1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Two blocks and a pully

  1. Aug 10, 2005 #1
    A block of mass m1 = 1 kg rests on a table with which it has a coefficient of friction µ = 0.77. A string attached to the block passes over a pulley to a block of mass m3 = 3 kg. The pulley is a uniform disk of mass m2 = 0.7 kg and radius 15 cm. As the mass m3 falls, the string does not slip on the pulley.

    With what acceleration does the mass m3 fall?

    i did the following:

    [A]sum of Force on mass 1: T(1) - f = m(1) a
    thus, T(1) = m(1)a + mu*m(1)g

    sum of torque on mass 2: RT(3) = I *alpha

    [C]sum of forces on mass 3: m(3)a = m(3)g - T(3)
    thus, T(3) = m(3)g - m(3)a

    and since alpha = a/R and I = .5 m(2) R^2

    I plug in equatiion for T(3) into , and got something like this:

    R m(3) g
    ------------------- =a
    .5 m(2) R + R m(3)

    and plug in numbers, didn't work out...any hints?
     

    Attached Files:

  2. jcsd
  3. Aug 10, 2005 #2

    Fermat

    User Avatar
    Homework Helper

    The problem is in .

    The torque on m2 results from the difference in tensions in the rope on either side of the pulley.
    That torque should be (T3 - T1)R
     
  4. Aug 10, 2005 #3
    i got it, thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Two blocks and a pully
  1. Pully and block (Replies: 4)

Loading...