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## Homework Statement

Two masses (mA = 2.5kg) and (mB = 4.0kg) are on inclines and are connected together by a string as shown in the figure . The coefficient of kinetic friction between each mass and its incline is = 0.30.

Picture: http://img209.imageshack.us/img209/1416/incline1.jpg [Broken]

**Q:**If mA moves up and mB moves down, determine the acceleration.

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## Homework Equations

Net F

_{y}= Fn-mg(cos theta)= ma

_{y}

Net F

_{x}= Fn-mg(sin theta)-uFn= ma

_{x}

Ffr= uFn

## The Attempt at a Solution

**1**. I just can't figure this out...here is what I have so far:

F(mA)= 2.5(g)*sin(51)-2.5(g)cos(51)(.3)= 19-4.6= 14.4N

The downward force of mA is 14.4N

For block B:

F(mB)= 4(g)sin(21)- 4(g)cos(21)(.3)= 14-11= 3N

Block B is falling downward at a force of 4N

So how in the world can the system be moving to the right? Even if it was, there would still be a tension of 14N, divide that be the weight (6.5) and you get 2.1 m/s^2...which is not the right answer.

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Any help is appreciated guys...I'm lost here. And also one more if your filling altruistic

## Homework Statement

A 3.0kg block sits on top of a 5.0kg block which is on a horizontal surface. The 5.0-kg block is pulled to the right with a force (F) as shown in the figure . The coefficient of static friction between all surfaces is 0.59 and the kinetic coefficient is 0.39.

Picture: http://img191.imageshack.us/img191/753/giancolich05p031.jpg [Broken]

**Q:**a) What is the minimum value of F needed to move the two blocks?

b)If the force is 10% greater than your answer for (a), what is the acceleration of each block?

## Homework Equations

Same equations.

## The Attempt at a Solution

Fn= 5kg+3kg(9.8)= 78.4

78.4*(.59)=46.3

...so the block should require 46.3 N of force to move. Add that to the tension of the string from the 3kg block on top:

3kg*9.8=17.3

...the top block should require 17.3N to move +46.3= 63.6N...which is also wrong.

## Homework Statement

## Homework Equations

## The Attempt at a Solution

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