# Tension in a rope placed on a cone

## Homework Statement

A rope of mass m forming a circle is placed over a smooth round cone with half angle θ. Find the tension in the rope .

## The Attempt at a Solution

Since the half angle is θ , the normal force N acts at an angle θ with the horizontal . The weight acts downwards.

Assuming (since I am not sure ) that the tension T acts towards the center of the circle formed by the rope , doing the force balance ,

Ncosθ = T
Nsinθ = mg

So,tension T = mgcotθ .

I don't have the answer key ,so is this the correct answer ? I somehow have the feeling that this is incorrect .

Many thanks .

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Orodruin
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Tension always acts along the rope, i.e., in the rope direction. You also cannot make an equality between the normal force and the tension in this fashion. If you want to make a force balance analysis, you must look at a small part of the rope, otherwise the horizontal forces will cancel out trivially due to symmetry.

Right .

After reading your response my analysis in the OP looks absolutely terrible .

If I do force balance on an infinitesimal length of the rope ,then I get $T = \frac{mgcotθ}{2\pi}$ . Do you get the same answer ?

Orodruin
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Thanks :)

Please check my understanding on this question .

i) The tension in the rope exists because the horizontal component of normal force from the cone pushes the rope radially outwards .

ii) The magnitude of the tension depends only on the weight of the rope and apex angle of the cone irrespective of the position of the rope on the cone OR the length of the rope .

Is my analysis of the problem correct ?

Orodruin
Staff Emeritus
Homework Helper
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Yes, this seems like a fair description of the situation.

However, note the requirement of a frictionless contact between the rope and cone. This might not be applicable in some situations I can think of.

Shameek Baranwal
Can you please explain how you got to the second answer? I'm a bit puzzled.

haruspex