1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
     

    Attached Files:

  4. Feb 11, 2010 #3

    Lok

    User Avatar

    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 ?
     

    Attached Files:

  6. Feb 11, 2010 #5

    Lok

    User Avatar

    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...
     

    Attached Files:

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

    Lok

    User Avatar

    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook