Question

Find the acceleration of $\text{B}$, if the acceleration of wedge $\text{A}$ is $a$ as shown in figure.

• A. $\frac{a\cos {\alpha }_1}{\cos {\alpha }_2}$
• B. $\frac{a\sin {\alpha }_1}{\cos {\alpha }_2}$
• C. $\frac{a\cos {\alpha }_2}{\cos {\alpha }_1}$
• D. $\frac{\cos {\alpha }_1}{\cos {\alpha }_2}$

Answer 1

The correct option is A $\frac{a\cos {\alpha }_1}{\cos {\alpha }_2}$


Let the acceleration of $\text{B}$ be $b$ and it will move up the inclined plane with inclination ${\alpha }_2$.

The direction perpendicular to contact for $\text{A}$ and $\text{B}$ is the horizontal direction, as shown in the diagram.

$\Rightarrow$ From wedge constraint, the component of acceleration of $\text{A}$ and $\text{B}$ along a direction perpendicular to contact surface will be equal.

Thus, equating component of $a$ and $b$ in horizontal direction.

$\Rightarrow a\cos {\alpha }_1=b\cos {\alpha }_2$

$\therefore b=\frac{a\cos {\alpha }_1}{\cos {\alpha }_2}$

$\begin{array}{c}Ans.A\end{array}$

$\begin{array}{c}\text{Why this question}?\\ \text{Tip: Observe the geometry and apply wedge}\\ \text{constraint relationship for acceleration}\\ \text{of A and B.}\end{array}$