Virtual displacement. A small imaginary displacement given to a particle or a point of a system, consistent with the constraints of the system, and made without any lapse of time (the forces are taken as unchanged during it).

Virtual work. The work done by a force acting on a particle during a virtual displacement: if $\vec F$ acts and the virtual displacement is $\delta\vec r$, the virtual work is $\vec F\cdot\delta\vec r$.

Principle of virtual work (coplanar forces on a particle). A particle acted on by coplanar forces $\vec F_1,\vec F_2,\dots$ is in equilibrium if and only if the total virtual work is zero for every small virtual displacement:

$\sum \vec F_i\cdot\delta\vec r=0 .$

(If it holds for displacements along both axes, $\sum X=0$ and $\sum Y=0$.)

Tension in the string. Let each rod have weight $w$ and length $a$. AB is fixed horizontally at the top and the hexagon hangs below it; M and N are the midpoints of AB and DE, joined by a string of tension $T$.

By symmetry the figure is symmetric about the vertical through M, with DE horizontal. Let BC and AF make angle $\theta$ with the horizontal. Since DE is horizontal and D lies below the midpoint line, CD and EF also make angle $\theta$ with the horizontal (closing the figure horizontally gives this). Measuring depths below AB:

• string length $MN = 2a\sin\theta$
• centres of gravity of BC, AF: depth $\tfrac12 a\sin\theta$
• centres of gravity of CD, EF: depth $\tfrac32 a\sin\theta$
• centre of gravity of DE: depth $2a\sin\theta$

Sum of the depths of the five movable rods:

$2\cdot\tfrac12 a\sin\theta+2\cdot\tfrac32 a\sin\theta+2a\sin\theta=6a\sin\theta .$

Give a virtual displacement $\delta\theta$. The weights do positive work (they descend) and the tension, which pulls N towards M, does negative work as the string lengthens. By the principle of virtual work:

$w\,\delta(6a\sin\theta)-T\,\delta(2a\sin\theta)=0$

$6wa\cos\theta\,\delta\theta-2Ta\cos\theta\,\delta\theta=0$

$\Rightarrow\; T=3w .$

So the tension in the string is three times the weight of one rod. With $\tfrac W2$ as the weight of each rod this is $T=\tfrac{3W}{2}$, which is the stated result; with $W$ as the weight of each rod it reads $T=3W$.

1. Definition of Common Catenary

A common catenary is the curve in which a uniform, flexible, inextensible string hangs freely under its own weight when suspended from two fixed points.
It is called “common” because the string is of uniform linear density.

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2. Equation of the Uniform Catenary in Intrinsic Form

Let:

• (C) be the lowest point of the chain,
• (P) be any point on the chain,
• (s) be the length of arc (CP),
• (\psi) be the angle which the tangent at (P) makes with the horizontal,
• (w) be the weight per unit length of the chain,
• (T_0) be the tension at the lowest point (C).

Consider the equilibrium of the portion (CP) of the chain.

The forces acting on (CP) are:

1. Tension (T) at (P), acting along the tangent,
2. Tension (T_0) at (C), acting horizontally,
3. Weight (ws) of the portion (CP), acting vertically downward.

Resolving horizontally:

[ T\cos\psi = T_0 ]

Resolving vertically:

[ T\sin\psi = ws ]

Dividing the second equation by the first,

[ \tan\psi = \frac{ws}{T_0} ]

Let

[ c=\frac{T_0}{w} ]

Then

[ \tan\psi = \frac{s}{c} ]

or,

[ \boxed{s=c\tan\psi} ]

This is the required intrinsic equation of the common catenary.

Also, from (T\cos\psi=T_0),

[ T=T_0\sec\psi=wc\sec\psi ]

and the Cartesian equation, taking the origin at the lowest point, is

[ y=c(\sec\psi-1)=c\left(\cosh\frac{x}{c}-1\right). ]

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3. Numerical Problem

Given:

Distance between supports (A) and (B):

[ AB=462\text{ ft} ]

Since the supports lie in a horizontal line, the lowest point (C) lies midway between them. Therefore,

[ AC=CB=\frac{462}{2}=231\text{ ft}. ]

At (A),

[ \tan\alpha=\frac{3}{4}. ]

Hence,

[ \sec\alpha=\sqrt{1+\tan^2\alpha} =\sqrt{1+\frac{9}{16}} =\sqrt{\frac{25}{16}} =\frac{5}{4}. ]

