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54

This is a statics problem. Assume the cable is static, perfectly straight and horizontal. Pick any point on the cable and the sum of the forces on that point must equal zero. There is a force, due to gravity, "downward". So, there must be an equal, opposing force "upward". This upward force must come from the tension in the cable. But, if the cable is ...


19

The tension of the rope is the shared magnitude of the two forces. Imagine cutting the rope at a point and inserting a spring scale in its place. The reading will show the tension. A rope with zero tension would be hanging loosely or laying on the ground, neglecting the rope's mass.


16

Here's a slightly more mathematical answer to go with Vortico's excellent physical answer. Tension is not a force, and doesn't have a magnitude and direction. This is confusing, because we frequently talk about the "force of tension", which is a different thing from the tension itself. The tension is actually an example of a mathematical object called a ...


14

Imagine a heavy chord raised off the ground between two blocks. Rather than consider all of the mass pieces of the rope, and the forces on them, we can simplify the problem a little bit by considering a slightly different one. The chord can be represented by a heavy ball (in the middle of the chord) connected by two massless strings to the blocks. From ...


12

This is the boring answer that your physics teacher would give you :-) The diagram above shows an idealised rope with all it's weight concentrated at the centre of the rope. If the rope mass is $m$ then the force downwards is just $mg$, where $g$ is the acceleration due to gravity. Suppose you're pulling on the rope with force $F$, and the angle of sag ...


8

There are three parts to this: In mechanical equilibrium, things go to their lowest energy state. A straight line is the shortest distance between two points. Whenever you've minimized something, it means that small deviations don't change its value (to first order). Let's start with the third point, which is mathematical, and then look at the physics of ...


7

The most interesting version of this question (I think) is this: Suppose that the rope has been attached long enough to bring the two galaxies to a stop relative to each other. (d[physical distance]/d[cosmic time] = 0). For subsequent times, does the rope need to maintain any tension to keep the galaxies from starting to accelerate away from each other? If ...


5

Waves on a string are transverse waves not longitudinal waves. They are not variations in pressure, but variations in the displacement of the string. The (average) displacement is greatest at the anti-nodes and zero at the nodes.


5

Strictly speaking, tension is not the same as force, although it is sometimes described as the magnitude of the 'pulling force' experienced by an element (such as a rope). The important thing to remember when resolving forces in classical mechanics and to understand tension is to apply Newton's three laws of motion. They are: 1st Law: an object with no ...


5

(a) Write down the potential energy of the rope as the function $y(x)$. You're almost right, up to a minus sign in the limits of the integral: $$V=\dfrac{mg}{l}\int_{-x_0}^{d-x_0}y(x)\sqrt{1+y'^2}\ dx$$ (b) Since the problem is static, interpret the potential energy as the Lagrangian and find the Lagrangian density. The Lagrangian is usually given ...


4

D'Alembert equation reads, for the considered case, $$\mu\frac{\partial^2 y}{\partial t^2} = T_0 \frac{\partial^2 y}{\partial x^2}\:.\tag{1}$$ It is nothing but $F=ma$ along the vertical direction ($y$). Here $y$ denotes the small deformation of the string along the vertical direction from the stationary (horizontal) configuration. We disregard horizontal ...


4

The equation you quote is an approximation that is only valid if the horizontal force on the string remains the same. In practice that is not the case - and your concern is valid. The increase in length $\Delta \ell =\ell(\frac{1}{\cos\theta}-1)$; how much additional force that generates depends on the unstretched length of the string (or equivalently on ...


4

I think you have misunderstood the meaning of the equation $$\Delta X \Delta P \geq \hbar / 2 \, .$$ This is not surprising given that the notation used here is really, really misleading. It should be written like this $$\sigma_X \sigma_P \geq \hbar / 2 \, .$$ To understand this we have to explain what $\sigma_X$ and $\sigma_P$ mean. Suppose you have a ...


4

The second condition is saying that there is no discontinuity in the slope of the rope at the junction. In other words, there is no "kink" in the rope. Imagine if this assumption were to fail in the following way: $$ \frac{\partial D_1}{\partial x}(0,t) = -1, \qquad \frac{\partial D_2}{\partial x}(0,t) = 1 $$ Then near the origin, the rope would look ...


