Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Acceleration and tension for two blocks connected to frictionless pulley

  1. Oct 7, 2008 #1
    I have tried to get this problem many different ways but still dont get it.

    A 1.00 kg aluminum block and a 8.00 kg copper block are connected by a light string over a frictionless pulley. The two blocks are allowed to move on a fixed steel block wedge (of angle θ = 40.0°) as shown in Figure P4.63.

    View attachment p4-63.bmp
    Figure P4.63

    Making use of Table 4.2, determine the following.
    (a) the acceleration of the two blocks

    Your answer differs from the correct answer by 10% to 100%. m/s2
    (b) the tension in the string N


    Coefficients of Frictiona µs µk
    Steel on steel 0.74 0.57
    Aluminum on steel 0.61 0.47
    Copper on steel 0.53 0.36
    Rubber on concrete 1.0 0.8
    Wood on wood 0.25-0.5 0.2
    Glass on glass 0.94 0.4
    Waxed wood on wet snow 0.14 0.1
    Waxed wood on dry snow - 0.04
    Metal on metal (lubricated) 0.15 0.06
    Ice on ice 0.1 0.03
    Teflon on Teflon 0.04 0.04
    Synovial joints in humans 0.01 0.003
    a All values are approximate.
     
  2. jcsd
  3. Oct 8, 2008 #2
    I need help with this too... I keep getting a negative acceleration and I'm pretty sure I did everything correctly.

    Cu block = 8 kg
    Al block = 1 kg
    theta = 30
    ∑Fx (Al) = ma = T - Fs ---> T = Fs + ma
    ∑Fx (Cu) = ma = sin 30 x mg - T - Fs

    Substitution yields...
    a (m + m) = sin 30 x mg - Fs (Al) - Fs (Cu)
    (m + m ) = mass of Cu + mass of Al

    Fs (Cu) = cos 30.0 x mg x 0.53 = 36.0 N
    Fs (Al) = mg x 0.61 = 5.98 N

    when I solve for a I get -0.31 m/s^2, which is impossible
    so I thought a would be zero, but according to the website thats also wrong
     
    Last edited: Oct 8, 2008
  4. Oct 8, 2008 #3
    anyone else that can help us with this
     
  5. Oct 9, 2008 #4
    I tried to do this problem like how you did, but ended up in the same place.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook