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Homework Help: Coefficient of kinetic friction, need today

  1. Apr 30, 2010 #1
    coefficient of kinetic friction, URGENT need today

    The system shown in the figure below has an acceleration of magnitude 0.51 m/s2, where m1 =
    4.30 kg and m2 = 6.00 kg. Assume that the coefficient of kinetic
    friction between block and incline is the same for both inclines.

    Link to figure:
    http://img9.imageshack.us/img9/3437/figureyh.jpg [Broken]
    m1 is box on the left, m2 is box on the right. both angles are 35 degrees.


    a) find the coefficient of kinetic friction

    Here's what I tried:
    coefficient of kinetic friction = (gsinΘ - a) / gcos Θ
    and got .63668

    when i input my answer in the system said "your answer differs from the correct answer by orders of magnitude."

    b) find the tension in the string
    T= m1a + m1gsinΘ
    = 26.36 N

    system said "your answer differs by 10% of correct answer..."


    Please help. I need this by tonight. Thank you!!!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 30, 2010 #2

    hotvette

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    Homework Helper

    Re: coefficient of kinetic friction, URGENT need today

    Let's start with the basics. What equation of motion (i.e. [tex]\Sigma[/tex]F = ma) did you use for each mass?
     
  4. Apr 30, 2010 #3
    Re: coefficient of kinetic friction, URGENT need today

    ^

    (a)
    ∑ Fx = ma --> mgsinΘ - fk = ma
    ∑ Fy = 0 --> n - mgcosΘ = 0
    n = mgcosΘ
    and fk = mukn

    so muk = (gsinΘ -a) / gcos Θ
     
  5. Apr 30, 2010 #4

    hotvette

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    Homework Helper

    Re: coefficient of kinetic friction, URGENT need today

    I presume that's for m2. But, you forgot the force of the tension. Also, you need a similiar equation for m1. You'll then have two equations in two unknowns (T and mu) that can be solved by substitution.
     
    Last edited: May 1, 2010
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