Question

A block of mass $m$ is placed on a wedge. The wedge can be accelerated in four manners marked as (1), (2), (3) and (4) as shown in the figure. If the normal reactions in the situation (1), (2), (3) and (4) are $N_1,N_2,N_3$ and $N_4$ respectively and acceleration with which the block slides on the wedge in situations are $b_1,b_2,b_3$ and $b_4$ respectively then:
(A) $N_3>N_1>N_2>N_4$
(B) $N_4>N_3>N_1>N_2$
(C) $b_2>b_3>b_4>b_1$
(D) $b_2>b_3>b_1>b_4$

Which of the above statements are correct?

• A. $A$ and $B$
• B. $B$ and $D$
• C. $C$ and $D$
• D. $A$ and $C$

Answer 1

The correct option is D $A$ and $C$

Using concept of pseudo force the given system can be reframed as shown
(1)

Balancing forces perpendicular to incline we have,
$N_1=mg\cos {37}^{\circ }+ma\sin {37}^{\circ }$
$N_1=\frac{4}{5}mg+\frac{3}{5}ma$
and, along incline we have
$mg\sin {37}^{\circ }-ma\cos {37}^{\circ }=m{b}_1$
$b_1=\frac{3}{5}g-\frac{4}{5}a$

(2)

Similarly for this case we get,
$N_2=mg\cos {37}^{\circ }-ma\sin {37}^{\circ }$
$\Rightarrow N_2=\frac{4}{5}mg-\frac{3}{5}ma$
and, $mg\sin {37}^{\circ }+ma\cos {37}^{\circ }=m{b}_2$
$\Rightarrow b_2=\frac{3}{5}g+\frac{4}{5}a$

(3)

Similarly, for this case we get,
$N_3=mg\cos {37}^{\circ }+ma\cos {37}^{\circ }$
$\Rightarrow N_3=\frac{4}{5}mg+\frac{4}{5}ma$
and, $mg\sin {37}^{\circ }+ma\sin {37}^{\circ }=m{b}_3$
$b_3=\frac{3}{5}g+\frac{3}{5}a$

(4)

Similarly for this case we get,
$N_4=mg\cos {37}^{\circ }-ma\cos {37}^{\circ }$
$\Rightarrow N_4=\frac{4}{5}mg-\frac{4}{5}ma$
and, $mg\sin {37}^{\circ }-ma\sin {37}^{\circ }=m{b}_4$
$\Rightarrow b_4=\frac{3}{5}g-\frac{3}{5}a$

Thus, comparing the values of normal forces and accelerations, we concluded that
$N_3>N_1>N_2>N_4$ and $b_2>b_3>b_4>b_1$