Pulley with 2 Blocks on an Inclined Plane. Help please

[BioCat]

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.

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!

 

[Chestermiller]

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

 

[BioCat]

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?

 

[Chestermiller]

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

 

[HalfLostAndSearching]

I would first recommend that you take a step back and review the basics of inclined planes, pulleys, and the equations you have listed. It is important to have a solid understanding of the concepts before attempting to solve a problem.

Now, to answer your questions specifically:

1. What must be the mass of block A if both blocks are to slide to the right at a constant velocity?
In order for both blocks to slide to the right at a constant velocity, the net force on both blocks must be zero. This means that the force of friction must be equal to the force of gravity pulling the blocks down the incline. Using the equation Ff=μFn, we can set this up as:
Ff = μmgcos(θ) = mg sin(θ)
Solving for the mass of block A:
mA = mB μcos(θ)/sin(θ)

2. What must be the mass of block A if both blocks are to slide to the left at a constant velocity?
Similarly, in order for both blocks to slide to the left at a constant velocity, the net force on both blocks must be zero. This means that the force of friction must be equal to the force of gravity pulling the blocks down the incline. Using the equation Ff=μFn, we can set this up as:
Ff = μmgcos(θ) = mg sin(θ)
However, in this case, the force of friction will act in the opposite direction, so we need to take the negative value:
Ff = -μmgcos(θ) = mg sin(θ)
Solving for the mass of block A:
mA = mB -μcos(θ)/sin(θ)

I highly recommend that you double check these equations and solutions, and also try to understand the underlying concepts behind them. Good luck!