Question

A block of mass $m$ is placed on a wedge. The wedge can be accelerated in four manners marked as $\left(1\right),\left(2\right),\left(3\right)$ and $\left(4\right)$. If the normal reactions (on $m$) in situation $\left(1\right),\left(2\right),\left(3\right)$ and $\left(4\right)$ are $N_1,N_2,N_3\text{and}{N}_4$ respectively and acceleration with which the block slides with respect to the wedge in situations are $b_1,b_2,b_3\text{and}{b}_4$ respectively, then $(\text{Given}a=\frac{g}{2}m{s}^{-2})$

• 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 $b_2>b_3>b_4>b_1$

FBD in case $\left(1\right)$


Perpendicular to incline, balancing the forces
$N_1=\frac{4mg}{5}+\frac{3ma}{5}=\frac{11mg}{10}$
Along the incline, from Newton's second law
$\frac{3mg}{5}-\frac{4ma}{5}=m{b}_1$
$\Rightarrow b_1=\frac{6g-4g}{10}=\frac{g}{5}$

FBD in case $\left(2\right)$

Perpendicular to incline, balancing the forces
$N_2=\frac{4mg}{5}-\frac{3ma}{5}=\frac{mg}{2}$
Along the incline, from Newton's second law
$\frac{3mg}{5}+\frac{4ma}{5}=m{b}_2$
$\Rightarrow b_2=g$

FBD in case $\left(3\right)$
Perpendicular to incline, balancing the forces
$N_3=m(g+a)\times \cos {37}^{\circ }$
$N_3=m\left(\frac{3g}{2}\right)\times \frac{4}{5}=\frac{6mg}{5}$
Along the incline, from Newton's second law
$m{b}_3=m(g+a)\sin {37}^{\circ }$
$b_3=(g+a)\sin {37}^{\circ }$
$b_3=\frac{3g}{2}\times \frac{3}{5}=\frac{9g}{10}$

FBD in case $\left(4\right)$


Perpendicular to incline, balancing the forces
$N_4=m(g-a)\cos {37}^{\circ }$
$N_4=\frac{mg}{2}\times \frac{4}{5}=\frac{2mg}{5}$
Along the incline, from Newton's second law
$m{b}_4=m(g-a)\sin {37}^{\circ }$
$b_4=(g-a)\sin {37}^{\circ }$
$b_4=\frac{g}{2}\times \frac{3}{5}=\frac{3g}{10}$


So, $N_3>N_1>N_2>N_4$ and $b_2>b_3>b_4>b_1$