Friction Force with Incline and Pulley

[Alex126]

Homework Statement

There is a mass m₁ on an incline (angle α), connected to a pulley with a string, and on the other side of the pulley, after another string, there is a mass m₂. See picture if it's unclear, I'm not sure how to express the problem.

Anyway the plane has a friction coefficient μ.

Need to find the Tension T of the string and the acceleration a of the system. Pretty sure I know how to solve the general problem, but I have an issue with the Friction Force.

Homework Equations

F = m*a
Friction = μ * Normal force

The Attempt at a Solution

As I said, my main problem is with the Friction Force. In particular, I don't know whether I should calculate the Normal force as the first or the second option:

1. Normal Force = Weight Force 1 * cos (α)
2. Normal Force = Weight Force 1 * cos (α) + Weight Force 2

In other words, I don't know if I should include the Weight Force 2. My first thought was to include only the first object, since it's the only one directly in contact with the plane, but since the second mass is also indirectly connected with the plane through the strings and the first object, I wonder if it should be included too.

Once that's found, the problem should be solved with F = m*a. So I would do:

(assume motion down the plane)
+Weight Force 1_x - Friction - Weight Force 2 = (m₁+m₂) * a
(Weight Force 1_x = m₁ * g * sin (α))

That gives a.

Then, focusing on the second body:
+Tension - Weight Force 2 = m₂*a

That gives T.

 

[Doc Al]

In other words, I don't know if I should include the Weight Force 2. My first thought was to include only the first object, since it's the only one directly in contact with the plane, but since the second mass is also indirectly connected with the plane through the strings and the first object, I wonder if it should be included too.

The weight of m2 acts only on m2, not on m1. But the weight of m2 does affect the tension in the string, which does act on m1. But that won't affect the normal force acting on m1.

 

[Alex126]

Ok. So the Normal force is just the Weight_1_Y after all, right?

 

[Doc Al]

Ok. So the Normal force is just the Weight_1_Y after all, right?

Just like you said here:

1. Normal Force = Weight Force 1 * cos (α)

 

[Alex126]

Ok, thanks!

 

[Merlin3189]

Homework Equations

F = m*a
Friction = μ * Normal force

The Attempt at a Solution

...
1. Normal Force = Weight Force 1 * cos (α)
2. Normal Force = Weight Force 1 * cos (α) + Weight Force 2
...

Neither of these seems to be simple substitution into one of your "relevant equations". So where do they come from?

If you draw a Diagram showing all the forces acting on M1, you should be able to calculate the normal force by using F=ma in a direction where a=0.
Then you can calculate the friction force using your other equation.

In other words, I don't know if I should include the Weight Force 2. My first thought was to include only the first object, since it's the only one directly in contact with the plane, but since the second mass is also indirectly connected with the plane through the strings and the first object, I wonder if it should be included too.

You need to look at the forces acting on M1 in order to find the normal force between it and the plane. A diagram showing these forces would help.

As Doc says, there is a force on M1 that would not be there if M2 were not there. So you should show this on your diagram. You do not yet know the magnitude of this force: it would be a mistake to assume that this force is equal to the weight of M2. But that is not a problem - if you look at the diagram of forces, you can see why.

 

[Alex126]

You need to look at the forces acting on M1 in order to find the normal force between it and the plane. A diagram showing these forces would help.

So I look at the forces acting "directly" on M1. In this case, they would be:

- T (tension)
- W_1x and W_1y maybe? Not sure if they count, since they have an effect on M1 (so they "act on M1" I presume), but they also "derive" from M1 in the first place...
- N (Normal Force) since it's the plane's reaction on M1
- F_Fric maybe?

As Doc says, there is a force on M1 that would not be there if M2 were not there. So you should show this on your diagram. You do not yet know the magnitude of this force: it would be a mistake to assume that this force is equal to the weight of M2. But that is not a problem - if you look at the diagram of forces, you can see why.

If you meant Tension T, yes, it's not equal to W₂. As I said earlier, I would calculate it by putting together (in a system) the forces acting on each body. In an extended version, this is what I would do:

M1
Y axis has no motion since N + W_1y = 0
X axis has (assuming motion down the plane and to the left): +W_1x - F_Fric - T = m₁*a

M2
X axis has no motion
Y axis has (assuming motion upwards): +T - W₂ = m₂*a

Put these together and you get a and then T.

 

[Doc Al]

Looks good.

- T (tension)
- W1x and W1y maybe? Not sure if they count, since they have an effect on M1 (so they "act on M1" I presume), but they also "derive" from M1 in the first place...
- N (Normal Force) since it's the plane's reaction on M1
- FFric maybe?

All of these are forces acting on M1 so they all count and should appear on a free body diagram of M1. (Why in the world would you doubt that the weight of M1 -- which is the gravitational force of the Earth pulling down on M1 -- would count? Without gravity exerting forces on the masses, nothing much would happen here.)

 

[Merlin3189]

Great. Just what I'd hoped for.

My only caveat is the direction of friction. The μN calculation tells you only the magnitude of friction. You could determine whether it is up or down the plane by comparing W1_x and W2, to decide which way it slides. Here you can't decide, because you don't know W1 or W2. So you will calculate the magnitude of a, but not which way it is. For the magnitude it will not matter, because
|(W2-W1_x)| - |F| = (M1 + M2) |a| whichever way you guess.

 

[Alex126]

the gravitational force of the Earth pulling down on M1

Mmh yea, when you put it that way (gravity is an external force after all) it's more obvious.

My only caveat is the direction of friction

True, but the exercise actually had numerical values (which were beside the point for what I needed to know) which confirmed that the motion was down and to the left along the plane.

Alright, thanks y'all.