# Friction Force with Incline and Pulley

• Alex126
In summary: But that is not a problem - if you look at the diagram of forces, you can see why.In summary, the homework statement is trying to find the normal force between two masses connected by a string. The problem is figuring out which force to include when calculating it.
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.

Alex126 said:
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.

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

Alex126 said:
Ok. So the Normal force is just the Weight_1_Y after all, right?
Just like you said here:
Alex126 said:
1. Normal Force = Weight Force 1 * cos (α)

Alex126
Ok, thanks!

Alex126 said:

## 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.

Merlin3189 said:
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)
- 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?
Merlin3189 said:
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 + W1y = 0
X axis has (assuming motion down the plane and to the left): +W1x - FFric - 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.

Looks good.

Alex126 said:
- 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.)

Alex126
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 W1x 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-W1x)| - |F| = (M1 + M2) |a| whichever way you guess.

Alex126
Doc Al said:
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.
Merlin3189 said:
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.

## 1. What is friction force with incline and pulley?

Friction force with incline and pulley is the force that opposes the motion of an object on an inclined plane, which is caused by the contact between the object and the surface it is moving on. The pulley adds an additional force due to its rotation that affects the overall friction force.

## 2. How is friction force calculated on an inclined plane?

The friction force on an inclined plane can be calculated using the formula Ff = μN, where μ is the coefficient of friction and N is the normal force acting on the object. The normal force can be calculated using the formula N = mgcosθ, where m is the mass of the object, g is the acceleration due to gravity, and θ is the angle of inclination.

## 3. What factors affect the friction force on an inclined plane?

The friction force on an inclined plane is affected by the coefficient of friction, the normal force, and the angle of inclination. The coefficient of friction depends on the materials and surface properties of the object and the inclined surface. The normal force is affected by the mass and angle of inclination, while the angle of inclination affects the component of the weight force that acts parallel to the inclined surface.

## 4. How does a pulley affect the friction force on an inclined plane?

A pulley adds an additional force to the friction force on an inclined plane due to its rotation. This force is known as the tension force and adds to the normal force acting on the object. The tension force can be calculated using the formula T = mgcosθ + ma, where a is the acceleration of the object down the inclined plane.

## 5. What is the purpose of calculating friction force with incline and pulley?

Calculating friction force with incline and pulley is important in understanding the motion and stability of objects on inclined planes. This knowledge is useful in various fields such as engineering, physics, and mechanics, as well as in everyday situations such as moving objects up or down a ramp or hill.

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