Imagine a rope of zero mass is tossed over a horizontal cylinder, and the rope ends are pulled downward in parallel with tension T on each end. Given no movement, there is an equal and opposite force of magnitude 2T on the cylinder. And the rope is in contact with the cylinder for half of the cylinder's circumference.

My questions - what is the normal force of the rope on the cylinder at each point around the half circumference? Does this force vary as a function of the angle?

I can imagine the force to reach a maximum value at the top of the cylinder and tapering to zero on each side where the rope leaves the cylinder, but certainly I would like to see an actual derived equation.

  • $\begingroup$ You're looking for an equation? What's your conceptual question? $\endgroup$
    – Bill N
    Aug 5, 2021 at 13:18

2 Answers 2


Well we can derive a formula for this if we make certain assumptions like :

a) The cylinder is perfectly rigid

b) The rope is inextensible and massless

c) the cylinder is perfectly smooth

Let us consider an elemental length of the rope dl at an angle $\theta$ with the horizontal. The angle subtended by this length is d$\theta$ such that dl=R.d$\theta$ where R is the radius of the cylinder. The tension force acts tangentially on either side of this length of rope.The cos component of tension force cancels out and the sin component is vectorially directed inward and has a net magnitude of 2T sin$\frac {d\theta}{2}$.The normal reaction has the same magnitude and is directed outwards since net force on the length of rope is zero due to equilibrium condition (Refer to fig.b) .Then we can write ,

(normal reaction)N=2Tsin$\frac {d\theta}{2}$

Since $\frac {d\theta}{2}$ is very small , sin$\frac {d\theta}{2}$~$\frac {d\theta}{2}$

So , N=2T$\frac {d\theta}{2}$

$$N = T d\theta$$

Which is our required formula

However instead of normal reaction at a point if we consider the normal reaction on a finitely large segment of the rope subtending an angle $\theta $ then net normal force on this segment is $$N = 2T sin \frac{\theta}{2}$$

Vectorially, normal reaction at a point is , $$\vec N = Td\theta(cos \theta\; \hat i + sin\theta\; \hat j)$$ refer to fig.c for direction of coordinate axis.

Now lets check if this formula is true.To do that I will use the simple concept that the net upward force of normal reaction must be equal to net downward force of tension i.e 2T.

In other words, 2T must be equal to $\int^\pi_0 T sin \theta d\theta$ which clearly evaluates to 2T proving that our formula is correct.

enter image description here


It is statically indeterminate.

Basically, the two requirements are: 1) the net force be 2$T$ acting downwards and 2) the net torque on the cylinder is zero.

Here are two possibilities for the traction (force per unit length) distribution: enter image description here

In any case, if we assume no friction, the tractions are directed normal to the cylinder surface. If we have friction, see the Capstan equation. In the first case, I've drawn constant magnitude tractions, whose vertical components must integrate over the circumference to yield $2T$. The distribution in the second case is more arbitrary but the same is true: the tractions must integrate to yield 2$T$ in the vertical direction and $0$ in the horizontal direction.

To actually solve this, you need to admit some additional information and relieve the constraints of the problem. For example, if you specify the elasticity of the cylinder and allow it to deform, it may be possible to solve for the traction distribution analytically.

  • $\begingroup$ It is not statically indeterminate for the simple case of no elasticity. $\endgroup$ Aug 3, 2021 at 18:53
  • $\begingroup$ I never claimed that. $\endgroup$
    – Evan
    Aug 3, 2021 at 18:57

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