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Newton Force Problem

  1. Feb 9, 2007 #1
    1. The problem statement, all variables and given/known data
    Two blocks attached by a string are at rest on an inclined surface. The Lower Block has a mass of m1=0.2Kg and a coefficient of static friction u=0.4. The upper block has a mass m2=0.1Kg and a coefficient of static friction u=0.6. The angle theta is slowly increased.

    Mass 2 is above Mass 1 in the picture!!!!
    Also theta is above the horizontal!!!


    2. Relevant equations
    Here are the questions it is asking for.
    a)At what angle theta do the blocks begin to slide?
    b)What is the tension in the string just before sliding begins?


    I have a few questions myself about the setup.
    Does the mass of the blocks have any revelence to the acceleration down the ramp? Or will there be tension in the string solely for the fact that the coefficent of friction for the upper block is higher making it move slower. I doubt a physics book would want do a problem where the rope is slack and the blocks collide. At least not in Newtons chapter about forces.

    3. The attempt at a solution

    I made the sum of forces for both i and j vectors for both blocks. I solved for the general equation of force normal and friction force for both blocks. Note that they are both the same equations.

    Now I have theta, acceleration, and tension as unknown variables. What do I do next?


    By the way thanks in advance for any help. And how do you guys actually use greek and mathmatical symbols in your text?
     
  2. jcsd
  3. Feb 9, 2007 #2

    PhanthomJay

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    Which block wants to move first if there were no string? It might help to visua;lize what happens when the string is aded to the system. Forget about the acceleration variable....at the instant just before sliding, they are still at rest.
     
  4. Feb 9, 2007 #3
    I worked that out but I don't see how to get any farther. With 3 unknowns I'm not sure where to start. Maybe if my teacher would teach. :)
     
  5. Feb 9, 2007 #4

    PhanthomJay

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    Why don't you show us the equations you are using. Did you take a free body diagram of each block. Don't worry about formula appearance for now ......instead of [tex] mgsin\theta [/tex] you can write mgsin theta....etc
     
  6. Feb 10, 2007 #5
    F=ma and f=un

    For mass 1:

    i: T + (F-friction) - M1*g*sin Theta = M1*a
    j: (F-Normal) - M1*g*cos Theta = 0

    for Mass 2:

    i: (F-Friction) - T - M2*g*sin theta = M2*a

    j: (F-Normal) - m2*g*cos Theta = 0

    Note that the 2 in m2 would be a subscipt

    Then I substituted F-normal into the friction equation for both masses. That is all I have so far.

    thanks
     
  7. Feb 10, 2007 #6

    Hootenanny

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    Your equations look good to me. The next step is to write an equation for the normal force for each case, i.e. (F-Normal) = .... and then substitute this into your (i) expressions for friction. Do you follow?
     
  8. Feb 10, 2007 #7
    Then I substituted F-normal into the friction equation for both masses. That is all I have so far.

    I don;t know if you missed that but I did solve for the general equations of Force normal and Force of friction for both mass one and two.
     
  9. Feb 10, 2007 #8

    Hootenanny

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    Sorry, didn't see that line. Okay, what can you tell me about the tension?
     
  10. Feb 10, 2007 #9
    I solved for T for mass 1 and then plugged it into the second mass equations T. So T cancelled out and I'm left with Theta and Acceleration. I'm pretty sure you can't just set acceleration to zero because there is an infinite amount of possibilities then. Still stuck. Sorry for all the questions and thanks even more.

    i got a= (4.578)cosTheta + (3.27)sinTheta
     
  11. Feb 10, 2007 #10

    PhanthomJay

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    The reason why you might be stuck is that you are insisting that there must be an acceleration! The blocks are at rest at the start...and they are still at rest at the instant they are about to slide. No velocity change implies no acceleration. Try solving those equations again. There is only one solution.
     
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