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In an ideal case, we assume strings and ropes to be massless and having same tension throughout. However, I was doing some questions which involved a rope with mass. The question was that if a uniform thick rope of length 5 m is resting on a horizontal smooth (assume frictionless) surface and is pulled by a force of 5N from one end, then what is the tension in the rope at the point 1 m away from the end where force is applied? (sorry for the poor image)

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I came to the conclusion that since rope is not fixed, tension anywhere should be zero. However, when I checked the answer, it was 4 N. I failed to understand how.

Then the next part of the question was that if the rope now hangs between two fixed points which are at same level, then what would be the tension in the rope at its lowest point?

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The answer has to be expressed in terms of mass of string m, acceleration due to gravity and $\theta$ where $\theta$ is the angle that tangents to the rope make with the horizontal.

I took that force acting on lowest point is mg, since it is center of mass and hence T = mg. However, according to the answer it depends upon $\theta$ as well. However, I failed to see what I was doing wrong?

I would appreciate any help in solving these questions, since I have no clue of where exactly I am wrong. I did search the Internet and a few books, but everywhere they assumed string/rope to be massless.

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  • $\begingroup$ What is the total weight of the rope (or the linear weight)? $\endgroup$ – ja72 Jul 15 '15 at 20:14
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In the first problem, you have to realize that the rope won't just move because you want it to; inertia is still a factor. When a force is applied to the rope, it accelerates but the mass of the rope gives it inertia and means the rope has to become taut in order to start moving in full.

Since the rope is uniform, you can divide the total mass by the total length and get the mass per unit length. at this point, wherever they want you to figure out the tension in the rope, you can effectively treat the rest of the rope beyond that point as you would a rigid box of with the same mass as in that remaining segment of rope. The rope ahead of that point is similarly a rigid box with the corresponding mass. Then this problem becomes equivalent to one where 2 boxes are connected by massless string and a force is applied to one box. You then find the tension in the string.

In the second problem, you once again need to remember that the rope has mass all along its length. You also need to keep in mind that in order to offset the mass of the rope, the tension in it, which acts along the rope, at the point the rope contacts the walls needs to have a vertical component equal to $mg/2$ (2 contact points and the wall needs to support the weight). The total tension in the rope at those points then also depends on $\theta$. From here, every successive point on the rope, as you work your way towards the middle, must support the weight of the remaining rope below it as well as balance the horizontal component of tension established at the walls. So, you'd find the mass of the rope below any given point (using the mass per unit length value you calculated) and make that weight the vertical component of tension at that point. Then the $\theta$ value at that point can be found by incorporating that with the horizontal tension along the rope.

As a sanity check, for the first problem, you should find the maximum tension at the end where the force is applied and zero tension at the opposite end. For the second problem, you should find the maximum tension at both ends where it meets the walls and zero vertical tension at the lowest point.

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  • $\begingroup$ Only the vertical component will be zero at the bottom, but then that's obvious as the tangent if the rope is horizontal. Thing is that you can just do what Daipayan wrote, the total force exerted on the right part of the rope should be zero, the wall exerts a horizontal force of $\frac{mg}{2}\tan(\theta)$, therefore the part of the rope to the left of the lowest point will exert the opposite force, which is therefore the tension at the lowest point. $\endgroup$ – Count Iblis Jul 15 '15 at 17:02
  • $\begingroup$ Yeah, that's what I wrote, but reading it over, I see how it wasn't very clear $\endgroup$ – Jim Jul 15 '15 at 17:05
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For the second question- COnsider the string to be made up two parts separated by a vertical line passing through the lowest point.

Now, consider the point where the string meets the wall.The string exerts a force on the wall(Normal force,tangential to the curve at that point) and in trun experiences a force in the opposite direction.

Now resolve these normal force on the string into its two components. The horizontal component is balanced by the tension force which the string experiences on the lowest point due to the pull of the other segment of the string.

