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Block $B$ lying on a table weighs W. The coefficient of static friction between the block and the table is $\mu$. Assume that the cord between $B$ and the knot is horizontal. The maximum weight of the block $A$ for which the system will be stationary is
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Verified Answer
The correct answer is:
$\mu W tan \theta$
Let weight of $A$ is $W^{\prime}$. From the free body diagram. For equilibrium of the system,
$$
\begin{aligned}
T \cos \theta &=\mu N=\mu W \\
T \sin \theta &=W^{\prime}
\end{aligned}
$$
where $T=$ tension in the thread lying between knot and the support.
On divindg, we get $\frac{T \sin \theta}{T \cos \theta}=\frac{W^{\prime}}{\mu W}$
$\Rightarrow \quad \tan \theta=\frac{W^{\prime}}{\mu W}$
$\Rightarrow W^{\prime}=\mu W \tan \theta$
$$
\begin{aligned}
T \cos \theta &=\mu N=\mu W \\
T \sin \theta &=W^{\prime}
\end{aligned}
$$
where $T=$ tension in the thread lying between knot and the support.
On divindg, we get $\frac{T \sin \theta}{T \cos \theta}=\frac{W^{\prime}}{\mu W}$
$\Rightarrow \quad \tan \theta=\frac{W^{\prime}}{\mu W}$
$\Rightarrow W^{\prime}=\mu W \tan \theta$
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