# Tension in a hanging rope

## Homework Statement

A flexible rope of length l and mass m hangs between two walls. The length of the rope is more than the distance between the walls, and the rope sags downward. At the ends, the rope makes an angle of $\alpha$ with the walls. At the middle, the rope approximately has the shape of an arc of a circle; the radius of the approximating (osculating) circle is R. What is the tension in the rope at its ends? What is the tension in the rope at its middle? What is the speed of transverse waves at the ends? At the middle?

## Homework Equations

$$v=\sqrt{\frac{T}{\mu}}$$

## The Attempt at a Solution

Well, at the ends, it's kind of easy I think. Each end needs to support half the weight of the rope, so the force acting straight down on the end is $\frac{1}{2}mg$. We want to know what force in the rope would create that vertical force, so...
$$\cos{\alpha}=\frac{\frac{1}{2}mg}{T_1}$$

$$T_1=\frac{mg}{2\cos{\alpha}}$$

The tension in the middle is where I have troubles. I started by looking at a segment of the rope of length s at the bottom of the rope. The force required to keep this up is equal to $m_s g$, which, with $\mu$ equal to the mass per unit length of the rope, is equal to

$$\mu sg = \frac{mgs}{l}$$

This force is exerted at an angle $\theta$ above the horizontal, so the force exerted on a segment s by both rope sections becomes:

$$F=\frac{mgs}{l\sin{\theta}}$$

The formula for arc length is $s=R\theta$... But, in this case, the angle from the center of the approximating circle that subtends the arc of length s is equal to twice the theta that the rope makes with the horizontal at the ends of the arc (I wish I had a diagram to show you...), so: $s=2R\theta$. The formula for the force then becomes:

$$F=\frac{2mgR\theta}{l\sin{\theta}}$$

The tension in the rope is equal to this force at the infinitesimal point at the very bottom, so...

$$T_{2}=\lim_{\theta\to0}\frac{2mgR\theta}{l\sin{\theta}}$$

$$T_{2}=\frac{2mgR}{l}\lim_{\theta\to0}\frac{\theta}{\sin{\theta}}$$

That limit is equal to 1, so,

$$T_{2}=\frac{2mgR}{l}$$

To find the velocities we just need to plug that into the equation

$$v=\sqrt{\frac{T}{\mu}}$$, with $$\mu=\frac{m}{l}$$

Getting:

$$v_1=\sqrt{\frac{gl}{2\cos{\alpha}}}$$ at the top

and

$$v_2=\sqrt{2gR}$$ at the bottom.

So, I got answers for everything, but I have absolutely no idea if they're any good. I'm just wondering if I did this right. Thanks!

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Bump.

Bump.
This problem sucks, I know. :\

You have found tension in the ends.
Let tension in the middle = T2
I do not know what is the meaning of tension in the middle. Does it mean tension exactly at the mid point? If so, then following will work: -

Consider the left half of the rope as the system under consideration. The forces on the system are
1. Weight mg/2 downward
2. Tension T1 at the left end at angle alpha with the vertical
3. Tension T2 at the bottom end towards right.

For horizontal equalibrium,
T2 = T1 sin(alpha)
T2 = mg/(2 cos(alpha)) * sin(alpha) = (mg/2)*tan(alpha)