Block and Pulley question

1. Jan 24, 2009

vpv

1. The problem statement, all variables and given/known data
Basically, one must find the acceleration of the system as well as the tension within the strings.

2. Relevant equations
EF = ma
Ff = muFn
Also the angle between the string and the top of the block is 78 degrees.

3. The attempt at a solution
Well I drew out the FBD for the block with 47 kg. I equated Fn to FTx (Tension in the x direction).

I am totally stuck now.
Help plz.

2. Jan 24, 2009

LowlyPion

Your top block has Tension and the m*a (a of the system) and the retarding force of friction that is determined by the normal force and the coefficient .

The bottom block is a little trickier because you have a force (tension) that has 2 components. That you can figure from your θ. In this case the horizontal force contributes to friction and the m*(g - a) is the vertical component of tension.

Good thing you know the angle.

3. Jan 24, 2009

vpv

I didn't understand the second part of your answer about the bottom block. And first, you have to find the acceleration and then the tension. My teacher said somehow the tension force from 40 kg block cancels the tension from 47 kg block when you look at the whole thing as a system. but I don't know how I would find the acceleration before tension...

4. Jan 24, 2009

LowlyPion

What you have to do is develop a set of equations to solve for your unknowns.

The Tension in the cable is the same for both blocks is what links the equations together.

The tension applied to the top block less the friction retarding it determines its acceleration doesn't it?

5. Jan 24, 2009

vpv

But tension is not known and neither is acceleration. You said that I should do Ft - Ff = ma but I dont know two of the variables.

6. Jan 24, 2009

LowlyPion

I understand.

Develop as many equations as you have unknowns is the point.

7. Jan 24, 2009

vpv

Thats what I am trying to do. I got Ftx = FN when I isolated 47kg block, since a = 0 in x direction. But am I suppose to assume that Fg = Fty + Ff? Then that would mean the acceleration in y direction is 0 which totally kill the purpose of finding the acceleration of the system. But If I say that Fg - Fty - Ff = ma, I have three different unknowns: Ft, Ff, and a in y direction while two unknowns in x direction, Ft and Fn.

8. Jan 24, 2009

LowlyPion

Personally I would just put numbers in for the values to keep track of what I already know.

For instance isn't the top block 40*(at + .2*9.8) = Tt

I use at because the top block moves horizontally at the same angle as the rope.
The bottom mass only moves an incremental distance Δd vertically by the distance of the top block times Sin78.
This suggests to me then that the bottom acceleration ab = Sin78*at

Tt = Tb

Now develop the equation for the bottom mass right?

9. Jan 24, 2009

nvn

vpv wrote: "But If I say Fg - Fty - Ff = ma, I have three different unknowns, Fty, Ff, and a in y direction, while two unknowns in x direction, Ftx and Fn."

Ff is a function of Fn, which is a function of Ftx, which is a function of T. Fty is a function of T. Therefore, I see only two unknowns in your above sentence so far: T and ab. But ab is a function of at.

10. Jan 24, 2009

vpv

Have no idea what you just said. Actually, I don't have any idea what anyone has explained. Please just show me how you would do it... I spent a whole day on this, none of my peers knew it...

11. Jan 25, 2009

nvn

Actually, you are starting to do quite well in post 7 and are formulating some good equations. It seemed to me you are understanding the excellent explanations by LowlyPion. The wording of my statement in post 9 might not be the best, so just ignore it. Instead, check out post 8, which is extremely helpful information. I think you are well on your way to solving the problem. I can tell from your equations in post 7. Just start substituting what you know into your equations, and I think it will all come together. And check out the advice by LowlyPion.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?