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In the diagram shown, the normal reaction force between $2 \mathrm{~kg}$ and $1 \mathrm{~kg}$ is (Given $g=10 \mathrm{~ms}^{-2}$) (Consider the surface, to be smooth):
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Verified Answer
The correct answer is:
$25 \mathrm{~N}$
Total force along inclined plane,
$\begin{aligned}
f & =-\mathrm{F}_1+6 g \sin 30^{\circ}+\mathrm{F}_2 \\
f & =-60+6 g \sin 30^{\circ}+18 \\
f & =-60+6 \times 10 \times \frac{1}{2}+18 \\
f & =12 \mathrm{~N} \text { (downwards) } \\
\text {acceleration, } a & =\frac{f}{m}=\frac{12}{6}=2 \mathrm{~m} / \mathrm{s}^2
\end{aligned}$
As per question, the normal force between $2 \mathrm{~kg}$ and $1 \mathrm{~kg}$ is,
$\begin{aligned}
\mathrm{N}-18-10 \sin 30^{\circ} & =m a \\
\mathrm{~N}-18-10 \times \frac{1}{2} & =1 \times 2 \\
\mathrm{~N} & =2+23 \\
\mathrm{~N} & =25 \mathrm{~N}
\end{aligned}$
$\begin{aligned}
f & =-\mathrm{F}_1+6 g \sin 30^{\circ}+\mathrm{F}_2 \\
f & =-60+6 g \sin 30^{\circ}+18 \\
f & =-60+6 \times 10 \times \frac{1}{2}+18 \\
f & =12 \mathrm{~N} \text { (downwards) } \\
\text {acceleration, } a & =\frac{f}{m}=\frac{12}{6}=2 \mathrm{~m} / \mathrm{s}^2
\end{aligned}$
As per question, the normal force between $2 \mathrm{~kg}$ and $1 \mathrm{~kg}$ is,
$\begin{aligned}
\mathrm{N}-18-10 \sin 30^{\circ} & =m a \\
\mathrm{~N}-18-10 \times \frac{1}{2} & =1 \times 2 \\
\mathrm{~N} & =2+23 \\
\mathrm{~N} & =25 \mathrm{~N}
\end{aligned}$
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