Why are the angles in rope equilibrium the same?

[bolzano95]

Homework Statement

Between two parallel walls we tighten a rope. One end of a rope is tied a little higher on the wall than the end on the opposite wall. On the rope we put a pulley with mass m which slides to equilibrium position.

Homework Equations

The Attempt at a Solution

I assumed that in equilibrium the angles with which the ends of the rope are connected with the wall are different. But in the solution states that in the equilibrium the inclined angles are the same.
Interesting.
How can this be explained?
I know that forces in x direction cancel each other out, but I wouldn't conclude that the forces in the rope are also the same.

 

[kuruman]

Is the tension on one side of the pulley different from the other side? Presumably this is an ideal pulley.

 

[bolzano95]

Yes, the pulley here is presumed ideal. The problem doesn't give anything except what is written above under problem statement.

 

[kuruman]

Yes, the pulley here is presumed ideal. The problem doesn't give anything except what is written above under problem statement.

You don't need more to answer the question. What is the most important property of an ideal pulley other than it is assumed massless and there is no friction at the bearings?

 

[bolzano95]

Ok, so in this case I disregard the mass of the pulley and the friction but what I still don't get is why in the equilibrium the inclined angles are the same.

 

[kuruman]

Because the most important property of an ideal pulley is that it changes the direction of the tension without changing its magnitude. Write down the horizontal components of the tension on each side of the pulley and see what must be necessary in order to have equilibrium.

 

[bolzano95]

Lets suppose the angles and therefore forces are different. I would write horizontal components for equilibrium as:
F_{1}cos\alpha= F_{2}cos\beta

and for vertical:
F_{g}= F_{1}sin\alpha + F_{2}sin\beta

I still don't get it, sorry.

 

[kuruman]

Lets suppose the angles and therefore forces are different. I would write horizontal components for equilibrium as:

You cannot suppose that the forces are different. This is an ideal pulley that has the property that I explained in post #6. What happens to $F_1 \cos\alpha=F_2 \cos\beta$ when $F_1=F_2$ as is the case here?

 

[bolzano95]

You are right. Then the angles are the same.

 

[kuruman]

Here is another way to look at it. The magnitudes of the tensions on either side are always equal. As long as the angles are different, the pulley will accelerate because the horizontal tension components are unbalanced. Once the angles become equal, the horizontal components are balanced, the pulley stops accelerating and is in equilibrium.