Question

A solid blcok 'A' weighting 'Q' kg is resting on a flat floor. A smooth cylindder 'B' weighting 'P' kg is placed between the solid A and the vertical wall as shown in figure. The friction between the cylinder, wall and the block A is negligible. The coefficient of friction between the block A and floor is $\mu$. The minimum weight P required to disturb the block A is

• A. $\frac{Q(1-Tan\theta )}{\mu Tan\theta }$
• B. $\frac{\mu QTan\theta }{(1-\mu Tan\theta )}$
• C. $\mu Qcos\theta$
• D. $\frac{\mu Q}{cos\theta }$

Answer 1

The correct option is B $\frac{\mu QTan\theta }{(1-\mu Tan\theta )}$

Draw the free body diagram of both the weight P and Q



Apply Lami's theorem for weight 'P'

$\frac{R_A}{sin90}=\frac{R_W}{sin(90+\theta )}=\frac{P}{sin(180-\theta )}$

$R_A=\frac{R_W}{cos\theta }=\frac{P}{sin\theta }$

$\therefore P=R_{A}sin\theta$........(1)

In the weight Q, take

$\Sigma F_x=0$

$\mu R=R_{A}cos\theta$

$\therefore$ $R_{A}cos\theta =\mu (Q+R_{A}sin\theta )$

$\frac{P}{sin\theta }\times cos\theta =\mu (Q+p)$

$P=\mu QTan\theta +P\mu Tan\theta$

$\therefore$ $P=\frac{\mu Tan\theta }{1-\mu Tan\theta }$