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Find The acceleration (Newton's Laws of Motion problem).

  1. Oct 28, 2012 #1
    1. The problem statement, all variables and given/known data


    4.3.jpg
    Fig. 4.3

    A 5.0-kg block and a 4.0-kg block are connected by a 0.6-kg rod. The links between the blocks and the rod are denoted by A and B. A force F is applied to the upper block.


    In Fig. 4.3, the force in link B is 40 N. The acceleration of the blocks and rod assembly, including direction, is closest to:

    1) zero
    2) 2.4 m/s2, downward
    3) 1.2 m/s2, downward
    4) 2.4 m/s2, upward
    5) 1.2 m/s2, upward

    2. Relevant equations

    we have to use Newton's Laws of Motion to find the answer.

    3. The attempt at a solution

    I tried to figure out the acceleration for each of 5.0-kg block, 4.0-kg block and 0.6-kg rod using these equations:

    for the 5.0-kg block \ ƩF=ma ==> F - m1g - Fa = m1a ==> F-Fa = m1(a+g)

    for 4.0-kg block \ ƩF=ma ==> Fb - m2g = m2a ==> Fb = m2(a+g)

    for the 0.6-kg rod \ Fa + Fb ==> (F(m1+m2+m3) - m1F)/(m1+m2+m3) + (m2F)/(m1+m2+m3)


    and i could do that for the 4.0-kg block and 0.6-kg rod but i couldn't with the 5.0-kg block because F is unknown for me!


    - I find Fa and Fb this way :

    First ƩF for the hole system is ƩF=ma ==> F-(m1+m2+m3)g = (m1+m2+m3)a
    so, a = F/(m1+m2+m3) -g

    for m1 ƩF=ma ==> F-Fa-m1g = m1a
    so, F-Fa = m(a+g) = m1F/(m1+m2+m3) ==> Fa = F(m1+m2+m3)-m1F/(m1+m2+m3) ==> Fa = 0.48N

    and Fb = 40N (Given in the question)



    m1 = mass of 5.0-kg block
    m2 = mass of 0.6-kg rod
    m3 = mass of 4.0-kg block

    Fa = Force of A
    Fb = Force of B
     
    Last edited: Oct 28, 2012
  2. jcsd
  3. Oct 28, 2012 #2
    Read the rules: you must attempt a solution before we can help
     
  4. Oct 28, 2012 #3
    sorry, i just put my attempt!
     
    Last edited: Oct 28, 2012
  5. Oct 28, 2012 #4
    Fa and Fb aren't the only forces acting on the rod.
     
  6. Oct 28, 2012 #5
    If you only want to answer the question, though, you only have to look at block B. The acceleration of the whole system is the same for each piece. You have an equation with only the acceleration as the unknown, so just solve for a there.
     
  7. Oct 29, 2012 #6
    I'm not sure what advantage i take from this information specifically!
     
  8. Oct 29, 2012 #7
    Here is my attempt:


    ƩF (for the block B) equals Fb - m2g = m2a ==> Fb = m2(a+g) ==> 40 = 4(a+9.8) ==> a=0.2 m/s2.
    So, it closest to zero ( if i substitute g with value 10, a will be 0 ).



    Is that correct ? or i should use the simple equation a = F/m ?
     
    Last edited: Oct 29, 2012
  9. Oct 29, 2012 #8
    Yes. I'm guessing this is for an exam? It's perfectly reasonable that they might have g=10m/s^2, especially if you're not allowed to use a calculator.
    a=F/m is only true when F is the net force.
     
  10. Oct 29, 2012 #9
    oh! That is way easier than i thought :)
    it is kind of online Assignment.


    Thanks.
     
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