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. $\theta$ ,

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