For the catenary,

[ x=c\ln(\sec\psi+\tan\psi). ]

At (A), (x=231) and (\psi=\alpha). Therefore,

[ 231=c\ln\left(\frac{5}{4}+\frac{3}{4}\right) ]

[ 231=c\ln 2 ]

[ c=\frac{231}{\ln 2}. ]

The length of arc from the lowest point (C) to (A) is

[ s_A=c\tan\alpha =c\cdot \frac{3}{4}. ]

Hence the total length of the chain is

[ L=2s_A =2c\cdot \frac{3}{4} =\frac{3}{2}c. ]

Substituting (c=\dfrac{231}{\ln 2}),

[ L=\frac{3}{2}\cdot \frac{231}{\ln 2} =\frac{693}{2\ln 2}. ]

[ L\approx 499.894\text{ ft}. ]

[ \boxed{L\approx 499.89\text{ ft}} ]

So, the length of the chain is approximately 499.89 feet.

1. Centre of gravity

The centre of gravity of a body is the point through which the resultant weight of the body acts, whatever be the orientation of the body in a uniform gravitational field.
For a uniform plane lamina, the centre of gravity coincides with its centroid.

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2. Centre of gravity of a quadrant of an ellipse

Let the ellipse be

[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 ]

Consider the quadrant (OAB) in the first quadrant, bounded by the (x)-axis, the (y)-axis and the ellipse.
Let the lamina have uniform surface density (\sigma).

Area of the quadrant:

[ A=\frac{\pi ab}{4} ]

Let ((\bar{x},\bar{y})) be the centre of gravity of the quadrant.

Finding (\bar{x})

Take a vertical strip at distance (x) from the (y)-axis, of width (dx).
The height of the strip is

[ y=b\sqrt{1-\frac{x^2}{a^2}} ]

So,

[ dA=y,dx=b\sqrt{1-\frac{x^2}{a^2}},dx ]

Now,

[ \bar{x}=\frac{1}{A}\int x,dA ]

[ \bar{x}=\frac{1}{A}\int_0^a x,b\sqrt{1-\frac{x^2}{a^2}},dx ]

Let

[ x=a\sin\theta ]

Then

[ dx=a\cos\theta,d\theta ]

and

[ \sqrt{1-\frac{x^2}{a^2}}=\cos\theta ]

When (x=0), (\theta=0); when (x=a), (\theta=\frac{\pi}{2}).

Thus,

[ \int_0^a x,b\sqrt{1-\frac{x^2}{a^2}},dx

a^2b\int_0^{\pi/2}\sin\theta\cos^2\theta,d\theta ]

[

a^2b\left[-\frac{\cos^3\theta}{3}\right]_0^{\pi/2}

\frac{a^2b}{3} ]

Therefore,

[ \bar{x}=\frac{a^2b/3}{\pi ab/4} ]

[ \boxed{\bar{x}=\frac{4a}{3\pi}} ]

Finding (\bar{y})

By symmetry, or by taking a horizontal strip,

[ \bar{y}=\frac{4b}{3\pi} ]

Hence the centre of gravity of the quadrant of the ellipse is

[ \boxed{\left(\bar{x},\bar{y}\right)=\left(\frac{4a}{3\pi},\frac{4b}{3\pi}\right)} ]

relative to the corner (O), i.e. the centre of the ellipse.

Solution:

Let the uniform rod of length $2a$ and mass $M$ be placed along the $x$-axis with its middle point at the origin. The ends of the rod are at $x=-a$ and $x=+a$.

Mass per unit length of the rod: [ \lambda=\frac{M}{2a} ]

Consider a small element of length $dx$ at a distance $x$ from the centre. Its mass is [ dm=\lambda,dx=\frac{M}{2a},dx ]

The moment of inertia of this element about the axis through the centre and perpendicular to the rod is [ dI=dm,x^2=\frac{M}{2a}x^2,dx ]

Total moment of inertia: [ I=\int_{-a}^{a} \frac{M}{2a}x^2,dx ] [ I=\frac{M}{2a}\left[\frac{x^3}{3}\right]_{-a}^{a} ] [ I=\frac{M}{2a}\left(\frac{a^3}{3}-\frac{-a^3}{3}\right) ] [ I=\frac{M}{2a}\cdot \frac{2a^3}{3} ] [ I=\frac{Ma^2}{3} ]

Since the length of the rod is $L=2a$, we have $a=\frac{L}{2}$. Hence [ I=\frac{M}{3}\left(\frac{L}{2}\right)^2=\frac{ML^2}{12} ]

[ \boxed{I=\frac{Ma^2}{3}=\frac{ML^2}{12}} ]

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OR

Motion of a particle falling from infinity through a hole in the earth

Let the earth be a uniform sphere of radius $R$ and mass $M$. Let $m$ be the mass of the particle. The particle starts from rest at infinity, so its total mechanical energy is zero.

