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Homework Help: Acceleration with two wooden crates

  1. Sep 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 24.0 kg and the larger bottom crate has a mass of m2 = 86.0 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is μs = 0.8 and the coefficient of kinetic friction between the two crates is μk = 0.63. A massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem).

    The tension is increased in the rope to 1122.0 N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?

    2. Relevant equations

    Fg = Fn = m*g
    Fk = μk * Fn
    Fnet,x = T-Fk

    3. The attempt at a solution

    Fg = 24*9.8 = 235.2N
    Fn = 235.2N
    Fk = .63 * Fn = 148.176N
    T = 1122N
    Fnet,x = 1122-148.176 = 973.824N
    acceleration in x-direction = 973.824/24 = 40.576 m/s^2 <- seems to be incorrect
  2. jcsd
  3. Sep 19, 2012 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    You are asked for the acceleration of the upper crate. This is the acceleration of the lower (86 kg) crate. Also, express your answer in proper significant figures.

  4. Sep 19, 2012 #3
    So how do you suggest that you find the upper crate's acceleration?
  5. Sep 19, 2012 #4
    To me the only force applied to the top block is just mgμk since it is sliding.
    Thus maximum acceleration equal to gμk.

    For no sliding, maximum acceleration is gμs.
  6. Sep 19, 2012 #5
    I just typed it in, & it's correct. Thanks for your help!
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