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Consider a situation like this There is a rope of uniform mass $M$ whose one end is attached to a wall.

enter image description here

Obviously, the gravitational force $mg$ and normal force are acting on it in vertical direction (that's not our concern).

My real concern is:

When I "push" on the rope, I basically apply a force on the end of the rope in the direction of the wall, and the rope slacks.

enter image description here

My simple question is, since I apply a force $F$ on the rope (or more specifically on the atom which is at the farthest end of the rope), what is the Newton's 3rd law pair/reaction force of my Force $F$?

Now, with my basic knowledge, I have deduced that reaction force can't be tension as the rope is slacked, and there is NO way a slacked rope can produce tension.

Also, even though this question seems to be not so good, but can someone explain why I don't possibly feel that force on my hand?

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This question is confusing in part because of the way ropes are approximated in beginning physics classes. You connect masses together with massless strings. That way you can ignore the forces on them and focus on the forces on the masses.

But what happens when you apply a force to just a massless string? You break the approximation.

It is better to think of a chain for this kind of problem. Each link is a mass that transmits force to the next link. Unlike a string, the tension isn't always the same all through the chain. If you accelerate the chain, it is like accelerating a series of masses connected with strings. The tension in each string is different.

You also think of a string as a thing that makes a straight line in the direction of the tension. The usual kind of problem doesn't consider what happens when the tension isn't there. What happens is you break the approximation of an ideal string. It matters how stiff the string is. A chain isn't helpful here. There are no forces that keep one link parallel to the next.

As a first step toward pushing a chain, consider what happens when you make some slack and allow the chain to hang quietly. It forms a curve called a catenary. Each link has 3 forces: two tensions from the next link and the weight of the link. The forces add to $0$. In advanced classes, you find the equation for the shape.

If you push a chain as you have drawn it, you are holding the end and exerting a force. It is pretty much the same as pulling on the end. What would happen is the end link would accelerate. The next link would follow behind, dragging the link after that. Soon you would be pulling the whole chain behind you.

There are some interesting problems where the forces between links of a chain or parts of a massive rope matter. Here is one from Veritasium - I Rented A Helicopter To Settle A Physics Debate.

Another fun one is the Mould effect, aka chain fountain.

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since I apply a force F on the rope...

I think you will find it very difficult to apply an arbitrary force on the end of the rope. It is very light, so it moves rapidly and does not apply much force back.

But the simple answer is $F=ma$. When you push the end of the rope, some mass of rope is accelerated. That acceleration requires a force (and via Newton's third, a paired force acting back on your hand).

The lighter the rope, the less the mass, the less the force possible (at human accelerations). If you were to try this with a length of thread, I doubt you would be able to produce any force that was detectable by your hand.

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  • $\begingroup$ So a reaction force of equal magnitude as my force does act on my hand?it's just that it's magnitude is very small for it's effect to be felt by my hand? $\endgroup$
    – Aakash
    Apr 27, 2023 at 15:01
  • $\begingroup$ Yes. Try pushing a bowling ball. Now try pushing a feather with the same force. You won't be able to. Your arm can't accelerate enough to make a big $F$ when you have a small $m$. Same with the end of the rope. $\endgroup$
    – BowlOfRed
    Apr 27, 2023 at 16:24
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Ropes are complex bodies that can undergo many deformations, so I will try brainstorming the situation in the following way.

-When you interact with the end of the rope, whether by pushing, pulling, strafing etc., you are exerting a force on the atoms that are in contact with you.
-Because we are considering net effects on a complex system, it is very hard to think about how every moving bit affects the next. I like to imagine starting from a very short rope and slowly making it longer so to imagine how my push would disperse.
-It is not important if we consider gravity to be the primary force, with you holding it to be the reaction force, or even the wall. There are still the same forces at play. That's the beauty of the 3rd law.

First, NO GRAVITY: the only forces at play are your push and the intra-molecular ones of the rope which resist motion and deformation.
Start by having it straightened out and by pushing on its end. Some parts of your force will distribute perpendicularly to its axis (and make it skid left and right) and will therefore not contribute to a reaction force
All components along its axis, pointing in your direction, will.
Remember you are both moving mass and deforming the structure of the chain that makes the rope up, so whether this deformation is elastic or not, the material will resist your push in some way, even simply because of Newton's first principle, and you start feeling it as soon as you start pushing.

Let's turn GRAVITY ON. In your example, you are holding an end so to keep it under tension. A subtlety about this is that the rope is ALWAYS under some tension, because gravity is pulling it down. For now, your role is the same as the hinge that holds the rope to the wall, you are both resisting gravity.
While you finish preparing your experiment, right before putting the rope under tension so to hold it in a straight line, the reaction force from holding its end is the friction between your skin and the atoms you are touching not to let it slip away. As soon as you pull it even more, tension will also contribute to the reaction force by pulling back even more.
Now, if you start from this position of gravity+tension, moving your end closer to the wall will simply alleviate the tension, reducing the resistive froces on both you and the hinge. The only deformation happening is the rope getting steeper and steeper between you, the lowest point at its center, and the wall.
The net force acting against you by holding it is indeed gravity (or you are acting against gravity the same way your floor does for you).

