Question

A body of mass $10\text{kg}$ is acted upon by two perpendicular forces, $6\text{N}$ and $8\text{N}$. The resultant acceleration of the body is

• A. $1{\text{ms}}^{–2},$ at an angle of ${\tan }^{–1}\left(\frac{4}{3}\right)$ with respect to $6\text{N}$ force.
• B. $0.2{\text{ms}}^{–2},$ at an angle of ${\tan }^{–1}\left(\frac{4}{3}\right)$ with respect to $6\text{N}$ force.
• C. $1{\text{ms}}^{–2},$ at an angle of ${\tan }^{–1}\left(\frac{3}{4}\right)$ with respect to $8\text{N}$ force.
• D. $0.2{\text{ms}}^{–2},$ at an angle of ${\tan }^{–1}\left(\frac{3}{4}\right)$ with respect to $8\text{N}$ force.

Answer 1

Answer: (A), (C)

$1{\text{ms}}^{–2},$ at an angle of ${\tan }^{–1}\left(\frac{3}{4}\right)$ with respect to $8\text{N}$ force.

We have given mass $m=10\text{kg},F_2=8\text{N},F_1=6\text{N}$.
Let's look at this figure to understand better.
Here, $R$ is the resultant force
$R=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10\text{N}$
Now, we know $F=ma$ or $a=\frac{F}{m}=\frac{R}{m}=\frac{10}{10}=1{\text{ms}}^{-2}\dots \dots \left(1\right)$
So, $\tan {\theta }_1=\frac{8}{6}=\frac{4}{3};{\theta }_1={\tan }^{-1}\frac{4}{3}\dots \dots \left(2\right)$
$\tan {\theta }_2=\frac{6}{8}=\frac{3}{4};{\theta }_2={\tan }^{-1}\frac{3}{4}\dots \dots \left(3\right)$

Hence from equations (1) & (2) option (a) is correct and from equations (1) & (3) option (c) also our acceleration is different hence options (b) & (d) are wrong.