2
$\begingroup$

A uniform rope of length $l$ is suspended from two hinges, making an angle of $\theta$ with the horizontal at the hinges. Find the depth $d$ of the lowest point of the rope.

Similar questions include a pulley-block system, where we find the acceleration "along" the rope by dividing the net "pulling force" by the total mass. We can always find the "acceleration along the curve" by dividing the net force "along the curve" by the mass.

In the above question in particular, we can easily find the depth $d$ by breaking the rope into two, and equating the net pulling force (the difference of tensions acting at the ends) to the summed component of gravitational force acting on each element along the rope.

enter image description here

However, what balances the summed component of gravitational force acting perpendicular to the rope (the net tension is along the rope)? What even is the mathematical basis for this hand-wavy method to solve problems?

$\endgroup$
10
  • $\begingroup$ Do you mean that at the lowest point nothing compensates for the gravitational force as the tension is horizontal? $\endgroup$
    – hasoart
    Jul 7, 2021 at 17:59
  • $\begingroup$ No.. this question refers to a specific method of solving problems; involving finding the net force along curves $\endgroup$
    – doobdoob
    Jul 7, 2021 at 18:00
  • $\begingroup$ What is the mathematical basis for this method, and what is the significance of the force perpendicular to the curve? And in this particular question, since the object is in equilibrium, what balances it? $\endgroup$
    – doobdoob
    Jul 7, 2021 at 18:01
  • $\begingroup$ I guess the mathematical basis you are looking for is the Newtons second law? The gravity on one half of the rope is balanced by the vertical component of tension at the hinge $\endgroup$
    – hasoart
    Jul 7, 2021 at 18:02
  • 1
    $\begingroup$ I don't think you understand my method.. I will try to illustrate it with a diagram. $\endgroup$
    – doobdoob
    Jul 7, 2021 at 18:08

1 Answer 1

1
$\begingroup$

enter image description here

The tension forces are acting at slightly different angles, which makes it possible for them to have a vertical component to counteract gravity. Now what goes for tensions, if we write Newton's second law in the direction of the rope element, we get $$T_1+ \text{d}mg\sin\theta=T_2\cos\text{d}\theta \approx T_2$$ From here we get your equation that $$T_2-T_1=\text{d}mg\sin\theta$$ This gives that $\text{d}T=\text{d}mg \sin\theta$, which after integrating gives the result you wanted.

PS. In my diagram $T_1$ and $T_2$ are the forces on the infinitesimally small rope piece, not the forces acting on the half of the rope.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks for the help! $\endgroup$
    – doobdoob
    Jul 7, 2021 at 19:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.