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A block $B$, lying on a table, weights $w$. 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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The correct answer is:
$\mu w \tan \theta$
Given that, weight of block $B=w$
Coefficient of friction between block and table $=\mu$
Let the maximum weight of block $A=w_A$
To keep the cord between block $B$ and knot horizontal, block must remains in equilibrium.
Now, from FBD of knot and block $B$, we get
$w_A=T \sin \theta$ ...(i)
and $\quad f=\mu w=T \cos \theta$ ...(ii)
By dividing Eq. (i) by Eq. (ii), we get
$\begin{aligned} \frac{w_A}{\mu w} & =\frac{T \sin \theta}{T \cos \theta} \\ w_A & =\mu w \tan \theta\end{aligned}$
Coefficient of friction between block and table $=\mu$
Let the maximum weight of block $A=w_A$
To keep the cord between block $B$ and knot horizontal, block must remains in equilibrium.
Now, from FBD of knot and block $B$, we get
$w_A=T \sin \theta$ ...(i)
and $\quad f=\mu w=T \cos \theta$ ...(ii)
By dividing Eq. (i) by Eq. (ii), we get
$\begin{aligned} \frac{w_A}{\mu w} & =\frac{T \sin \theta}{T \cos \theta} \\ w_A & =\mu w \tan \theta\end{aligned}$
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