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

• A

  $\frac{w \tan \theta}{\mu}$
• Correct answer
  B

  $\mu w \tan \theta$
• C

  $\mu w \sqrt{1+\tan ^2 \theta}$
• D

  $\mu w \sin \theta$

Solution

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 $\begin{aligned} \frac{w_A}{\mu w} & =\frac{T \sin \theta}{T \cos \theta} \\ w_A & =\mu w \tan \theta\end{aligned}$

Hence, the correct answer is option: 'B'.