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Tensions and inclines question

  1. Oct 2, 2012 #1
    1. The problem statement, all variables and given/known data



    blocks A B AND C are connected as show. block A and B have the same mass "m" and the coefficient of kinetic friction between each block is uk. block C descends with constant velocity. use 30 degrees for the angle of the incline.

    given m, uk
    find

    a.
    tension in rope between blocks A AND B

    b.

    mass of block C

    c.

    if the rope connected to block A is cut, determine the accel of block C


    3. The attempt at a solution

    http://i.imgur.com/KJsro.png

    FAx = T1 - uk*N1 = 0
    FAy = N1 - mg = 0

    FBx = T2 - T1 - uk*N2 = 0
    FBy = N - mg*cos(30) = 0

    FCy = mc*g - T2 = 0


    a.

    N1 = mg -> T1 = uk*mg


    b.

    N2 = mg*cos(30) - > T2 = uk*mg + uk*mg*cos(30)

    mc*g - ( uk*mg + uk*mg*cos(30) ) = 0

    mc = uk*m + uk*m*sqrt(3)/2

    I think these are right(confirm pls)


    c is what confuses me but my attempt:

    FBx = T2 - uk*N2 - g*sin(30) = 0
    FBy = N2 = mg*cos(30)

    FCy = mc*g - T2 = mc*a


    T2 = uk*mg*cos(30) + g*sin(30)

    a = (mc * g - T2) / mc * g


    (uk*m + uk*m*sqrt(3)/2 - uk*m*sqrt(3)/2 + 1/2 ) / uk*m + uk*m*sqrt(3)/2 (g should cancel out)

    is this all right? thanks for any help.
     
  2. jcsd
  3. Oct 3, 2012 #2
    a. Correct t1=mgμk
    b. t2 is balanced by 3 forces, t1, friction and the weight of the mass.
    c. t3 is balance by weight of mass c.
     
  4. Oct 3, 2012 #3

    PhanthomJay

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    Note omission in red. Also, block B is accelerating, correct? Why have you set the sum of forces equal to 0??
    yes
    yes
    please correct
    where did the g come from?
    Make appropriate corrections.
     
  5. Oct 4, 2012 #4
    T1 = uk*mg

    T2 - T1 - Ff - mgsin(30) = 0

    T2 = uk*mg + uk*mg*cos(30) + mg*sin(30)

    mcg - T2 = 0

    mc = uk*m + uk*m*sqrt(3)/2 + m/2


    c)

    FBx = T2 - uk*mg*cos(30) - mg*sin(30) = ma

    FCy = mc*g - T2 = mc*a

    T2 = mc*a + mc*g


    mc*a + mc*g - uk*mg*cos(30) - mg*sin(30) = ma

    mc*g - uk*mg*cos(30) - mg*sin(30) = ma - mc*a

    mc*g - uk*mg*cos(30) - mg*sin(30) = a(m - mc)


    a = (mc*g - uk*mg*sqrt(3)/2 - mg/2) / (m - mc)

    a = (uk*mg + uk*mg*sqrt(3)/2 + mg/2 - uk*mg*sqrt(3)/2 - mg/2) /( m - mc)


    a = (uk*mg) /( m - uk*m - uk*m*sqrt(3)/2 - m/2)
     
  6. Oct 4, 2012 #5

    PhanthomJay

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    OK
    You have a signage error here. Check algebra.
    this should be OK once you make the correction on your algebra/signage error noted above when determining T2.
    These last 2 steps are OK where you make the substitution for mc
     
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