Laws Of Motion Question 312
Question: A triangular block of mass M with angles$ 30{}^\circ $ ,$ ~60{}^\circ $ , and $ 90{}^\circ $ rests with its $ 30{}^\circ -90{}^\circ $ side on a horizontal table. A cubical block of mass m rests on the $ 60{}^\circ -30{}^\circ $ side. The acceleration which M must have relative to the table to keep m stationary relative to the triangular block assuming frictionless contact is
Options:
A) $ g $
B) $ \frac{g}{\sqrt{2}} $
C) (c)$ \frac{g}{\sqrt{3}} $
D) $ \frac{g}{\sqrt{5}} $
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Answer:
Correct Answer: C
Solution:
[c] $ ma\cos \theta 30{}^\circ =mg\sin 30{}^\circ $
$ \therefore a=\frac{g}{\sqrt{3}} $
