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!

Block over another block, with friction force

  1. Aug 17, 2010 #1
    1. The problem statement, all variables and given/known data

    An A block of mass 8 kg, is lying over an horizontal table, joint to a bucket of mass 1 kg, by a rope which passes by a pulley. The masses of the rope and the pulley are insignificant. Over the A block is lying a B block, of mass 2 kg. The coefficient of static friction, and kinetic friction, between the A and B blocks, and between the A block and the table, are static µ = 0.5, and kinetic µ = 0.4
    1) If the bucket is loaded with 10 kg of sand, calculate the acceleration of the blocks and the bucket.
    2) Show that for this load of sand, the B block does not slide over the A block.
    3) What is the maximum load of sand that you can put in the bucket, without making the B block slide over the A block?

    [PLAIN]http://a.imageshack.us/img825/7114/problemax.jpg [Broken]

    2. Relevant equations

    F = ma, Frk = µk N, Frs max = µs N

    3. The attempt at a solution

    1) First, making a free body diagram for the blocks (where m blocks is the mass of the 2 blocks, and m bucket is the mass of the bucket loaded with sand(:

    i) ΣFy = 0 = N - m blocks * g
    N = m blocks * g

    ii) ΣFx = T - Frk = m blocks * a (T = tension of the rope, Frk = kinetic friction force, between the table and the blocks)

    And for the bucket:

    ii) ΣFy = m bucket * g - T = m bucket a

    So, T = -m bucket a + m bucket * g

    Now, replacing T in ii)

    -m bucket * g + mbucket * g - Frk = m blocks * a

    -m bucket * g + mbucket * g - Frk - m blocks * a = 0

    Frk is µk N, and N = m blocks * g, so:

    -m bucket * a + m bucket * g - µk * m blocks * g - m blocks * a = 0

    a (-m bucket - m blocks) + g (m bucket - µk * m blocks) = 0

    a = (-g (m bucket - µk * m blocks)/(-m bucket - m blocks))

    Now, m bucket is 11 kg, µk is 0.4, m blocks is 10 kg, and g = 9.81 m/s², so

    a = (-9.81 m/s² (11 kg - 0.4 * 10 kg)/(-11 kg - 10kg)) = 3.27 m/s²

    2) Now, making a free body diagram for the B block (m b block is the mass of the B block):

    ΣFy = 0 = n - m B block * g
    n = m B block * g

    ΣFx = Fr s = m B block * a

    m B block is 2 kg, and a = 3.27 m/s², so

    Fr s = 2 kg * 3.27 m/s² = 6.54 N

    Now, the maximum static friction force that can move the B block, is given by µs * n, where n is the normal force between the B block and the A block, and is equal to m B block * g.

    So: Frs max = µs m B block * g = 0.5 * 2 kg * g = 9.81 N

    The maximum static friction force, 9.81 N, is greater than the static force that moves the B block, which is 6.54 N. So, the B block does not slide over the A block.

    3) Given that Frs = m B block * a, there is a maximum acceleration (a max) that the blocks can have, without making the B block slide over the A block.
    So: Frs max = m B block * a max
    µs * n = m B block * a max
    µs * m B block * g = m B block * a max
    µs * g = a max

    Now, using this acceleration, and making a free body diagram for the bucket:

    i) ΣFy = m bucket * g - T = m bucket * a max

    And for the blocks:

    ii) ΣFx = T - Frk = m blocks * a max
    iii) ΣFy = N - m blocks * g = 0

    T = m blocks * a max + Frk

    Replacing T in i)

    m bucket * g - m blocks * a max - Frk = m bucket * a max

    m bucket * g - m blocks * a max - Frk - m bucket * a max = 0

    m bucket * g - m blocks * a max - µk * m blocks * g - m bucket * a max = 0

    m bucket (g - a max) - m blocks * a max - µk * m blocks * g = 0

    m bucket = (+µk * m blocks * g + m blocks * a max) / (g - a max)

    m bucket = (+0.4 * 10 kg * g + 10 kg * a max) / (g - a max)

    m bucket = (+0.4 * 10 kg * g + 10 kg * µs * g) / (g - µs * g)

    m bucket = (+0.4 * 10 kg * g + 10 kg * 0.5 * g) / (g - 0.5 * g) = 18 kg

    Now, 18 kg is the maximum mass of the bucket allowed. The bucket alone has a mass of 1 kg, so, the maximum load of sand, would be 17 kg.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 18, 2010 #2

    collinsmark

    User Avatar
    Homework Helper
    Gold Member

    Hello Emil233,

    Welcome to Physics Forums!

    Your solution looks okay to me. :approve:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook