# Finding the coefficient of friction

I need help with my homework.

http://img36.imageshack.us/img36/8717/friction.png [Broken]

I don't how to figure this out because you only get velocity :/

## The Attempt at a Solution

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rock.freak667
Homework Helper
Well on the horizontal block, what are the forces acting?

On the vertical block, what are the forces acting?

Find the resultant force on each block. What do you get?

Well its not at equilibrium so i can't just sum the forces to get zero ...?

rock.freak667
Homework Helper
Well its not at equilibrium so i can't just sum the forces to get zero ...?

That is why you put the resultant force as 'ma'

so Force normal = 2.94N
and force moving horizontal would be (0.3)(a)

then
μ(2.94) + (0.3)(a) = (0.3)(a) ? ... crap i'm not getting something here :/

rock.freak667
Homework Helper
so Force normal = 2.94N
and force moving horizontal would be (0.3)(a)

then
μ(2.94) + (0.3)(a) = (0.3)(a) ? ... crap i'm not getting something here :/

Consider the horizontal and vertical blocks separately.

On the horizontal block you have a tension T

so T-μ(0.3*9.81) = 0.3a

now do the same for the vertical block.

You will now have two equations in T and a. Eliminate 'T' from the two, leaving one equation in 'a'.

Oh like net force?

The vertical would then be..

T - 0.3a = 0.3a ?

rock.freak667
Homework Helper
Oh like net force?

The vertical would then be..

T - 0.3a = 0.3a ?

Vertically, you'd get mg-T, since the block moves down.

So you would get

T-μ(0.3*9.81) = 0.3a
(0.3*9.81) - T = 0.3a

and substitute ?..

T-μ(0.3*9.81) = (0.3*9.81) - T

that doesn't cancel out the T's ?

rock.freak667
Homework Helper
So you would get

T-μ(0.3*9.81) = 0.3a
(0.3*9.81) - T = 0.3a

and substitute ?..

T-μ(0.3*9.81) = (0.3*9.81) - T

that doesn't cancel out the T's ?

No. If you add the two equations together, what do you get?

You get

μ(2.94) + (2.94) = 0.6a ?

rock.freak667
Homework Helper
right, so you can find 'a' in terms of 'μ'. You also know that 'a' is constant. Now use a kinematic equation involving distance, time and acceleration.

You should now be able to get the value of μ.