Pulley with 2 Blocks on an Inclined Plane. Help please

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SUMMARY

The discussion focuses on solving a physics problem involving two blocks on inclined planes connected by a rope over a frictionless pulley. The blocks are inclined at 50 degrees, with block B having a mass of 4.5 kg and a coefficient of friction of 0.75. To determine the mass of block A for both blocks to slide at constant velocity, users are advised to draw free body diagrams for each block, resolving forces into components parallel and perpendicular to the inclines. Key equations include F=ma, Ff=μFn, and Fgx=mgsin(θ).

PREREQUISITES
  • Understanding of Newton's laws of motion (F=ma)
  • Knowledge of frictional forces and coefficients (Ff=μFn)
  • Ability to resolve forces into components (Fgx and Fgy)
  • Familiarity with free body diagrams for analyzing forces
NEXT STEPS
  • Learn how to draw and analyze free body diagrams for inclined planes
  • Study the effects of friction on motion in physics
  • Explore the concept of tension in ropes over pulleys
  • Investigate the relationship between mass, force, and acceleration in different scenarios
USEFUL FOR

Students studying physics, particularly those focusing on mechanics involving inclined planes, pulleys, and forces. This discussion is beneficial for anyone needing to understand the dynamics of connected masses on inclines.

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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