
    So in your free body diagram, add the force due to friction. Then proceed as you did above and you will get the answer.

    P.S: How did you get that symbol for mu without latex? What's the ascii value for that?
     
    Last edited: Jul 14, 2005
  7. Jul 14, 2005 #6
    Sid, there is no friction.
    Since the pulley and string are massless, the tension is uniform and is equal to the sum of the components along the slope viz. mg sin 18° and mg sin 72°. The acceleration will be in the direction of greater force. Force along the slope will be given by (mg sin 72° - mg sin 18°); by drawing free body diagrams. Now accelration will be force/mass i.e.
    g(sin 72°-sin 18°)

    P.S. How did you get the ° sign? I used copy+paste from the first post.
     
  8. Jul 14, 2005 #7
    I also have another problem that will be good practice.
    In the same situation, only take the angles to be 30° and 60°, let the mass of block on 30° be 3 times that of the one on 60°. Also, let the pulley be massive and the coefficients of static friction between 30° be half of that on the 60° side.
     
  9. Jul 14, 2005 #8

    siddharth

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    There is no friction in the first question. (Which huskydc solved). There is friction in the new question he asked.
     
  10. Jul 14, 2005 #9
    ok, little problem.

    here's what i've done so far, based on what I did on the first post.

    block 1
    sum of forces on x comp: T + f(1) - mg sin (q1) = ma
    y comp: N(1) = mg cos (q1)

    block 2
    x comp: mg sin(q2) - T - f(2) = ma
    y comp: N(2) = mg cos (q2)

    then I plug tension for another, solve for a, i get about 6.08, and its wrong, i'm not sure if i did the calculations wrong
     
  11. Jul 14, 2005 #10

    siddharth

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    For the first block, which way does friction act? It acts in the direction opposite to the direction of motion. While solving the problem, you assumed that the acceleration of the first block is up the incline. So how friction will act? What will it's sign be?
     
  12. Jul 14, 2005 #11
    Yes, i initially i assume block 1 is going up the incline,
    so friction goes down the incline for block 1
    and friction goes up the incline for block 2.

    and i solve for a and i got 3.03! whew...finally...thanks a lot
     
    Last edited: Jul 14, 2005
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