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. If the normal reactions in 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. $b_2>b_3>b_1>b_4$

Answer 1

Answer: (A), (C)

The correct options are
A $N_3>N_1>N_2>N_4$
C $b_2>b_3>b_4>b_1$

(1)
Balancing forces perpendicular to incline $N_1=mg\cos {37}^{\circ }+ma\sin {37}^{\circ }$
$N_1=\frac{4}{5}mg+\frac{3}{5}ma$
and along incline
$mg\sin {37}^{\circ }-ma\cos {37}^{\circ }=m{b}_1$
$b_1=\frac{3}{5}g-\frac{4}{5}a$
(2)

Similarly for this case get $N_2=\frac{4}{5}mg-\frac{3}{5}ma$
and $b_2=\frac{3}{5}g+\frac{4}{5}a$

(3)
Similarly $N_3=\frac{4}{5}mg+\frac{4}{5}ma$
and $b_3=\frac{3}{5}g+\frac{3}{5}a$
(4)

Similarly $N_4=\frac{4}{5}mg-\frac{4}{5}ma$
$b_4=\frac{3}{5}g-\frac{3}{5}a$