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53

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 ...


17

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.


15

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 ...


7

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 ...


6

Strings are not described very accurately in popular science, because much of the physics of strings was only understood long after the mathematical theory was somewhat advanced, and an accurate classical analog for the string wasn't available until relatively recently. The classical analog people often use is a vibrating band of energy, but this is mostly ...


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

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

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 ...


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 ...


3

Since $\mathbf{h} + \mathbf{s} = \mathbf{r}$, there's no need to use $\mathbf{h}$ or any of its time derivatives. Physics doesn't care about something like $\dot{m}$. It cares about things like, "What's the energy?" and "What's the force?". But the kinetic energy of the bob is $\frac{1}{2}mv^2$ regardless of $\dot{m}$. The energy in the stretched spring is ...


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 ...


3

When you pluck a string it does not start out like the fundamental above. The string is pulled into a bent shape of two straight lines and an angle and it may not be bent at the middle. Releasing the bent string causes a bunch of harmonics of various amplitudes depending on how far off-center it was bent. (It can not return to the bent angle shape and the ...


3

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 ...


3

It turns out that this is a subject of a recent paper in Proceedings of the Royal Society A and the paper is open access so you can read it for free. The gist is that the chain isn't really best modeled as a chain but as a linked set of rods. The authors conclude that not only is the chain pulled down by gravity, but it is also being pushed up by the ...


3

Your intuition is right: the density of the string goes down a little bit when you increase the tension. HOWEVER: the wave in a string is a transverse wave which depends on the tension and the mass per unit length. If you double the tension the mass per unit length goes down by a small amount (the string gets a bit "thinner" because it gets longer) . Both ...


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

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

It's pretty simple: the forces at the anchor points would be infinite because of the 90° angle ;-) An example: Imagine two pillars with the same height. If you attach a rope on both of them and try to tighten it, you will slowly increase the pulling force at the top of the pillars while increasing the angle between rope and pillar. To fully straighten the ...


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 ...



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