Now for the final complication. Depending on the specifications of the rope you choose and how fast you push it, there might be compressive forces that fight you back, some of which even reach the wall, and therefore contribute to the resistive force you feel.
Because Newton's principles are formulated upon the TOTAL forces at play, I don't see a problem with the simple answer reaction force=push, because your push is NOT the only force at play.

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You can't push a rope. No seriously, you can't because its weight is going to make it hang down in a catenary shape and you need to always pull to keep it in place.

The amount you need to pull is given by the catenary equations. Consider the hanging cable with dimensions shown below:

fig1

Actually given the pull force $H$, the remaining dimensions and values are derived from the catenary equations. You need a measure of the weight of the rope, and it is given as $w = \frac{m g}{L}$ usually, that is $w$ is weight per unit length. Also you must know $S$, the span where the rope hangs from.

I have summarized the catenary equations below, as well as their parabolic approximations

Quantity Catenary Approximation
Catenary Constant, $a$ $ a = \frac{H}{w} $
Catenary Shape $y = a \left( \cosh \left( \frac{ x-S/2}{a} \right) - \cosh \left( \frac{S}{2 a} \right) \right) $ $y = -\frac{w x (S-x)}{2 H}$
Hanging Length, $L$ $L = 2 a \,\sinh \left( \frac{S}{2 a} \right)$ $L = S + \frac{w^2 S^3}{24 H^2}$
Maximum Sag, $D$ $D = a \left( \cosh \left( \frac{S}{2 a} \right) -1 \right)$ $D = \frac{w S^2}{8 H}$
Support Force, $V$ $V = H \sinh \left( \frac{S}{2 a} \right) $ $V = \frac{w S}{2} + \frac{w^3 S^3}{48 H^2}$
Total Force, $T=\sqrt{H^2+V^2}$ $T = H \cosh \left( \frac{S}{2 a} \right) $ $T =H + \frac{w^2 S^2}{8 H}$

The above assume and inextensible rope.

So for example for a rope of length $L$ and mass $m$, you have $w= \frac{m g}{L}$ and using the parabolic approximation for $L$, you can solve for the pull force

$$ H = \sqrt{ \frac{ m^2 g^2 S^3 }{24 L^2 (L-S)} } $$

If you wanted to find the exact answer, you would need a numerical method to solve the catenary equations. Usually bisection works the fastest, but you can employ single point iteration

$$ H \leftarrow \frac{w S}{2} {\rm asinh}\left( \frac{w L}{2 H} \right) $$

The convergence of the the above isn't great, as after 40 iterations the error was still over 0.5% in a test case I did.

Some additional examinations and you will discover than $V = w \frac{L}{2}$ for both the exact and approx. equations. This makes sense as each $V$ holds up half the total weight $W = m g = w L = 2 V$

Also sag $D$ and total support force $T$ are related with $D = \frac{T-H}{w}$


Now specifically if you want to find the effective mass as seen from then end, what you can do assume a velocity profile for each part of the catenary that results in a given end velocity $v_0$ on the handle and equate the total kinetic energy to that of a lumped mass $m_{\rm eff}$ with velocity $v_0$

$$ KE = \tfrac{1}{2} m_{\rm eff}\,v_0^2 $$

You can consider the distribution of velocities as even along the rope, such that $$v(x) = v_0 \frac{ s(x) }{L} $$ where $s(x) = \int \limits_0^x \sqrt{1 + \left( \frac{ {\rm d}y}{{\rm d}x}\right)^2}\,{\rm d}x$ is the arc length of the rope at position $x$ from the left end.

The you assemble to total kinetic energy as

$$ KE = \int \limits_0^S \tfrac{m}{2 L} v(x)^2 \sqrt{1 + \left( \frac{ {\rm d}y}{{\rm d}x}\right)^2}\,{\rm d}x = \tfrac{1}{3} m v_0^2 $$

So the effective mass felt at the end of the rope under a catenary shape is

$$ \boxed{ m_{\rm eff} = \tfrac{1}{3} m } $$

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  • $\begingroup$ What if the rope is kept on a table and then pushed???? $\endgroup$
    – Aakash
    Apr 27, 2023 at 17:43
  • $\begingroup$ So then the effective mass depends on the shape it takes. And the shape isn't defined in this problem. If you do have a shape, then as you move the end with velocity $v_0$, each slice ${\rm d}x$ of the rope must have some velocity $v(x)$ and you find the effective mass from the total kinetic energy of the system $$ \tfrac{1}{2} m_{\rm eff} v_0^2 = \tfrac{1}{2} \int \rho A v(x)^2 {\rm d}x $$ where $\rho$ is the density and $A$ is the section area. Also note that $\rho A = \tfrac{m}{L}$ with mass $m$ and length $L$. $\endgroup$
    – JAlex
    Apr 27, 2023 at 17:51
  • $\begingroup$ @Aakash - I did the effective mass calculation for the catenary shape and you find $1/3$ is the mass fraction at the end of the rope. $\endgroup$
    – JAlex
    Apr 27, 2023 at 18:37

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