Question

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

• A. $\frac{W \tan\theta}{\mu}$
• B. $\mu W \tan\,\theta$
• C. $\mu W\sqrt{1+ \tan^{2}\,\theta }$
• D. $\mu W\, \sin\,\theta$

Answer 1

Answer: (B)

Let weight of $A$ is $W'$. From the free body diagram, For equilibrium of the system,

$T \cos \theta=\mu N=\mu W$ ...(i)
$T \sin \theta=W'$ ...(ii)
where, $T=$ tension in the thread lying between knot and the support.
On divindg Eq. (ii) by Eq. (i), we get
$\frac{T \sin \theta}{T \cos \theta} =\frac{W'}{\mu W}$
$\Rightarrow \tan \theta =\frac{W'}{\mu W}$
$\Rightarrow W'=\mu W \tan \theta$