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 $W^{\prime }$. From the free body diagram, For equilibrium of the system,

$T\cos \theta =\mu N=\mu W$ ...(i)
$T\sin \theta =W^{\prime }$ ...(ii)
where, $T=$ tension in the thread lying between knot and the support.
On divindg Eq. (ii) by Eq. (i), we get
$\frac{T\sin \theta }{T\cos \theta }=\frac{W^{\prime }}{\mu W}$
$\Rightarrow \tan \theta =\frac{W^{\prime }}{\mu W}$
$\Rightarrow W^{\prime }=\mu W\tan \theta$