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Homework Help: Block movement

  1. Feb 11, 2010 #1
    1. The problem statement, all variables and given/known data
    Blocks A and B have mass of Ga and Gb respectively. Wire has no mass, it starts moving so that B has constant speed downward


    2. Relevant equations
    1.Find μ (frictionkoeficient) regarding GA og GB.
    2.When system start moving a cat with weight Ga jums on block A and system stops ( falls into equilabration ) Show that acceleration is proven by this formula :
    a= - (μGA / (2GA + 2GB) ) * g

    3. The attempt at a solution
     
  2. jcsd
  3. Feb 11, 2010 #2
    heres the pic to describe prob
     

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  4. Feb 11, 2010 #3

    Lok

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    Start by enumerating the forces that act upon the two blocks. Ignore the cat for now.
     
  5. Feb 11, 2010 #4
    You mean like this ?
     

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  6. Feb 11, 2010 #5

    Lok

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    I mean Block A: gravity, friction ....... and later a cat
    Block B : gravity

    In order to unify them you place them in an equation:

    Ffriction=Fgravity

    Ffriction=μGag

    Fgravity=Gbg

    Have fun
     
  7. Feb 11, 2010 #6
    ok here is how I attempt to solve the case
    Take a look at the pic

    step 1 regarding 3)
    Since there is no mass on "wheel" F(res) = m(wheel) * a = 0*a = 0
    step 2 regarding 3)
    K = [tex]\sqrt{}((SxS)+(SxS)[/tex] = [tex]\sqrt{}2[/tex] * S
    step 3 regarding 1)
    newtons 2. in
    x direction gives S - R = Ma
    y direction gives N = Mg
    step 4 regarding 2)
    y direction mg - S = ma

    if we add S-R = Ma and mg - S = ma we end up with
    (M+m)a = mg - R
    since R = μ * N and N = Mg
    (M+m)a = mg - ( μ * Mg)
    -μ * mg - mg - Mg = a (M+m)
    μ*mg = Ma + ma - Mg * mg hence
    μ = (Ma + ma - Mg * mg)/ mg

    whatever I try it does not leave me with Mg or/and mg alone in equasion...
     

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  8. Feb 11, 2010 #7
    can someone please help me need to do this by tomorrow :(
     
  9. Feb 11, 2010 #8

    Lok

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    I will not read all of the above because it is much too much....
    The problem is simple ... not complicated.
    B pulls A at a constant velocity which means that the friction of A is equal to the pulling force of B which is a gravitational pull.
    That means that μGag=Gbg from which we get μ=Gb/Ga just as simple as that.
    About the second part I don't get what acceleration it is all about but maybe some else undestands.
     
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