Friction on an inclined plane/pulley

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Homework Help Overview

The discussion revolves around two physics problems involving forces, friction, and motion on inclined planes and horizontal surfaces. The first problem involves two masses connected by a string on inclines, while the second problem concerns two blocks on a horizontal surface affected by a pulling force.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the forces acting on the masses, including gravitational, frictional, and tension forces. There is uncertainty about the direction of acceleration and the role of tension in the system.

Discussion Status

Some participants have provided insights regarding the need to consider tension in the equations, while others question the wording of the problems and the assumptions about the direction of motion. The discussion is ongoing with various interpretations being explored.

Contextual Notes

There are indications of confusion regarding the setup of the problems, particularly in relation to the forces acting on the blocks and the expected direction of acceleration. Participants express a need for clarification on these points.

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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

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

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.
 
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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.
 
Can you get the system to accelerate to the left? If so, what is the acceleration? (I don't believe the problem is worded correctly).
 
the system could be moving to the right and decelerating.
 
if you resolved this question, can u please post the answer. i want to see if i solved it correctly. thanks.
 

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