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A body of weight $50 \mathrm{~N}$ is placed on a horizontal surface as shown in the figure. The minimum force required to move the body is $28.28 \mathrm{~N}$. The frictional force and the normal reaction are respectively.
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Verified Answer
The correct answer is:
$20 \mathrm{~N}, 30 \mathrm{~N}$
From the free body diagram
Frictional force, $f=F \cos 45$
$$
\begin{aligned}
& =28.28 \cos 45 \\
& =28.28 \times \frac{1}{\sqrt{2}}=20 \mathrm{~N}
\end{aligned}
$$
Normal reaction, $\mathrm{R}=50-\mathrm{F} \sin 45$
$$
\begin{aligned}
& =50-28.28 \times \frac{1}{\sqrt{2}} \\
& =30 \mathrm{~N}
\end{aligned}
$$
Frictional force, $f=F \cos 45$
$$
\begin{aligned}
& =28.28 \cos 45 \\
& =28.28 \times \frac{1}{\sqrt{2}}=20 \mathrm{~N}
\end{aligned}
$$
Normal reaction, $\mathrm{R}=50-\mathrm{F} \sin 45$
$$
\begin{aligned}
& =50-28.28 \times \frac{1}{\sqrt{2}} \\
& =30 \mathrm{~N}
\end{aligned}
$$
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