# Homework Help: A mass sliding down an incline plane with a pulley attached

1. Sep 18, 2008

### Benzoate

1. The problem statement, all variables and given/known data
A mass m slides on a plane. It is attached to an inextensible massless string, the other end of which is attach to an equal mass. you can assume the pulley is frictionless , but that the surface has a coefficient of static friction. and the frictional force has the form F(f). As you increase the angle theta, at what angle will the block stop moving?

2. Relevant equations

3. The attempt at a solution

Relevant forces: Frictional force, normal force, gravitational force and tension force. I think that I need to set my acceleration to zero since the block stops moving. Forces for both masses will be in the x and y directions:

x-component: For m1: T-m*g*mu*cos(theta)=m*dv/dt
For m2: ma=0, since m2 hangs vertically from the pulley

y-component: For m1: I don't think there is a tension force in the y direction , therefore, the forces in the y direction are just the normal force exerted by the inclined plane and the gravitional force:ma = 0 since the mass is not accelerating in the y-direction: N-m*g*mu*cos(theta)=0 ==> N=mg*mu*cos(theta)
For m2, I'm not sure whether to include the tension acting on m1 and m2 or just the tension acting on m2. I know both masses have the same tension, since the pulley is massless. I think both masses would have the same acceleration since both masses are attached to the same string. well here is what I think is the correct force equation for m2:
for m2: T-m2*g=m*a

2. Sep 18, 2008

### tiny-tim

Hi Benzoate!

First, you can assume that everything is stationary!

So forget acceleration … it's zero!

Look at the forces on m2 … there's only the weight W and the tension T. So T = … ?

Now look at the forces on m1. T is the same (because the pulley is frictionless), and the only other forces are W N and Ff.

Carry on from there.

3. Sep 18, 2008

### Benzoate

Am I assuming everything is stationary because the blocks stops moving?

The reason why I am supposed to assume all the net forces of each of the masses is zero is because I want to find theta when blocks stopped moving completely right?

I 'm sorry, which part of my statement did you find confusing?

Okay. Your saying that all the netforces acting upon each of the two masses are zero? If thats the case, wouldn't the net forces acting upon each of the two masses be equal to each other?

I will rewrite my forces equations:

y component

m1: N=mg*cos(theta)
m2: T=mg

x component:

m1: T-m*g*mu*cos(theta)= 0
m2: ma=0

I 'm sorry, which part of my statement did you find confusing?

4. Sep 19, 2008

### tiny-tim

Hi Benzoate!
That's right … the question is designed so as to avoid any calculation of accleration.
You're being confusing again

the net force on each mass is zero, so yes they're equal because they're both zero …

if you meant is the tension force the same for both, then yes … as I said, T is the same for both masses because the pulley is frictionless (not "massless", btw).
hmm … your x and y components for m2 are horizontal and vertical, but for m1 they're along and perpendicular to the slope … that's ok, but you should say so.

your equations are correct except for T-m*g*mu*cos(theta) … you've left out the component of the weight.
Mostly your repeated references to acceleration, but also your use of different x and y directions for the two masses, your omission of one of the weight terms, and your general inclination to ramble!

5. Sep 21, 2008

### Benzoate

How did I live out a component for weight, if T-m*g*mu*cos(theta) is a force equation going along the x-axis

Never mind I think I got it; my force equation would look like this

T-m*g*mu*cos(theta)-m*g*sin(theta)=0

6. Sep 21, 2008

### tiny-tim

That's the component of the friction, µN.

I'm talking about the weight, mg.

7. Sep 21, 2008

### Almanzo

The two masses are attached to each other by an unstretcheable rope, so any force acting on one of the masses will accelerate both of them. To find the acceleration, the force must be divided by the sum of the masses. (That is, if the force would tend to accelerate them away from each other if they were not attached. A rope can absorb tension, but not compression.)

I take it that one of the masses (m2) is hanging vertically from the rope, while the other mass (m1) is lying on an inclined plane. In that case, the gravity gm2 on the hanging mass is, so to speak, distributed over the two masses, and a m1/(m1+m2) part of it tries to accelerate the lying mass (m1) upwards along the plane. The frictional force is opposed to this acceleration (but can never be larger). The masses will start moving only if the {m1/(m1+m2)} part of gm2 exceeds the maximum possible friction on m1.

8. Sep 21, 2008

### Benzoate

IF the block stops moving , and I need to find the angle at which it stops moving, then the acceleration of the block is zerio.

T+wx-f=0, wx being the weight in the x-direction, f being friction

T=m2*g

m2*g+m1*g*sin(theta)-mu*g*m1*cos(theta)=0

m2+m1*sin(theta)-mu*m1*cos(theta)=0

m1*(sin(theta)-mu*cos(theta)=-m2 ==> sin(theta)-mu*cos(theta)=(-m2/m1) ==> tan(theta)-mu=sec(theta)*m2/m1==>

tan(theta)+sec(theta)*m2/m1=mu

Is my final equation correct?

9. Sep 21, 2008

### Benzoate

IF the block stops moving , and I need to find the angle at which it stops moving, then the acceleration of the block is zerio.

T+wx-f=0, wx being the weight in the x-direction, f being friction

T=m2*g

m2*g+m1*g*sin(theta)-mu*g*m1*cos(theta)=0

m2+m1*sin(theta)-mu*m1*cos(theta)=0

m1*(sin(theta)-mu*cos(theta)=-m2 ==> sin(theta)-mu*cos(theta)=(-m2/m1) ==> tan(theta)-mu=sec(theta)*m2/m1==>

tan(theta)+sec(theta)*m2/m1=mu

Does my final equation look alright? I not sure how to extract theta by itself, where I am able to set theta equal to everything.

10. Sep 22, 2008

### Almanzo

I think it might be well to define what you mean by T, w, N, x, and so on, because that would make the problem easier to understand.

The way I understand it, there is an inclined plane, and theta is the angle between the plane and the horizontal. Mass m1 is lying on the plane, while mass mass m2 is suspended from a rope hanging vertically, but running around a fixed wheel, and being attached to m1. If m2 moves downward over a distance D, m1 will move upward over a distance D*sin(theta), and to the right over a distance D*cos(theta).

The forces from the rope on m1 and m2 I take to be equal in magnitude. These forces are directed along the rope, hence upwards in the case of m2, and parallel to the plane in the case of m1. If everything is not moving, these forces must be equal in magnitude to gm2, the force of gravity on m2.

Assuming that m1 is not moving, the actual friction force on m1 (which is also parallel to the plane) must be equal to gm2. This actual friction force is smaller than the maximum friction force, or equal to it.

The maximum friction force is equal to the friction coefficient multiplied by the force on m1 normal to the plane (that is, in right angles to the plane).

Four forces are acting on m1. The force from the rope, the force from gravity, the normal force, with magnitude N, and the friction force, with magnitude F.

These forces are in equilibrium, so their components parallel to the plane add up to zero, and their components normal to the plane add up to zero. The force of gravity on m1, with magnitude gm1, is the only one with two nonzero components.

Parallel to the plane: gm2 = F + gm1* sin(theta)
Normal to the plane: N = gm1 * cos(theta)

So F/N = {gm2 - gm1*sin(theta)}/{gm1*cos(theta)}

If m1 = m2, F/N = {1- sin(theta)}/cos(theta)

If F is maximal (m1 is on the brink of starting to slide upwards), F/N is the friction coefficient, which is therefore 1/cos(theta) - tan(theta)

So, by tilting the plane, one can find the value of theta where the masses will start to move, and for this value, the friction coefficient will be 1/cos(theta) - tan(theta). This looks like your final formula, so it is probably right.