Question

A block of mass m is placed on a triangular block of mass M, which in turn is placed on a horizontal surface as shown in

figure (9-E21). Assuming frictionless

• A. g
• B. $\dfrac{g}{{\sqrt 2 }}$
• C. $\dfrac{g}{{\sqrt 3 }}$
• D. $\dfrac{g}{{\sqrt 5 }}$

Answer 1

Answer: (C)

$\dfrac{g}{{\sqrt 3 }}$

Answer 2

To keep the block of mass m stationary relative to the wedge accelerating horizontally with acceleration $a$, the net force on the block in the frame of the wedge must be zero. In the inertial frame, the block has horizontal acceleration $a$ and zero vertical acceleration. Resolving forces on the block (gravity $mg$ and normal force $N$) in the inertial frame, the horizontal equation of motion is $N \sin \theta = ma$ and the vertical equation of motion is $N \cos \theta = mg$. Dividing the first equation by the second gives $\tan \theta = a/g$, so $a = g \tan \theta$. Assuming the angle of inclination is $30^\circ$, the required acceleration is $a = g \tan 30^\circ = g/\sqrt{3}$.