Question

A triangular block of mass $M$ with angles ${30}^o,{60}^o,{90}^o$ rests with its ${30}^o-{90}^o$ side on a horizontal smooth fixed table. A cubical block of mass $m$ rests on the ${60}^o-{30}^o$ side of the triangular block. What horizontal acceleration a must $M$ have relative to the stationary table so that $m$ remains stationary with respect to the triangular block $[M=9kg,m=1kg]$?

• A. $2.8m/s^2$
• B. $5.6m/s^2$
• C. $8.4m/s^2$
• D. $zero$

Answer 1

The correct option is B $5.6m/s^2$

Ground Frame:
The FBD is shown.
The block is moving towards $x-axis$ with acceleration $a$.
$\Rightarrow \sum F_y=0$
$\Rightarrow N\cos \theta =mg$ ....(1)
$\Rightarrow \sum F_x=ma$
$\Rightarrow N\sin \theta =ma$ .....(2)
Dividing (2) by (1) $\frac{\sin \theta }{\cos \theta }=\frac{a}{g}$
$\Rightarrow a=g\tan \theta =9.8\times \tan {30}^0=9.8\times \frac{1}{\sqrt{3}}=5.66m/s^2$