Question

Two forces $3\mathrm{N}$ and $2\mathrm{N}$ are at an angle $\mathrm{\theta }$ such that the resultant is $\mathrm{R}$. The first force is now increased to $6\mathrm{N}$ and the resultant become $2\mathrm{R}$. The value of $\mathrm{\theta }$ is

• A. $30°$
• B. $60°$
• C. $90°$
• D. $120°$

Answer 1

The correct option is D

$120°$

Explanation for correct option:
Option (d) $120°$

Step 1: Given data

First force at initial stage = $3\mathrm{N}$

Second force at initial stage = $2\mathrm{N}$

Resultant force at initial stage = $\mathrm{R}$

First force at later stage = $6\mathrm{N}$

Resultant force at later stage = $2\mathrm{R}$

Step 2: Formula used

Parallelogram rule:

The parallelogram rule says that if we place two vectors so they have the same initial point, and then complete the vectors into a parallelogram, then the sum of the vectors is the directed diagonal that starts at the same point as the vectors.

So, $\mathrm{F}=\sqrt{{\mathrm{F}}_1^2+{\mathrm{F}}_2^2+2{\mathrm{F}}_1{\mathrm{F}}_2\mathrm{cos\theta }}$

Step 3: Find Value of $\mathrm{\theta }$

Now, by substituting the given values for initial stage in the above-mentioned formula, we get:

${\mathrm{F}}_1=3\mathrm{N},{\mathrm{F}}_2=2\mathrm{N}$

$\mathrm{R}=\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2+2\mathrm{ABcos\theta }}$
$\mathrm{R}=\sqrt{3^2+2^2+12\mathrm{cos\theta }}$ ………………………..(1)

$$\begin{gathered} \mathrm{R}=\sqrt{3^2+2^2+12\mathrm{cos\theta }} \\ R=\sqrt{9+4+12\mathrm{cos\theta }} \\ R=\sqrt{13+12\mathrm{cos\theta }} \\ {R}^2=13+12\cos \theta \end{gathered}$$

Now, we will substitute the given values for later stage in the formula:

${\mathrm{F}}_1=6\mathrm{N},{\mathrm{F}}_2=2\mathrm{N}$
$2\mathrm{R}=\sqrt{6^2+2^2+24\mathrm{cos\theta }}$ ……………………….(2)
$$\begin{gathered} 2R=\sqrt{36+4+24\cos \theta } \\ 2R=\sqrt{40+24\cos \theta } \\ 4R^2=40+24\cos \theta \\ {R}^2=10+6\cos \theta \end{gathered}$$
From doing (1) - (2) we get,

$$\begin{gathered} 13-10+12\cos \theta -6\cos \theta =0 \\ 3+6\cos \theta =0 \\ \cos \theta =-\frac{6}{3} \\ \cos \theta =-\frac{1}{2} \end{gathered}$$
$\mathrm{\theta }={120}^0$

Thus, option (d) is correct.