Is there no contact force applied between the pulley and rope directly in-line and on top of the pulley?

Question

Diagram

Suppose I have a simple pulley system: a massless rope over a frictionless pulley with 10 N of weight on either side.

The tension is shown with black arrows (only one side is shown to reduce the complexity of the diagram).

The pulley is exerting a net total force of 20 N to hold up the rope and its weights, as shown by an orange arrow (formed by the contact force, purple arrows).

Observation

Now, directly in-line and on top of the pulley, the rope is lying exactly horizontal. This means that there can't be a vertical component of force.

My question: Does this mean that directly in-line and on top of the pulley there is no contact force applied between the pulley and rope? My reasoning seems to suggest it, but I could have made a mistake.

Answer 1

As you look at smaller segments of the rope, the force exerted by that smaller segment is less. The force from any infinitesimal point of the rope is ... infinitesimal (not just the point at the top).

This is the same as saying that if you have a pressure exerted on an area, then although the total force over the area is finite, the force at any mathematical point is zero.

If you look at any finite length of the rope, then there is finite curvature and you can assume a finite force due to that curvature (even at the top).

Answer 2

An illustration showing BowlOfRed's explanation may help. See below:

Answer 3

If you have equal and opposite forces of $10$ newtons in each direction acting tangentially on a segment of the pulley of angular size $\delta \theta$ then the component of these forces perpendicular to the pulley at the mid-point of the segment is $20 \sin \left( \frac {\delta \theta} 2 \right)$ newtons. In the limit this becomes $10 \space d \theta$ newtons. This applies wherever the rope is in contact with the pulley. If we measure $\theta$ as an angle from the vertical then we have a constant normal force of $10 \space d \theta$ newtons exerted on each angular element $d \theta$ of the pulley from $\theta = -\frac {\pi} 2$ to $\theta = \frac {\pi} 2$.

At an angle $\theta$ to the vertical the vertical component of this normal force is $10 \cos \theta \space d \theta$ newtons so the total vertical force exerted on the pulley by the rope is

$\displaystyle \int_{-\frac {\pi} 2}^{\frac {\pi} 2} 10 \cos \theta \space d \theta = \left[ 10 \sin \theta\right]_{-\frac {\pi} 2}^{\frac {\pi} 2} = 20$ newtons (as expected)

and it is clear from symmetry (or by integrating $10 \sin \theta \space d \theta$) that the horizontal component of the normal force nets to zero.