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Homework Help: Moving blocks on an incline - relative motion

  1. Sep 14, 2010 #1

    JJBladester

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    1. The problem statement, all variables and given/known data
    Block A starts moving downward to the left at a constant acceleration = 80mm/s2. At the same time block B moves along block A to the right at an acceleration of 120mm/s2 relative to block A.

    a) Determine the acceleration of block B. Answer: aB=52.5m/s2
    b) Determine the velocity of B at t = 3.0 seconds. Answer: vB=157.5m/s2

    relative%20motion%20blocks.png

    Note: I think the professor didn't convert between mm's and m's when he obtained the answers. Looks like he wrote the answers in meters by accident.

    2. Relevant equations

    aB/A = acceleration of B with respect to A
    aB = acceleration of B (w.r.t. fixed origin)
    aA = acceleration of A (w.r.t. fixed origin)

    v(t) = v0 + at

    3. The attempt at a solution
    [tex]\left|a_{A}\right|=80mm/s^{2}=.08m/s^{2}[/tex]
    [tex]a_{A_{x}}=-a_{A}cos(\theta)\approx-.0274m/s^{2}\hat{i}[/tex]
    [tex]a_{A_{y}}=-a_{A}sin(\theta)\approx-.0752m/s^{2}\hat{j}[/tex]

    [tex]a_{B/A}=.12m/s^{2}\hat{i}+0\hat{j}[/tex]

    [tex]a_{B}=a_{A}+a_{B/A}=(-.0274+.12)\hat{i}+(-.0752+0)\hat{j}[/tex]
    [tex]\left|a_{B}\right|=\sqrt{(-.0274+.12)^{2}+(-.0752+0)^{2}}\approx.119m/s^{2}[/tex]

    My answer of .119m/s2 is not the correct answer of 52.5m/ss. Where did I go wrong?

    I know the answer to the second part (b) of the question is a simple application of v(t) = v0 + at.
     
    Last edited: Sep 14, 2010
  2. jcsd
  3. Sep 14, 2010 #2

    collinsmark

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    Homework Helper
    Gold Member

    Given the way the problem statement was written above, I think your answer is correct.
     
    Last edited: Sep 14, 2010
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