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. $\frac{Wtan\theta }{\mu }$
• B. $\mu Wtan\theta$
• C. $\mu W\sqrt{1+tan\theta }$
• D. $\mu Wsin\theta$

Answer 1

The correct option is B $\mu Wtan\theta$

The block B is under equilibrium by action of tension force on it rightwards(say $T^{\prime }$), and force of static friction of it leftwards($f$).

Hence $f=\mu N=\mu W=T^{\prime }$

Consider the forces acting on the knot..

Balancing the forces on knot horizontally,

$T^{\prime }=Tcos\theta$

Also balancing the forces on knot vertically,

$Tsin\theta =W$

$⟹W=\frac{T^{\prime }}{cos\theta }sin\theta$

$=T^{\prime }tan\theta$

$=\mu Wtan\theta$