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Homework Help: Newtons Law Involving Friction

  1. Oct 1, 2005 #1
    I have a block conected to another block by a string in an inclined surface, with a frictionless pulley. Both blocks are the same weight so i already figure the acceleration of the object that its being pulled up.

    What I don't know is how can I calculate the minimum kinetic friction that will keep the system from accelerating.

    I tried setting the acceleration to zero for this formula:

    [Tex] a= 1\ frac{m2*g-m1*g*sin\theta - \mu*m1*g*cos\theta} {m1+m2} [\tex]

    after that i really don't get the answer. Please any sugestions.
  2. jcsd
  3. Oct 1, 2005 #2
    what no help??
  4. Oct 1, 2005 #3


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    Your description of the set up is a bit unclear, but if it's like what I have in the attachment, then I get the same expression as yourself for the acceleration.

    If the system is not accelerating, then a = 0.
    Set the numerator in your expression to zero and solve for [tex]\mu[/tex].

    Attached Files:

    • isu.jpg
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  5. Oct 1, 2005 #4
    Complete Problem

    thats exactly what i did but it doesnt match the answer of the book.

    this is the problem:

    A block (mass m1) lying on a frictionless inclined plane is connected to a mass m2 by a massless cord passing over a pulley, as shown in Fig. 4-53.
    (a) Determine a formula for the acceleration of the system in terms of m1, m2, , and g.
    (b) What conditions apply to masses m1 and m2 for the acceleration to be in one direction (say, m1 down the plane), or in the opposite direction?

    What smallest value of ┬Ák will keep the system of the previous problem from accelerating?
  6. Oct 1, 2005 #5


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    was your answer for the formula for acceleration, part (a), the same as in the book ?
  7. Oct 1, 2005 #6
    if i try to see the smallest possible value of [tex] \mu [/tex] kinetic i get twp wrong answers "1.46" and "-0.61". It doesnt match my webassign page.
  8. Oct 1, 2005 #7


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    For [tex]\mu _k[/tex] I got

    [tex]\mu = \left( \frac{m_2}{m_1}\right)sec\theta - tan\theta[/tex]

    bit i can't see how I would get two answers ??
  9. Oct 1, 2005 #8
    Fermet I got it. You were great help. I was wrong because my algebra procedure was wrong. I will be asking you more questions some of these days. For now I want to thank you for your help. Have a good one.

    Best Regards,

  10. Oct 1, 2005 #9
    The answer is 0.63719 [tex] \approx [/tex] 0.64.
  11. Oct 1, 2005 #10


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