# Friction on an inclined plane/pulley

## 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 Fy= Fn-mg(cos theta)= may
Net Fx= Fn-mg(sin theta)-uFn= max
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?

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.

## The Attempt at a Solution

Last edited by a moderator:

PhanthomJay
Homework Helper
Gold Member
For question 1, you appear to have omitted the tension force in your equations. Either both blocks move to the right, with the same magnitude of acceleration, or both move to the left, with the same magnitude of acceleration. You really don't know until you plug in the numbers, to solve two equations with 2 unknowns.

But wouldn't they both be pulling in opposite directions, making the block on the left lead the acceleration?

I'm not sure how you can get the entire system to accelerate to the right since there is more force on the left side.

PhanthomJay