Question

The mass $M$ of the hanging block in figure which will prevent the small block from slipping over the triangular block, if all the surfaces are frictionless and the strings and the pulleys are light is

• A. $\frac{M^{\prime }+m}{\cot \theta -1}$
• B. $\frac{M^{\prime }-m}{\cot \theta +1}$
• C. $\frac{M^{\prime }+m}{\tan \theta -1}$
• D. $\frac{M^{\prime }-m}{\tan \theta +1}$

Answer 1

The correct option is A $\frac{M^{\prime }+m}{\cot \theta -1}$

FBD of hanging block

$Mg-T=Ma$ --- (1)

FBD of inclined plane


$T-N\sin \theta =M^{\prime }a$ --- (2)

FBD of small block


$mg=N\cos \theta$ --- (3)
$ma=N\sin \theta$ --- (4)

Equation (4)/(3) we get
$\frac{a}{g}=\tan \theta$
$a=g\tan \theta$

$N=\frac{mg}{\cos \theta }$ (from (3) )

Add equations (1) and (2) and put the values of $a$, $N$. We get

$Mg-N\sin \theta =Ma+M^{\prime }a$

$Mg-\frac{mg}{\cos \theta }\sin \theta =Mg\tan \theta +M^{\prime }g\tan \theta$

$⟹M=\frac{M^{\prime }+m}{\cot \theta -1}$