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A body of mass $10 \mathrm{~kg}$ is acted upon by two perpendicular forces, $6 \mathrm{~N}$ and $8 \mathrm{~N}$. The resultant acceleration of the body is
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Verified Answer
The correct answers are:
$1 \mathrm{~ms}^{-2}$ at an angle of $\tan ^{-1}\left(\frac{4}{3}\right)$ w.r.t. $6 \mathrm{~N}$ force
,
$1 \mathrm{~ms}^{-2}$ at an angle of $\tan ^{-1}\left(\frac{3}{4}\right)$ w.r.t. $8 \mathrm{~N}$ force
$1 \mathrm{~ms}^{-2}$ at an angle of $\tan ^{-1}\left(\frac{4}{3}\right)$ w.r.t. $6 \mathrm{~N}$ force
,
$1 \mathrm{~ms}^{-2}$ at an angle of $\tan ^{-1}\left(\frac{3}{4}\right)$ w.r.t. $8 \mathrm{~N}$ force
As given that, mass $=m=10 \mathrm{~kg}$ $F_1=6 \mathrm{~N}, F_2=8 \mathrm{~N}$
Resultant force $(F)$
$$
=\sqrt{F_1^2+F_2^2}=\sqrt{36+64}=10 \mathrm{~N}
$$
We know that $F=m a$,
$$
\Rightarrow \quad a=\frac{F}{m}=\frac{10}{10}=1 \mathrm{~m} / \mathrm{s}^2 \text {; along } R \text {. }
$$
If angle between $F$ and $F_1$ is $\theta_1$ then
$$
\begin{aligned}
&\tan \theta_1=\frac{8}{6}=\frac{4}{3} \\
&\theta_1=\tan ^{-1}\left(\frac{4}{3}\right)
\end{aligned}
$$
Now, If the angle between $F_2$ and $F$ is $\theta_2$ then
$$
\begin{aligned}
&\tan \theta_2=\frac{6}{8}=\frac{3}{4} \\
&\theta_2=\tan ^{-1}\left(\frac{3}{4}\right)
\end{aligned}
$$
Hence option (a), (c) verifies.
Resultant force $(F)$
$$
=\sqrt{F_1^2+F_2^2}=\sqrt{36+64}=10 \mathrm{~N}
$$
We know that $F=m a$,
$$
\Rightarrow \quad a=\frac{F}{m}=\frac{10}{10}=1 \mathrm{~m} / \mathrm{s}^2 \text {; along } R \text {. }
$$
If angle between $F$ and $F_1$ is $\theta_1$ then
$$
\begin{aligned}
&\tan \theta_1=\frac{8}{6}=\frac{4}{3} \\
&\theta_1=\tan ^{-1}\left(\frac{4}{3}\right)
\end{aligned}
$$
Now, If the angle between $F_2$ and $F$ is $\theta_2$ then
$$
\begin{aligned}
&\tan \theta_2=\frac{6}{8}=\frac{3}{4} \\
&\theta_2=\tan ^{-1}\left(\frac{3}{4}\right)
\end{aligned}
$$
Hence option (a), (c) verifies.
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