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Homework Help: Frictional Forces question

  1. Jul 8, 2011 #1
    Hello, I'm new here. I've browsed before and saw that this is a very helpful community, so I came here for assistance.

    1. The problem statement, all variables and given/known data

    Blocks A, B, and C are placed as in the figure and connected by ropes of negligible mass. Both A and B weigh 27.5N each, and the coefficient of kinetic friction between each block and the surface is 0.40. Block C descends with constant velocity.

    Image attached:

    2. Relevant equations

    What is the weight of block C?

    3. The attempt at a solution

    I've found the first part of this problem, which is the tension holding A and B, which is 11N. Part C has me confused. Here's my attempt:
    Block B

    NB=27.5Nsin(36.9) = 16.5115562

    fK= µNB = .40(16.5115562)= 6.604622479

    T2-T1-fK- 27.5cos(36.9)= 0

    T2-T1-6.604622479- 21.99132811= 0

    T1= 11N

    T2-11-6.604622479- 21.99132811= 0

    T2= 17.59595059N.

    Block C

    ∑Fx= T2-mCg = 0
    T2 = mCg = 17.59595059N

    Is this correct?

    Thanks in advanced.
    Last edited: Jul 8, 2011
  2. jcsd
  3. Jul 8, 2011 #2
    Since the system is not accelerating, [itex]\sum[/itex]F=0.


    F2=.4*mg*sin(36.9 degrees) = .4*27.5*sin(36.9) N


    F1 + F2 = mg (m = mass of block 3)

    .4*27.5(1+sin(36.9)) = mg (the weight of block 3)
  4. Jul 8, 2011 #3
    O M = opposite to the movement

    Block a:
    F in a = F friction in a = N.u = 27,5u O M

    Block b:

    F in b = Pb sin36,9° and Pb cos36,9°u O M = 27,5.3/5 and 27,5.4/5.u

    F friction = 27,5 . 3,6/5 > 27,5.3/5 so they have same direction

    27,5.6,6/5 = C = 5,5.6,6 = 36,3N
  5. Jul 10, 2011 #4
    Thanks, guys.
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