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Homework Help: Is it possible to solve for kinetic friction without given theta?

  1. Oct 15, 2012 #1
    1. The problem statement, all variables and given/known data
    A block of mass " lying on a ramp tilted at an
    angle  with the horizontal is connected to
    another block of mass # by a light string passing
    over a frictionless pulley as shown. If the block
    slides up the ramp with a constant velocity ),
    what is the coefficient of kinetic friction $%?

    2. Relevant equations

    3. The attempt at a solution
    ƩF1xx = m1a
    Ft - μkFn = m1a

    ƩF1y = m1a1y
    Fn - mgcosθ = m1a1y

    I can figure out the Ft from the other mass. But where I am stuck is I have 3 unknowns and 2 equations. I cant find a way to solve for μk when I dont have θ.

    Thanks for the help
  2. jcsd
  3. Oct 15, 2012 #2
    I cannot see the second mass in the equations.
  4. Oct 15, 2012 #3
    I cannot see the second mass in the equations.
  5. Oct 15, 2012 #4
    ƩF2y = m2a
    FT - m2g = m2a

    I solve this for Ft and plug that into the tension force on mass1
  6. Oct 15, 2012 #5
    So what are the three unknowns you have?
  7. Oct 15, 2012 #6
    μk,Fn, and θ
  8. Oct 15, 2012 #7
    It is not clear from the initial statement of the problem what is given and what is not. The initial statement sounds to me as if the angle were on an equal footing with masses which I assume you consider given. Could you copy the problem exactly?
  9. Oct 15, 2012 #8
    What I have in the original post is the whole question but I have attached a picture of the problem for clarity.

    Attached Files:

  10. Oct 15, 2012 #9
    I think you can consider the angle as a given.
  11. Oct 15, 2012 #10
    The assignment says that the block moves with a constant velocity - Newton's 1st law says that the end result Force affecting the mass is 0 if the body is still or is moving with a constant velocity and without changing direction - acceleration is 0.

    This is the part that I am doubting at the moment -is it correct to assume the mass m2 is pulling the mass m1 by its weight? If so, the weight is m2g - and as the block on the ramp isn t accelerating, the mass on the other side also isn't accelerating.

    So the resulting force to the block on the ramp is m2g.
    The frictional force or force of friction or ..:/ is Fh = μm1gcosθ

    The ramp directed gravitational pull is F = m1g sinθ

    The resulting force is m2g + (-m1g sinθ - μm1gcosθ) <- these are vectors - i don't know how to denote vectors :(
    the weight of m2 has to come level with both the friction and the gravitational pull.
    So after a bit of simplification:
    (m2g)² + (-m1g(sinθ + μcosθ))² = 0

    E: I meant you must know the value of all 3 variables - even the assignment given agrees with me.
    Last edited: Oct 15, 2012
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