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A thin plank of mass $m$ and length $I$ is pivoted at one end and it is held stationary in horizontal position by means of a light thread as shown in the figure then find out the force on the pivot.
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Verified Answer
The correct answer is:
$\mathrm{mg} / 2$
Free body diagram of the plank is shown in figure.
$\therefore$ Plank is in equilibrium condition
So $F_{\text {net }} \& T_{\text {net }}$ on the plank is zero
(i) from $F$
$\begin{aligned}
& F_{\text {net }}=O \\
& \Rightarrow F_{\text {net } \times}=0 \\
& N_1=O
\end{aligned}$
$\begin{aligned}
& \text { Now } \mathrm{F}_{\text {net, }}=\mathrm{O} \\
& \quad \Rightarrow \mathrm{N}_2+\mathrm{T}=\mathrm{mg} \ldots \text { (i) } \\
& \text { from } \tau_{\text {net }}=\mathrm{O} \\
& \Rightarrow \tau_{\text {net }} \text { about point } \mathrm{A} \text { is zero } \\
& \text { so } \mathrm{N}_2 \cdot \ell=\mathrm{mg} \cdot \ell / 2 \\
& \quad \Rightarrow \quad \mathrm{N}_2=\frac{\mathrm{mg}}{2}
\end{aligned}$
$\therefore$ Plank is in equilibrium condition
So $F_{\text {net }} \& T_{\text {net }}$ on the plank is zero
(i) from $F$
$\begin{aligned}
& F_{\text {net }}=O \\
& \Rightarrow F_{\text {net } \times}=0 \\
& N_1=O
\end{aligned}$
$\begin{aligned}
& \text { Now } \mathrm{F}_{\text {net, }}=\mathrm{O} \\
& \quad \Rightarrow \mathrm{N}_2+\mathrm{T}=\mathrm{mg} \ldots \text { (i) } \\
& \text { from } \tau_{\text {net }}=\mathrm{O} \\
& \Rightarrow \tau_{\text {net }} \text { about point } \mathrm{A} \text { is zero } \\
& \text { so } \mathrm{N}_2 \cdot \ell=\mathrm{mg} \cdot \ell / 2 \\
& \quad \Rightarrow \quad \mathrm{N}_2=\frac{\mathrm{mg}}{2}
\end{aligned}$
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