Pulley with 2 Blocks on an Inclined Plane. Help please

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Homework Help Overview

The problem involves two blocks positioned on inclined planes, each at an angle of 50 degrees. They are connected by a rope over a frictionless pulley. The inquiry focuses on determining the mass of block A required for both blocks to slide at constant velocity in either direction, given the mass of block B and the coefficient of friction.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply fundamental physics equations and concepts related to forces, tension, and friction. Questions arise about the relationship between tension and mass, as well as the correct setup for analyzing the forces acting on each block.

Discussion Status

Participants are engaged in clarifying the relationships between the forces acting on the blocks and the implications of their directions. Some guidance has been offered regarding the use of free body diagrams to visualize the forces, but there is no explicit consensus on the next steps or resolution of the problem.

Contextual Notes

Participants note the importance of understanding the direction of frictional forces and the need to clarify assumptions about tension and mass relationships. The original poster expresses uncertainty about the correct application of equations and the setup of the problem.

BioCat
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Homework Statement


I've been stuck on this problem for hours now. I know it has to be somewhat simple, but I am not too great in physics, so I am asking for help on how to complete this problem.

Two blocks are positioned on surfaces, each inclined at the same angle 50 degrees with respect to the horizontal. The blocks are connected by a rope which rests on a frictionless pulley at the top of the inclines as shown, so the blocks can slide together. The mass of block B is 4.5kg, and the coefficient of friction for both blocks and inclines is .75. Assume gravity is g=10m/s^{2}

Here is the closest picture I can find of it. Its basically the same thing except it is an isosceles triangle.

e3923ac4-5917-484c-afb5-f43f704edec8.jpe


What must be the mass of block A if both blocks are to slide to the right at a constant velocity?
What must be the mass of block A if both blocks are to slide to the left at a constant velocity?

Homework Equations



From all of the info I found online and in books, these are equations that I thought might be needed.

F=ma
Ff=μFn
T=mg? (I am not exactly sure how to get tension..)
Assuming the acceleration is 0, i got 0=Fg+Fn+Ff+T
Fgx=mgsin(θ)
Fgy=mgcos(θ)
So according to above, does Fgy=Fn?
And is T equal for both sides? I am unsure how to set up an equation where I can find which side the blocks are sliding based on the mass of block A. Please help!
 
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Hi Biocat. Welcome to Physics Forums!

For a rope passing over a frictionless pulley, how does the tension of the rope on mass A compare with the tension of the rope on mass B?

In a problem like this, it is best to resolve all the forces into components in the directions parallel and perpendicular to the inclines. This will make it easier to get the normal and frictional forces. Do a free body diagram on mass A, and another free body diagram on mass B. In these free body diagrams, show all the components of the forces acting on the masses in the directions parallel and perpendicular to the inclines. If the blocks are sliding to the right, what direction are the frictional forces on the blocks acting? If the blocks are sliding to the left, what direction are the frictional forces on the blocks acting?

Chet
 
Well, I thought that tension was equal on both sides, is that correct? So if the equation T=mg is right, then wouldn't that mean the mass is the same on both sides? That is obviously not that right answer though, so I am not sure what to do.

For the directions, would i have to change the signs of the coefficient of friction? So to the left would be negative, and the right, positive?
 
BioCat said:
Well, I thought that tension was equal on both sides, is that correct?
Yes.
So if the equation T=mg is right, then wouldn't that mean the mass is the same on both sides?
Yes. But the equation is not correct.
That is obviously not that right answer though, so I am not sure what to do.
You haven't followed my suggestion of drawing a free body diagram for each of the masses, and showing the forces acting on each of them. That is what to do.
For the directions, would i have to change the signs of the coefficient of friction? So to the left would be negative, and the right, positive?
No. The coefficient of friction is always positive. But, the direction of the frictional force exerted by the inclines on each of the masses would change. Until you draw your free body diagrams, you are going to continue to remain confused.

Chet
 

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