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Accleration of 2blocks

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data



    2. Relevant equations
    a= F/m

    F= mgsinθ-Ffr

    Ffr = [itex]\mu[/itex]mgcosθ

    d= 1/2gt^2

    3. The attempt at a solution

    a = (m1gsinθ+m2gsinθ)-([itex]\mu[/itex]m1gcosθ+[itex]\mu[/itex]m2gcosθ)/(m1+m2)

    a = 3.72 m/s^2

    t = 1.03s

    I need someone to verify if this is correct, no anwser key was provided

    d=
     

    Attached Files:

  2. jcsd
  3. Oct 12, 2011 #2

    ehild

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    Take care with the formulation. The first term has to be divided by (m1+m2), too. Your result is not correct.

    ehild
     
    Last edited: Oct 12, 2011
  4. Oct 12, 2011 #3
    a=[gsinθ-(μ1m1gcosθ+μm2gcosθ)]
    a=.298m/s^2

    t=3.66

    how about now?
     
  5. Oct 12, 2011 #4

    ehild

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    Still wrong. Did you divide the second term with m1+m2?

    ehild
     
  6. Oct 12, 2011 #5
    a=[gsinθ-g(μ1cosθ+μcosθ)]
    a = 2.35 m/s^2

    t = 1.30s

    ck
     
  7. Oct 12, 2011 #6

    ehild

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    Correct the red part.

    ehild
     
  8. Oct 12, 2011 #7
    [gsinθ-μ1gcosθ-μ2gcosθ]
    ?

    ehild i really appreciate the help

    is the contact force the difference between the forces of the blocks? or the sum
     
  9. Oct 12, 2011 #8

    ehild

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    The contact force is the magnitude of the force one block exerts to the other.

    The common acceleration is the sum of all forces on the system divided by the sum of the masses. The internal forces (the force of block 1 on block 2 and the one block2 exerts to block 1 are opposite and of equal magnitude) cancel, so a=∑Fi(external)]/mi.

    If T is the contact force, Newton's second law applied to both blocks says that

    m1a1=m1gsin(θ)-T-m1μ1gcos(θ) and

    m2a1=m2gsin(θ)+T-m2μ2gcos(θ) .

    If the blocks move together, a1=a2.
    Add the equations:

    a(m1+m2)=(m1+m2)g sin(θ)-(m1μ1+m2μ2)gcos(θ),
    that is,

    [tex]a=gsin(\theta)-\frac{m_1 \mu_1+m_2\mu_2}{m_1 +m_2}g cos(\theta)[/tex]

    ehild
     
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