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!

Force and motion problem with two tensions and 3 masses and an ideal pulley

  1. Sep 24, 2011 #1
    1. The problem statement, all variables and given/known data

    Two clay pots (m1 = 6 kg, m2 = 3 kg) joined together by a light string rest on a table. The frictional coefficient between the pots and the table is 0.35. The pots are also joined to a suspended mass (m3 = 4kg) by a string of negligible mass passed over an ideal pulley as shown. Calculate the tensions in strings and the acceleration of the system when the suspended mass is released.
    2. Relevant equations

    Ffr=uN
    SumF=ma
    w=9.81*m


    3. The attempt at a solution

    sumFm3=T2 + Wm3, m*a=T2- (4kg*9.81), ------> 4a=T2-39.2 (eq1)
    sumFm1+m2 y= FNy+FGy, N-w=m*a , N-w=0, w=N, 9.81*9kg= N, 88.29=N
    Ffriction=uN, Ffr=0.35 * 88.29N, Ffr=30.9015
    sumFm1+m2 x= T2-Ffr, 9*a=T2-30.9015 (eq2.)

    Subtracting eq1 from eq 2 gives me 5a=8.2985, a =1.6597
    plugging a back in to equation 1 gives me 45.839 for T2
    sumFm2=T1-Ffr, m*a= T1- N, N=(.35 * 6 * 9.81), 6(1.6597)=T1-20.601, T1=30.6692 N

    The answers are: .64m/sˆ2 for a and T1=24N and T2=37N.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 24, 2011 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You have a sign problem in your equations. If the acceleration of m1 & m2 is 'a' to the right, then the acceleration of m3 is 'a' downward. Give those accelerations the proper signs.
     
  4. Sep 24, 2011 #3
    [PLAIN]http://dl.dropbox.com/u/14756133/physics%20two%20clay%20pots.pngSorry. [Broken] I accidentally typed plus. If you look at the following equation, though, I have it as a minus. I'm subtracting the force due to gravity from the force due to tension in the downward motion of mass 3. and my positive direction is right for m1 and m2. There's an image of the problem!
     
    Last edited by a moderator: May 5, 2017
  5. Sep 24, 2011 #4
    Wait. I just understood what you were talking about. So it should read -4a = T2-39.2! Thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Force and motion problem with two tensions and 3 masses and an ideal pulley
Loading...