Finding Friction Coefficient of Masses GA and GB

[MisterOL]

Homework Statement

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

Homework 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

The Attempt at a Solution

 

[MisterOL]

heres the pic to describe prob

 

[Lok]

Start by enumerating the forces that act upon the two blocks. Ignore the cat for now.

 

[MisterOL]

You mean like this ?

 

[Lok]

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

 

[MisterOL]

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 = $$\sqrt{}((SxS)+(SxS)$$ = $$\sqrt{}2$$ * 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...

 

[MisterOL]

can someone please help me need to do this by tomorrow :(

 

[Lok]

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.