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.
1) g
2) $\dfrac{g}{{\sqrt 2 }}$
3) $\dfrac{g}{{\sqrt 3 }}$
4) $\dfrac{g}{{\sqrt 5 }}$

Answer 1

Hint:This is a very similar question of a block mass “m” on a hill, the only difference is that the hill is moving, here instead of a hill a triangular block of mass “M” is given. Draw a free body diagram in which all the vertical and horizontal forces are mentioned and divide the vertical net force by the horizontal net force.

Complete step by step solution:

The net force in the vertical direction will be zero because there is no acceleration.
$\sum {{F_y}} = 0$ ;
$N\cos \theta = mg$;
Now, in the horizontal direction the surface is frictionless so:
$mg\sin \theta = 0$ ;

Therefore, the force is acting on the forward side in the horizontal direction:
$N\sin \theta = {F_x}$;
Now, according to Newton’s second law force equals mass times acceleration:
$N\sin \theta = ma$;
Divide the vertical force ${F_y}$ by horizontal force${F_x}$:
$\dfrac{{N\sin \theta }}{{N\cos \theta }} = \dfrac{{ma}}{{mg}}$;
$ \Rightarrow \dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{a}{g}$;
Write the above equation in terms of a:
$ \Rightarrow \tan \theta = \dfrac{a}{g}$;
$ \Rightarrow a = g\tan \theta $;
Here, the cube is inclined at angle of $30^\circ $ with the base of the triangle so:
$ \Rightarrow a = g\tan 30$;
$ \Rightarrow a = \dfrac{g}{{\sqrt 3 }}$;

Final Answer:Option “3” is correct. Therefore, the acceleration, which M must have relative to the table, to keep m stationary relative to the triangular block is $\dfrac{g}{{\sqrt 3 }}$;

Note:Here, be careful while resolving the horizontal and vertical forces. In equating vertical forces, the normal resolved force “$N\cos \theta $” is equal to mg and not equal to $mg\cos \theta $ and similarly in equating the horizontal forces $mg\sin \theta $ will be equal to the frictional force and since the frictional force is zero therefore $mg\sin \theta $ will be zero. Equate the horizontal force ${F_x}$ with the normal resolved horizontal force $N\sin \theta $.