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 $Fig.6.364$. If the normal reactions in situations (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 on the 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)

$N_3>N_1>N_2>N_4$
$b_2>b_3>b_4>b_1$

From the FBD of situation 1,

$N_1=\mathrm{mgcos}{37}^{\circ }+\mathrm{masin}{37}^{\circ }-\left(a\right)$

$\mathrm{mgsin}{37}^{\circ }=m{b}_1+\mathrm{macos}{37}^{\circ }-$(b)

solving above, we get (I)

$N_1=\frac{m(4g+3a)}{5}$

$b_1=\frac{(3g-4a)}{5}$

From the F B D of situation 2,

$N_2+\mathrm{masin}{37}^{\circ }=\mathrm{mg}\cos {37}^{\circ }-C$ \\

$\mathrm{mgsin}{37}^{\circ }+\mathrm{macos}{37}^{\circ }=m{b}_2-$d

solving above, we get (ii)

$N_2=\frac{m(4g-3a)}{5}$

$b_2=\frac{(3g+4a)}{5}$

From the $FBD$ of situation 3

$

$N_3=\mathrm{mac}\theta {337}^{\circ }+\mathrm{mg}\cos$ ${37}^{\circ }-\left(e\right)$

$\mathrm{mg}\sin {37}^{\circ }+\mathrm{cmasin}{37}^{\circ }=m{b}_3$ -(f) \\

solving above, we get (iii)

$N_3=\frac{m(4g+4a)}{5}$

$b_3=\frac{(3g+3a)}{5}$

From the FBD of situation 4,

$N_4+\mathrm{macos}{37}^{\circ }=\mathrm{mgcos}{37}^{\circ }$

$\mathrm{mg}\sin {37}^{\circ }=\mathrm{masin}{37}^{\circ }+m{b}_4$

solving above, we get (iv)

$N_4=\frac{m(4g-4a)}{5}$

$b_4=\frac{(4g-4a)}{5}$

$Fromi,ii,iii,and,iv,weget$

$N_3>N_1>N_2>N_4\&b_2>b_3>b_4>b_1$

$

$\therefore option$

(A) & (c) are correct.