Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Pully/block system on an incline

  1. Apr 29, 2010 #1
    1. The problem statement, all variables and given/known data
    A block of mass m = 10 Kg is attached to a cable that wraps around the pulley. The pulley is pivoted at the center, and the angle of incline is θ = 30°. The radius of the pulley is R = 0.5 m and its moment of inertia is I = 0.4 Kgm2. The block is released from rest and slides down the ramp, and the coefficient of friction is μk= 0.2 between the block and the ramp.
    (a) Find the acceleration of the block. a = _______________________
    (b) Find the tension in the cable. T = ______________________

    I have no idea where to start.
     
  2. jcsd
  3. Apr 29, 2010 #2

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    1. Draw a free body diagram for the mass on the incline.
    2. Use your FBD to apply Newton's Law to the mass.
    3. Draw a free body diagram for the pulley.
    4. Use this FBD to apply Newton's Second Law (for rotations) to the pulley. Note that the tangential acceleration of a point on the pulley's rim is the linear acceleration of the rope and, of course, the mass.
    5. Steps 2 and 4 will give two equations and two unknowns, the acceleration a and the tension T.
    6. Solve the system of two equations and two unknowns to find what the problem asks.
     
  4. Apr 30, 2010 #3
    For my equation, I got mgsin(theta)-I(a/r)-umgcos(theta)=ma however this does not give me the correct answer. Any ideas?
     
  5. Apr 30, 2010 #4

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Unless you show what answer you got and how, I have no ideas because I don't know what you did. The equation seems correct.
     
  6. Apr 30, 2010 #5
    For the block
    Fx=mgsin(theta)-T-Ff=ma
    Fy=-mgcos(theta)+N=0

    For the pully
    T=Ialpha
    T=I(a/r)

    10*9.8*sin30-.8a-.2*10*9.8*cos30=10a

    a=2.72 ?
     
  7. Apr 30, 2010 #6

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    In the above equations "T" stands for "Torque" not "Tension". You treat it as if it were tension. What is the torque τ if the tension is T and the lever arm is r?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook