Question

$\text{Two rectangular blocks of weight}{W}_1\text{and}{W}_2\text{are connected by a flexiable string and rest upon a horizontal and an inclined plane respectively with the cord passing over a pulley as shown , n figure. If}{W}_1=W_2\text{and coefficient of static frictionn is}\mu \text{for all contiguous surface. What should be the inclination of the inclined plane at which motion of system will impend?}$

• A. $\tan \frac{\alpha }{2}=\mu$
• B. $\sin \alpha =\sqrt{1-{\mu }^2}$
• C. $\tan \alpha =\mu$
• D. $\tan \alpha =(1-\mu )$

Answer 1

The correct option is A

$\tan \frac{\alpha }{2}=\mu$

(a)

For block 1 equilibrium,

$N_1=W_2\cos \alpha$

$W_1\sin \alpha =R_1+\mu N_1$

$W_1\sin \alpha =R_1+\mu W_1\cos \alpha ...\left(i\right)$

$\text{For block 2}:N_2=W_2$

$\text{and}{R}_1=\mu N_2$

$R_1=\mu W_2...\left(ii\right)$

By equation (i) and (ii),

$W_1\sin \alpha =\mu W_2+\mu W_1\cos \alpha$

$\sin \alpha =\mu +\mu \cos \alpha (W_1=W_2=W)\text{(Given)}$

$2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}=\mu [1+2{\cos }^2\left(\frac{\alpha }{2}\right)-1]$

$\tan \frac{\alpha }{2}=\mu$