4

I think the answer is that the second diagram you drew won't happen. I just picked up a string and tried this. What happened is that the first diagram is easy. For the second, I have to twirl the string faster, and I can't quite get it to stay above my hand. The best I can do is to get the mass to swing in a plane almost even with my hand. Note: it's a ...


4

Imagine a finite segment of the rope, say on the left side. Suppose the tension at the top of the segment is $T_t$ and the tension at the bottom of the segment is $T_b$. Then the segment of rope feels a force $T_t$ up and $T_b$ down, or a net force $T_t - T_b$ up. Since the rope is massless there is no gravitational force so that $T_t - T_b$ is the net force ...


4

Nonzero fluxes are required because of some equations of motion linking them to a nonzero Euler character. Once they're there, they induce a superpotential that stabilizes some moduli, usually the complex structure moduli (the very "stabilizes" means that the allowed values of these moduli at which the total potential has a local minimum is discrete, ...


4

I did a bit of discussion on this subject in this thread on Music.SE. The fundamental doesn't necessarily have the strongest amplitude. As said by Alfred Centauri, it depends on the initial configuration: ideally, the string returns to exactly that configuration after each $\tfrac1{\nu_1}$, and the amplitude of each harmonic in frequency space is ...


4

Dear James, there is no reason why $\psi$ should be periodic. First, if you have a problem, imagine that $\psi$ are just auxiliary variables but the true ones are the bilinears $\psi_i \psi_j$ which are still periodic. Only things such as the world sheet stress-energy tensor $T_{++}$ and $T_{--}$ have to be periodic and they are because they're bilinear in ...


4

Concerning the second question, the Planck length is the Planck length and not Newton's length (yes, the OP has asked this question). Newton didn't know Planck's constant which was only discovered 2+ centuries later so he could discuss neither Planck's constant nor the Planck length and other natural units which are functions of Planck's constant. Max ...


3

You can't pull a piece of string perfectly straight - you just can't see the sag. (excluding the limit where variations in the thickness of the string are greater than the sag) Two simple ways of looking at it: Mathematically - As you pull tighter the sag gets less, but it's a function of 1/force, so to get zero sag you need infinite force. ...


3

You ask: In String Theory it is predicted that as a result of the closed strings we have spin-2 gravitons. 1) How do we know there must be an excitation of spin-2 particles? String theories have been chosen and are being extensively studied because they can have a representation of spin two particles . One searches for such theories because when ...


3

In this fourth Lecture of his string course , Lenny Susskind explains in a slightly technical and very accessible manner, why there is a spin 2 excitation for closed strings which can be interpreted a graviton. To explain this, he writes down all the lowest possible excitations of a closed string.There can be left and right moving waves of different ...


3

Here is a lecture to CERN summer students on "what is string theory" so physicists should be able to get a gist of the idea. There are more lectures in the "summer student lecture program" if somebody is really interested. I used "strings" to search. String scattering is a "simple" extension from Feynman diagrams, for those who know Feynman diagrams, as ...


3

Yes, of course, there would be some tension in the rope. The rope would eventually break, and maybe it would be slowing the galaxies motion if it were a really tight rope (you can't get rope with the required rigidity to stop the motion of galaxies in Nature). If one only considers a pair of galaxies only, the Hubble expansion doesn't really differ from the ...


3

Depending on the mass region of $\Phi$, either A or B can be taken as source and the corresponding response (vev). If $B\neq 0$ when $A=0$, it means that the system can spontaneously have a nontrivial vev even without any source. That indicates a phase transition. In the case both $A\neq 0$ and $B\neq 0$, it doesn't mean any phase transition. If we treat ...


3

There are many quantum field theory models which are exactly solvable in the Large $N$ limit, such that the $\mathbb{C}P^N$ model, the Thirring model, the $O(N)$ vector model etc. Please see the following review by Moshe Moshe and Jean Zinn-Justin covering many of these models. The main idea is that Feynman diagrams (for example the vacuum diagrams in the ...


3

The equation represents power = force times velocity. The force used is the restoring force of a small string segment ${\rm d}x$ which is equal to $-T \tan \theta = -T \frac{\partial y}{\partial x}$. The velocity of the segment is obviously $\frac{{\rm d}y}{{\rm d}t}$. You can do the math yourself to balance the forces for a small segment, and will arrive ...



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