Where is the equilibrium point of a mass sliding down slacking cord?

[yucheng]

Homework Statement

The ends of a cord length 3.5 m are attached to points A and B as shown in figure 3.2 a. A small smooth pulley carrying a body, mass m = 10 kg is placed on the cord and allowed to reach a point of equilibrium at C. Find the horizontal distance of this point from A and the tension in the cord.
(Basic Engineering Mechanics, J. H. Hughes and K. F. Martin)

Relevant Equations

N/A

My two questions:

1. The author claims that $T_1=T_2$ and $\alpha = \beta$, and this is supposed to be clear the force triangle. Why is this so?
2. Is it possible to use calculus of variation to find the lowest point C? That is, by maximizing the triangle ABC (Area of ABC = $\frac{1}{2}(line AB)(perpendicular of C to line AB)$), since by maximizing the perpendicular of C to line AB, we also minimize the height of the weight.
3. How about finding the lowest point of an ellipse, with A and B as its foci?

P.S. I have not studied the calculus of variations, but if this is possible, then I might as well study it later.

 

[Orodruin]

1. The pulley is smooth (edit: or there would be a net torque on the pulley). This means that $T_1 = T_2$. If $T_1 = T_2$ it is necessary that $\alpha = \beta$ for the horizontal forces to balance out.

2. It is not true that maximising the perpendicular necessarily minimises the height of the weight as the height where the perpendicular crosses AB varies.

3. That sounds like more trouble than what its worth.

 

[yucheng]

1. The pulley is smooth (edit: or there would be a net torque on the pulley). This means that $T_1 = T_2$. If $T_1 = T_2$ it is necessary that $\alpha = \beta$ for the horizontal forces to balance out.

Oh wait, this problem is supposed to be similar in principle to this simple pulley? Literally invert the picture below?

Also, what if its replaced by a hook instead? What is the justification for $T_1=T_2$? That the net force i.e. tension on the hook is zero?

 

[Orodruin]

Also, what if its replaced by a hook instead?

Depends on what you assume for the hook. If there is friction on the hook, then there will be several possible equilibrium points. If you assume that the hook surface is smooth and the hook glides along the rope frictionlessly, then it is equivalent to a smooth pulley.

 

[yucheng]

Depends on what you assume for the hook. If there is friction on the hook, then there will be several possible equilibrium points.

Even if there's friction (supposed it's not too great), won't the mass and hook slide to the lowest equilibrium point of the cord i.e. lowest gravitational potential energy?

 

[Orodruin]

Even if there's friction (supposed it's not too great), won't the mass and hook slide to the lowest equilibrium point of the cord i.e. lowest gravitational potential energy?

No. If there is friction it may stop earlier, much like a box sliding on an inclined slope may come to a stop before reaching the bottom of the slope.

 

[haruspex]

No. If there is friction it may stop earlier, much like a box sliding on an inclined slope may come to a stop before reaching the bottom of the slope.

To clarify, I believe you intend a slope that reduces in gradient towards the bottom.

 

[Orodruin]

To clarify, I believe you intend a slope that reduces in gradient towards the bottom.

I did not as it is not necessary, as long as the friction is larger than the gravitational force component parallel to the surface, the block will stop. The requirement for this is that
$$
mg\sin\theta < mg \mu \cos\theta\qquad \Longrightarrow \qquad \mu > \tan\theta,
$$
where $\theta$ is the inclination of the plane and $\mu$ the coefficient of friction. If this is true it is just a question of whether the slope is long enough or not.

 

[haruspex]

I did not as it is not necessary, as long as the friction is larger than the gravitational force component parallel to the surface, the block will stop. The requirement for this is that
$$
mg\sin\theta < mg \mu \cos\theta\qquad \Longrightarrow \qquad \mu > \tan\theta,
$$
where $\theta$ is the inclination of the plane and $\mu$ the coefficient of friction. If this is true it is just a question of whether the slope is long enough or not.

Ok, so you are supposing an initial push instead.

 

[Orodruin]

Ok, so you are supposing an initial push instead.

That was my thought, both work of course. The changing slope may be closer to the situation at hand here.

 

[Lnewqban]

Trying to respond your question #1 above:
The tension of the string is the same on both sides of the frictionless object that forces it to change its direction (pulley or hook).
For summation of forces in each of x-axis and y-axis to be zero, we should have a symmetrical geometry.
An isosceles triangle will form if you extend the portion of the string a until intersecting the horizontal line crossing B.

Because it is free to do it, naturally the mass will move towards the spatial point of minimum potential energy.

 

[yucheng]

1. The pulley is smooth (edit: or there would be a net torque on the pulley). This means that $T_1 = T_2$. If $T_1 = T_2$ it is necessary that $\alpha = \beta$ for the horizontal forces to balance out.

Hi Oroduin. Thinking about it, we're assuming that the pulley is a point particle. And it's smooth. So, how can there be a torque? Of course we still have the normal force acting on the pulley, by the rope.

Trying to respond your question #1 above:
The tension of the string is the same on both sides of the frictionless object that forces it to change its direction (pulley or hook).

By tension is the same, do you mean this (a cross section of the rope, forces i.e. tension acting on both sides of the cross section are equal in magnitude, at least locally. If we further assume massless rope, then it'll hold globally i.e. across the whole rope)

Thanks and sorry for the really late reply

 

[haruspex]

So, how can there be a torque?

That is exactly the point @Orodruin made. There can be no torque, so the tensions each side are equal in magnitude .

By tension is the same, do you mean this (a cross section of the rope, forces i.e. tension acting on both sides of the cross section are equal in magnitude, at least locally. If we further assume massless rope, then it'll hold globally i.e. across the whole rope)

Yes.

 

[yucheng]

That is exactly the point @Orodruin made. There can be no torque, so the tensions each side are equal in magnitude .

Yes you caught me on that, silly me; what I meant is that how is torque relevant here? Actuality, wait, @Orodruin mangroves (oops, mentions) that there is no resultant torque right? I.e. both forces give rise to torque, but they are in opposite directions and they cancel, but how do they give rise to torques that act on the pulley?

 

[haruspex]

Yes you caught me on that, silly me; what I meant is that how is torque relevant here?

Normally, when we are told that a pulley is frictionless it means there is no friction at the axle. Generally one assumes rolling contact between the pulley and the rope. On that basis, the fact that the torques must balance can be used to prove the tensions are the same.
But the result is still true even if it's just a smooth ring on the rope instead of a pulley. If the tensions were unequal then some of the rope would slide through the ring to equalise them.
Did anyone say "inextensible"? There is no such thing.

 

[yucheng]

Normally, when we are told that a pulley is frictionless it means there is no friction at the axle. Generally one assumes rolling contact between the pulley and the rope. On that basis, the fact that the torques must balance can be used to prove the tensions are the same.

Oh I see! Thanks!