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23

What is wrong with your argument is this paragraph: If we imagine the chain as having many small segments, then the potential energy of each segment is $E_p=mgh$. As the number of small segments approaches infinity, their masses equalize because the difference in mass between any two segments goes to $0$ as the number of segments goes to infinity. ...


22

To put the accepted answer in mathematical terms, if you have a curve $y(x)$, hanging fixed at $x_0$ and $x_L$ at an height $h=y(x_0)=y(x_L)$, of total length $L$ and mass $M$ then then linear mass density is going to be $\lambda = M/L$. The length of the curve is given by the integral of the arc-length $$L=\int_{x_0}^{x_L} \sqrt{ 1+\left({dy \over dx}\right)...


4

This limit of a sum is by definition the integral of the chain curve. There's no such thing as "the integral of the chain curve". Curves don't have integrals. For an integral, you need three things: an integrand, measure function, and a set over which those are defined (in basic integrals, these are the function you're integrating, the ...


4

The point being made is that as you approach the bottom of the rope, tension approaches 0. So we say that the tension at the very bottom of the rope is 0. In other words, the differential mass $\mathrm{d}M$ is infinitesimal, so for the purposes of computing tension, we treat $\mathrm{d}M\to0$.


3

The pulley is at rest, so the net force must be zero. Since there are two tensions $T$ trying to push it down, the tension of the upper string must be equal to twice this value, so that the vector sum is zero: \begin{equation} F_{\text{net}}=F'-T-T=0\to F'=2T \end{equation} If the pulley has a mass $M_p$, then \begin{equation} F_{\text{net}}=F''-T-T-M_{p}g=0 ...


3

Imagine the balance is attached to the ceiling by a string and a $10$ N mass is hung from the other end. You would say the reading on the balance is $10$ N. Neither the balance nor the mass is not accelerating. Consider a free-body diagram of the balance. The string attached to the ceiling is pulling up with some magnitude force, $F_1$. The string attached ...


3

This problem depends quite a bit on knowing how the balance is calibrated. Take the force diagram you've drawn here and compare it to the "normal" case in which the balance is attached to the ceiling and has a weight hanging on the bottom, which is a situation in which we know that the balance is supposed to read the weight of the object.


2

Imagine removing the black string and masses and instead just grabbing the pulley with your arms. If you pull with a force of $T$ with each arm, shouldn't the green string in the top then hold back against both? The green string tension should be $2T$. Back to your scenario, the situation is the same. The green string is holding up two masses all in all ...


2

If pulley had mass $M_{pulley}$ and it is in equilibrium in the frame in which you are working, then assume that tension in string is $T_{g}$ $$\sum \vec{F}_{net,pulley}=0$$ $$T_{g}-T-T-M_{pulley}g=0$$ $$T_{g}=M_{pulley}g+2T$$ If pulley is ideal, then $M_{pulley}=0$, so, $T_{g}=2T$. If there is friction in pulley and strings have mass then scenario becomes ...


2

If you solve the wave equation, you will find that the general solution can be written as a linear superposition of the normal modes or harmonics $\sin(k_n x)$ as: $$y(x,t) = \sum_{n} c_n \sin(k_n x) \cos(\omega_n t),$$ This solution is for the case when the string has some initial displacement (i.e., it is plucked) at $t=0$, and $k_n = n \pi/L$. Now, the $...


2

All of the points you raise can be important when considering pulley systems in real life, but they are often ignored in basic dynamics problems in order to simplify the analysis of the system. For example: If the string has mass then the mass on each side of the pulley will depend on the length of string on that side of the pulley, so it is no longer ...


1

during the $0.5$ seconds the following steps are needed $A$ drops under freefall. Find the distance dropped in terms of $g$ and time $t$. $B$ moves upwards. Find the distance that $B$ has moved up in terms of $g$, the velocity that $B$ began to move upwards ($v$) and time $t$. When the string goes taut again, the answer to 1) equals the answer to 2) and ...


1

In both cases the harmonics are the standing waves, which can be thought of as waves reflected multiple times from the ends of the string/pipe and interfering constructively with themselves. The only such a waves are those with the integer number of half-wavelengths fitting between the two ends of the string/pipe. What might seem confusing here is that the ...


1

Where the nodes are in the case of standing sound waves in a pipe depend on what whether you are thinking of displacement nodes or pressure nodes. The diagrams in the text to which you refer show the displacement nodes and antinodes which closely parallels the examples of waves in strings that it presented earlier. In this case at the closed end of a pipe ...


1

The (half-)wave length of the oscillations excited in a string is commensurate with (i.e., a fraction of) the string length, which for the main harmonics is much bigger than the width of the finger. Thus, if the perturbation of the string by the finger is expanded in Fourier series in terms of the available modes/harmonics, it will necessarily contain many ...


1

Before the string is cut the mass is in equilibrium, so the tension in the string is based on the sum of the vertical and horizontal forces being zero. After the string is cut the mass is no longer in equilibrium and will experience both a horizontal and vertical acceleration. That means there has to be a net force on the mass. For that to occur the tension ...


1

I think this might help. Always make draw a picture of the situation and go from there.


1

Your working for both the cases are correct. The tension changes instantaneously when the string is cut. I will try and explain it best I can. The case you are referring to is when there is a spring attached instead of a string. When we have a spring and the right most string is cut, the force applied by spring does not change instantly, and thus remains $\...


1

The centre of mass of the whole system is not accelerating at $\frac g 7$. The $4$ kg mass is accelerating downwards at $\frac g 7$ but the $3$ kg mass is accelerating upwards at $\frac g 7$. So the acceleration of the centre of mass is $\displaystyle \frac {4 \times \frac g 7 - 3 \times \frac g 7}{4+3} = \frac g {49}$ and so $\displaystyle P = 7g - \frac {...


1

this is Dr. Kaz. The tension in the string will be the same all over. It is understood that the string is longer than the distance between the points it is attached to. With the ring on the string, we will have a triangle whose base is the horizontal line between the attachment points. It is assumed that the ring has a weight. Without external force, the ...


1

If we suppose a real situation, with some air drag and/or friction between the disk and the string, it will rest in the middle, with or without a initial force. That is the point of smaller gravitational potential. But while the disk is moving to that final point, the angles of each side of the string with the horizontal are not equal. As the tension is the ...


1

The tension force is applied to this element of the string by the next element of the string. Since the next element exists along the length of the string and not off to the side of the string, then along the length is the only direction that it can apply a force.


1

Why is the mass of that point being discarded as if the point itself wouldn't count? This is because we are working with infinitesimals. Essentially the mass of an infinitesimal slice of rope is vanishingly small. When we integrate these slices then they count infinitesimally and we get in toto the total mass of the rope.


1

If you consider both objects and the string as your system then the tension in the string is an internal force, so it does no work. If you consider one object on its own, then the tension in the string is an external force and does work, which may be positive or negative, depending on the direction in which the object moves. If one object moves up the ...


1

Let me offer an entirely intuitive understanding, based on Newton's law of action and reaction. Pay attention to what happens when the wave, as depicted in the answer by Accidental Taylor Expansion, first strikes the rightmost RIGID support. Because of the shape of the wave, having the string ABOVE the neutral position of the support, the wave wants to ...


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