Day: October 11, 2021

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202110111120 Solution to 2012-DSE-PHY-1A-5

There are two forces \mathbf{F_1} and \mathbf{F_2}, of constant magnitudes (i.e., F_1=|\mathbf{F_1}|=\textrm{Const.}; F_2=|\mathbf{F_2}|=\textrm{Const.}), acting at the same point. The angle \theta between \mathbf{F_1} and \mathbf{F_2} increases from 0^\circ to 180^\circ.Apparently from the figure, F_1>F_2.

Let the direction of \mathbf{F_2} be fixed due east.

Then,

\begin{aligned} \mathbf{F_1} & = F_1\cos\theta\,\hat{\mathbf{i}}+F_1\sin\theta\,\hat{\mathbf{j}} \\ \mathbf{F_2} & = F_2\,\hat{\mathbf{i}} + 0\,\hat{\mathbf{j}} \\ \mathbf{F_3} & = \mathbf{F_1} + \mathbf{F_2} \\ & = ( F_1\cos\theta + F_2 )\,\hat{\mathbf{i}} + F_1\sin\theta\,\hat{\mathbf{j}} \\ \end{aligned}

\begin{aligned} F_3 & = |\mathbf{F_3}| \\ & = \sqrt{(F_1\cos\theta + F_2)^2+(F_1\sin\theta )^2} \\ & = \sqrt{(F_1)^2\cos^2\theta + 2F_1F_2\cos\theta +(F_2)^2+(F_1)^2\sin^2\theta} \\ & = \sqrt{(F_1)^2+(F_2)^2+2F_1F_2\cos\theta} \\ \end{aligned}

If \theta =0^\circ, then \cos\theta =1 and F_3=F_1+F_2;

if \theta =90^\circ, then \cos\theta =0 and F_3=\sqrt{(F_1)^2+(F_2)^2};

if \theta =180^\circ, then \cos\theta =-1 and F_3=F_1-F_2.

By the triangle inequality,

F_1+F_2>F_3=\sqrt{(F_1)^2+(F_2)^2}.

So the magnitude F_3 of the resultant force \mathbf{F_3} decreases throughout.

(Countercheck)

Differentiating F_3 w.r.t. \theta,

\begin{aligned} \quad \frac{\mathrm{d}}{\mathrm{d}\theta}(F_3) & = \frac{\mathrm{d}}{\mathrm{d}\theta} \sqrt{(F_1)^2+(F_2)^2+2F_1F_2\cos\theta} \\ & = \frac{-F_1F_2\sin\theta}{\sqrt{(F_1)^2+(F_2)^2+2F_1F_2\cos\theta}}\\ \end{aligned}

As \sin\theta \ge 0 for 0\le \theta \le \pi, and \sqrt{[\cdots ]}\ge 0, we have

\displaystyle{\frac{\mathrm{d}F_3}{\mathrm{d}\theta}\le 0}.

And the answer is A.

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202110110958 Sidenote (Section 3.8.2)

Given that the binomial coefficient is so defined as

\begin{pmatrix} N \\ n \end{pmatrix} \stackrel{\textrm{def}}{=} \displaystyle{\frac{N!}{n!(N-n)!}}.

Prove the identities (3.49), (3.50), and (3.51) in Section 3.8.2 Useful Identities for the Binomial Coefficients. Namely,

Eq. (3.49):

\begin{pmatrix} N \\ 0 \end{pmatrix} = \begin{pmatrix} N \\ N \end{pmatrix} = 1

Eq. (3.50):

\begin{pmatrix} N - 1 \\ n \end{pmatrix} + \begin{pmatrix} N - 1 \\ n-1 \end{pmatrix} = \begin{pmatrix} N \\ n \end{pmatrix}

Eq. (3.51):

\begin{pmatrix} N \\ n+1 \end{pmatrix} = \displaystyle{\frac{N-n}{n+1}}\begin{pmatrix} N \\ n \end{pmatrix}

R. H. Swendsen. (2012). An Introduction to Statistical Mechanics and Thermodynamics

Proof of Eq. (3.49).

\begin{aligned} \begin{pmatrix} N \\ 0 \end{pmatrix} & = \frac{N!}{0!(N-0)!} \\ & = \frac{N!}{1\times N!} \\ & = 1 \end{aligned}
\begin{aligned} \begin{pmatrix} N \\ N \end{pmatrix} & = \frac{N!}{N!(N-N)!} \\ & = \frac{N!}{N!\times 1} \\ & = 1 \end{aligned}

Proof of Eq. (3.50).

\begin{aligned} & \quad \begin{pmatrix} N-1 \\ n \end{pmatrix} + \begin{pmatrix} N-1 \\ n-1 \end{pmatrix} \\ & = \frac{(N-1)!}{(n)!(N-1-n)!} + \frac{(N-1)!}{(n-1)!\big((N-1)-(n-1)\big)!} \\ & = \frac{(N-1)!}{(n)!(N-1-n)!} + \frac{(N-1)!}{(n-1)!(N-n)!} \\ & = \frac{(N-n)(N-1)!+(n)(N-1)!}{(n)(n-1)!(N-n)(N-n-1)!} \\ & = \frac{N(N-1)!}{n!(N-n)!} \\ & = \frac{N!}{n!(N-n)!} \\ & = \begin{pmatrix} N \\ n \end{pmatrix} \\ \end{aligned}

Proof of Eq. (3.51).

\begin{aligned} &\quad \begin{pmatrix} N \\ n+1 \end{pmatrix} \\ & = \frac{N!}{(n+1)!\big(N-(n+1)\big) !} \\ & = \frac{(N!)(N-n)}{\big( (n+1)(n!)\big) \big( (N-n)(N-n-1)! \big)} \\ & = \bigg(\frac{N-n}{n+1}\bigg)\bigg(\frac{N!}{n!(N-n)!}\bigg) \\ & = \frac{N-n}{n+1}\begin{pmatrix} N \\n \end{pmatrix} \\ \end{aligned}

QED