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Forces and Friction of sliding blocks

  1. Oct 16, 2005 #1
    I've been looking at this question for the last two hours, here's the question and what i've done so far

    The picture shows 3 blocks, A B and C, block C is ontop of block A. Block B is connected to block A by a rope, and is hanging over the edge.

    here it is,

    Blocks A and B have weights of 44N and 22N, respectively. (a) Determine the minimum weight of block C to keep A from sliding if s between A and the table is 0.20. (b) Block C suddenly is liffted off A. What is the acceleration of block A if k between A and the table is 0.15?


    This is what i've done

    Block A + C
    mg= 44N
    m = 44/g
    = 4.49 + X

    Fnet(y) = N - Mg = ma

    N = Mg
    =(4.49 + x)g

    Fnet(x) = T - f = ma
    T - f = (4.49 + x)a

    Block B
    mg = 22N
    m= 22/g
    = 2.24

    Fnet = T - mg = ma
    T - (2.24)(9.8) = 0
    T = 22N

    From this i assumed that when there is no acceleration the Tension in the rope is 22N, therefore the static friction when there is no movement in Block A+C is 22N

    Fmax = s Fn
    22N = 0.2 (mg)
    22N = 0.2 (4.49 + x)(9.8)
    22 = 8.8 + 1.96x
    x= 6.7 Kg Therefore the min. weight of C is 6.7kg times 9.8m/s^2 = 66N

    im not sure if I did this right, so I would appreciate if someone could let me know. Thanks:smile:
     
  2. jcsd
  3. Oct 17, 2005 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    It is correct.

    ehild
     
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