1. Outside the earth $(r\ge R)$:

The gravitational force is [ F=-\frac{GMm}{r^2} ] By conservation of energy, at the earth’s surface $r=R$, [ \frac12 mv_s^2-\frac{GMm}{R}=0 ] [ v_s=\sqrt{\frac{2GM}{R}} ] Thus the particle reaches the surface with the escape velocity.

2. Inside the earth $(r<R)$:

Assuming uniform density, the mass enclosed within radius $r$ is [ M_r=M\frac{r^3}{R^3} ] The gravitational force on the particle is [ F=-\frac{GM_rm}{r^2} =-\frac{GMm}{R^3}r=-kr ] where [ k=\frac{GMm}{R^3} ] This is a restoring force proportional to displacement, so the particle executes simple harmonic motion about the centre of the earth. The angular frequency is [ \omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{GM}{R^3}} ] The equation of motion is [ \ddot{r}+\omega^2 r=0 ] with solution [ r=A\cos(\omega t+\phi) ]

At the surface $r=R$, the velocity is $v_s=\sqrt{2GM/R}$. For SHM, [ v^2=\omega^2(A^2-r^2) ] Substituting $r=R$ and $v=v_s$, [ \frac{2GM}{R}=\frac{GM}{R^3}(A^2-R^2) ] [ A^2-R^2=2R^2 ] [ A=\sqrt{3}R ] Thus the amplitude of oscillation is $\sqrt{3}R$, which is greater than $R$. Therefore the particle passes through the centre and reaches the opposite surface of the earth. At that point its speed is again $\sqrt{2GM/R}$, so it escapes to infinity.

Conclusion:
The particle falls from infinity to the earth’s surface with increasing speed, enters the hole, performs simple harmonic motion inside the earth with amplitude $\sqrt{3}R$, crosses the centre, emerges from the opposite side, and then escapes to infinity. It does not oscillate indefinitely because its total energy is zero. If it were dropped from rest at the surface, it would oscillate through the earth with amplitude $R$.

Tangential and normal components of acceleration

When a particle moves along a plane curve with speed $v$, its acceleration is resolved along two perpendicular directions:

• Tangential component $a_t$: the component along the tangent to the path, in the direction of motion. It measures the rate of change of the magnitude of velocity: $a_t=\frac{dv}{dt}=v\frac{dv}{ds}.$
• Normal component $a_n$: the component along the normal, directed towards the centre of curvature. It measures the rate of change of the direction of velocity: $a_n=\frac{v^2}{\rho}=v\frac{d\psi}{dt},$ where $\rho=\dfrac{ds}{d\psi}$ is the radius of curvature and $\psi$ is the angle the tangent makes with a fixed line.

Proof. Tangential acceleration is constant: $\frac{dv}{dt}=f\;\Rightarrow\; v=ft+c_1 .$

Tangential velocity and normal acceleration are in a constant ratio: $\frac{v}{a_n}=k\;\Rightarrow\; a_n=\frac vk .$

Since $a_n=v\dfrac{d\psi}{dt}$, $v\frac{d\psi}{dt}=\frac vk\;\Rightarrow\;\frac{d\psi}{dt}=\frac1k\;\Rightarrow\;\psi=\frac tk+c_2,\qquad t=k(\psi-c_2).$

Then $v=fk(\psi-c_2)+c_1$, and $dt=k\,d\psi$, so $dS=v\,dt=k\big[fk(\psi-c_2)+c_1\big]d\psi .$

Integrating, $S=k\left[fk\left(\frac{\psi^2}{2}-c_2\psi\right)+c_1\psi\right]+C .$

This has the form $\boxed{S=A\psi^2+B\psi+C}$ with $A=\dfrac{fk^2}{2}$, $B=k(c_1-fkc_2)$ and $C$ constant. This is the intrinsic equation of the path; the printed form $S=A\psi+B+C$ should read $S=A\psi^2+B\psi+C$ (a term in $\psi^2$ is needed).