0
$\begingroup$

If we have a rope fixed to the ceiling. How much gravitational force will the rope experience? Let's suppose the mass is $m$.

Well the obvious answer is $mg$.

But doesn't the part that is attached to the ceiling experience the weight of the whole of the wire whereas the bottom free hanging element experiences only its own weight which is negligible. How does that add up to give the complete force $mg$?

Suppose we have a question that wants to calculate the stress on the rope due to gravity, won't the force differential have an effect?

$\endgroup$
0
$\begingroup$

The total gravitational force on the whole rope will be $mg$, as you said.

You can get this by integrating the net force acting on each infinitesimal mass element $dm$ over the length of the rope. Let downwards be positive, and the origin be at the top of the rope. Assuming uniform linear mass density, $dm = \lambda\, dx$. An element at position $x$ will experience $g\,dm=g\lambda\,dx$ (its own weight) and on top of that a tension $T(x+dx)$ downwards (from the element below it) and a tension $-T(x)$ upwards (from the element above it).

For the total gravitational force on the rope, just integrate $\int_0^Lg\,dm$, and you get the expected answer $mg$.

To find tension, we need to do more work. Since the rope is at equilibrium, $g\lambda\,dx + T(x+dx)-T(x) = 0$. Rearranging:

$$g\lambda = \frac{T(x+dx)-T(x)}{dx} = \frac{dT}{dx}$$

which you can then integrate and solve for the tension (after applying some logical boundary condition).

So conceptually, the total gravitational force on the rope is nonzero. However, it is at equilibrium because it is suspended (and presumably fixed, so there is a second external force acting on it.) Each small mass element is at equilibrium because the tension in the rope balances the mass element's weight.

$\endgroup$
  • $\begingroup$ So if we want to calculate the stress on the rope, we will use mg or the the net Tension? $\endgroup$ – HitchHiker224 Jan 25 at 10:46
  • $\begingroup$ Using your definition of differential Tension, If we integrate over the length of the rope, we get T=mg. You said the rope is at equilibrium so it has an external force equal to mg, is that force Tension itself? $\endgroup$ – HitchHiker224 Jan 25 at 10:51
  • $\begingroup$ The external force is not tension. It's the ceiling (or a hook, or something) exerting a force on the rope. Tension is within the rope itself. $\endgroup$ – user194422 Jan 25 at 16:27
  • $\begingroup$ Oh sorry, my bad. That was a really stupid thing for me to say. But thank you for help. It cleared a lot of confusion. $\endgroup$ – HitchHiker224 Jan 25 at 16:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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