Hot answers tagged

56

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 ...


23

When you pluck the string you excite many many overtones, not just the fundamental. You can observe this by suppressing the fundamental. Pluck the string while holding a finger lightly at the center of the string. That point is an antinode for the fundamental and all odd harmonics, but a node for the even harmonics. Putting your finger at that point ...


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.


17

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 ...


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 ...


13

Because they are too long for the confined space. Reasoning: A very short string can't clot (say 1mm). So it needs a certain length before it can start clotting. If they could be placed neatly (say big pocket), they wouldn't clot because the ends are too far apart. They must be flexible. A steel bar doesn't clot. So you need certain elements to be just ...


13

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 ...


8

Earbud cords form coils in your pocket. Even if you crumple the cord, it eventually finds its way into coils as you move and the cord is agitated. The coils lie in planes that are stacked. It's been found that agitated coils spontaneously braid and knot themselves. See this paper: http://www.pnas.org/content/104/42/16432.full, which found that formation ...


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 ...


7

What would happen if I were to allow one end of a rope to fall past the event horizon of a black hole while I held the other end? As usual, this is in the context of a Schwarzschild black hole. First, outside the horizon, a object with constant radial coordinate 'feels' a constant proper acceleration, i.e., an accelerometer (think of a weight scale) ...


6

To understand the physics here we should first consider the assumptions made, and perhaps try to justify them with some physical arguments. Assumptions of this simple pendulum: Newtonian physics apply. (ie any non-classical effects are negligible) The "string", or perhaps more accurately rod, has a fixed length $l$. (That is the string remains under ...


5

This isn't exactly an answer to your question, because as it stands your question can't be answered, but I thought I'd post this because the answer really surprised me. Firstly, the reason your question can't be answered is that you can never get your rope below the event horizon. From the perspective of an observer stationary with respect to the black hole ...


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

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

(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

When you release the plucked string, its shape is momentarily triangular: tied down at the ends and pointed at the location of your finger. But the solutions to the wave equation are not triangle functions, but sinusoidal functions, whose displacements from rest obey $$y_n(x) \propto \sin \frac{2\pi x}{\lambda_0 / n},$$ where $\lambda_0$ is twice the ...


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

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

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. ...


4

When talking about black holes, you need to take into account time dilation. As you lower a rope into an event horizon, you will see time for the end of the rope slow down. You will not be able to say at some point: "Now the rope has crossed the event horizon", because you would need to wait indefinitely. The rope, on the other hand (or some observer you ...


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

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 ...



Only top voted, non community-wiki answers of a minimum length are eligible