Also use the fact that the vertical component balances the weight of the half-segment of the string.

Solve for tension.

As for your first question, the tension at a pint 1m away from the end is the force that pulls on the remaining string(the mass of which you can calculate by - *linear mass density times length) to move it with the common acceleration, which would be given by external force force divided by total mass.Use this.

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  • $\begingroup$ Thanks for the explanation. I understood why $N \sin \theta = (mg)/2$ but why isn't the horizontal component of normal force N by wall equal to the sum of all the horizontal components of the tensions at all the points? Why is the(horizontal) tension at the lowest point alone equal to the horizontal component of normal force? $\endgroup$ – Swapnil Rustagi Jul 16 '15 at 10:51
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    $\begingroup$ when we consider the half-segment as a an individual body.. the force on it due to an external object,ie the wall, needs to be balanced by another external force, which is in this case the pull of the other half segment(which is external wrt the segment in question) on the segment in question.The internal tensions cannot balance the external force .can it?So, i think it would be incorrect to equate an internal force with an external force.. $\endgroup$ – Daipayan Mukherjee Jul 16 '15 at 13:03
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On the first problem consider a small section of the rope ${\rm d}x$, which is accelerating by $\ddot{u}$. The section mass is ${\rm d}m = \frac{m}{\ell}{\rm d}x$ and now consider the change in tension along the rope.

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$$ (T+{\rm d}T)-(T) = \ddot{u} {\rm d}m \\ {\rm d}T = \frac{m}{\ell}\ddot{u} {\rm d}x $$

$$ T(x) = \frac{m}{\ell} \ddot{u} x $$

At the end where the load is applied $T_A=5$ and $x=\ell=5$ which yields $\ddot{u}=\frac{T_A}{m}$ and the tension at $x=4$ $$T = \frac{x}{\ell} T_A$$

The second problem is slightly more complex. I suggest you look up a derivation of the catenary equation. The shape of the rope in a span $S=5$ is defined by the general function

$$ y(x) = a \left( \cosh \left(\frac{x-S/2}{a}\right) - \cosh \left(\frac{S/2}{a}\right) \right) $$

where $a$ is the catenary constant and the mid point tension is $T_0=a \frac{m g}{\ell}$

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  • $\begingroup$ Why in a catenary is the horizontal tension constant throughout the length ? $\endgroup$ – Shashaank May 12 '17 at 11:32
  • $\begingroup$ Because there is no external load on the horizontal load to change it. All the horizontal loading in a catenary is applied at the end points. A small section of rope will have the same horizontal component of tension to the left as the right because of the force balance in this direction. Caveat: This is not true of the rope is acceleratring in the horizontal direction. $\endgroup$ – ja72 May 12 '17 at 11:38
  • $\begingroup$ Ok thank you I understood. Please just check what I am saying is correct or not (if I am correct , then I think I have got the Correct reasoning.) What you said ( that the horizontal component of tension is constant) will also be true if one end of the catenary is fixed and at the other end there is a load which is being pulled horizontally such that the entire system is in equilibrium. Now also horizontal component of tension will be constant throughout the length just that its value will be something else compared to the previous case when both ends were fixed.... continuing..... $\endgroup$ – Shashaank May 12 '17 at 11:43
  • $\begingroup$ If the string is accelerating the system is accelerating then the horizontal component of tension cannot be constant because the string is accelerating. Just like a normal block attached to a string which is being accelerated type of a question. Is that right ? If the acceleration is in the Y direction then the change in tension in the Y direction throughout the length of the string would be different than compared to the case when the string was in equilibrium because here in addition to the weight of the string the string is also accelerating . Is it right ? $\endgroup$ – Shashaank May 12 '17 at 11:46
  • $\begingroup$ If it is right then if the string is mass less and is accelerating then will the tension very throughout the length of the string or not $\endgroup$ – Shashaank May 12 '17 at 11:47

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