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A stationary uniform rod of mass $m=1 \mathrm{~kg}$ and length $I=2 \mathrm{~m}$ leans against a smooth vertical wall making an angle $\theta=45^{\circ}$ with rough horizontal floor. Find the frictional force that is exerted by the floor on the rod.
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Verified Answer
The correct answer is:
$5 \mathrm{~N}$
The correct option is A $5 \mathrm{~N}$
As the rod is stationary so the linear acceleration and angular acceleration of rod is zero.
$\begin{aligned}
& \text { i.e., } a_{\text {com }}=0 ; a=0 \\
& \mathrm{~N}_2=\mathrm{f} \\
& \left.\mathrm{N}_1=\mathrm{mg}\right\} \because \mathrm{a}_{\mathrm{com}}=0 \\
&
\end{aligned}$
Torque about any point of the rod should also be zero
$\begin{aligned}
& \therefore a=0 \\
& \tau_A=0 \Rightarrow m g \cos \theta \frac{1}{2}+f I \sin \theta=N_1 \cdot I \cos \theta \\
& \Rightarrow N_1 \cos \theta=f \sin \theta+\frac{m g \cos \theta}{2} \\
& \Rightarrow f=\frac{m g \cos }{2 \sin \theta} \frac{\theta \mathrm{mg} \cot \theta}{2} \\
& \Rightarrow f=\frac{1 \times 10 \times \cot 45^{\circ}}{2} \\
& \Rightarrow f=\frac{10}{2}=5 \mathrm{~N}
\end{aligned}$
So, option (a) is correct.
As the rod is stationary so the linear acceleration and angular acceleration of rod is zero.
$\begin{aligned}
& \text { i.e., } a_{\text {com }}=0 ; a=0 \\
& \mathrm{~N}_2=\mathrm{f} \\
& \left.\mathrm{N}_1=\mathrm{mg}\right\} \because \mathrm{a}_{\mathrm{com}}=0 \\
&
\end{aligned}$
Torque about any point of the rod should also be zero
$\begin{aligned}
& \therefore a=0 \\
& \tau_A=0 \Rightarrow m g \cos \theta \frac{1}{2}+f I \sin \theta=N_1 \cdot I \cos \theta \\
& \Rightarrow N_1 \cos \theta=f \sin \theta+\frac{m g \cos \theta}{2} \\
& \Rightarrow f=\frac{m g \cos }{2 \sin \theta} \frac{\theta \mathrm{mg} \cot \theta}{2} \\
& \Rightarrow f=\frac{1 \times 10 \times \cot 45^{\circ}}{2} \\
& \Rightarrow f=\frac{10}{2}=5 \mathrm{~N}
\end{aligned}$
So, option (a) is correct.
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