Question

In the figure shown the acceleration of $A$ is ${\overrightarrow{a}}_A=15\hat{i}+15\hat{j}$, then the acceleration of $B$ is ($A$ remains in contact with $B$ ).

• A. $6\hat{i}$
• B. $-15\hat{i}$
• C. $-10\hat{i}$
• D. $-5\hat{i}$

Answer 1

The correct option is D $-5\hat{i}$

Given, ${\overrightarrow{a}}_A=15\hat{i}+15\hat{j}$
$\Rightarrow a_{AX}=15\text{and}{a}_{AY}=15$



From wedge constaint, the acceleration of $A$ and $B$ in the direction normal to the contact surface should be equal.
Also, the wedge $B$ have to be in contact with lower surface so, it has acceleration along horizontal direction only as shown.
So, from the figure shown above we have,
$(a_A{)}_{\perp }=(a_B{)}_{\perp }$
$\Rightarrow a_{AX}\cos {53}^{\circ }-a_{AY}\cos {37}^{\circ }=a_B\cos {53}^{\circ }$
$\Rightarrow 15\left(\frac{3}{5}\right)-15\left(\frac{4}{5}\right)=a_B\left(\frac{3}{5}\right)$
$\Rightarrow (9-12)=a_B\left(\frac{3}{5}\right)$
$\Rightarrow a_B=\frac{-15}{3}$
$\Rightarrow a_B=-5{\text{m/s}}^2$
Thus, ${\overrightarrow{a}}_B=-5\hat{i}$