Question

Two fixed frictionless inclined plane making the angles ${30}^{\circ }$ and ${60}^{\circ }$ with the horizontal are shown in the figure. Two blocks $\text{A}$ and $\text{B}$ are placed on the two planes. What is the relative vertical acceleration of $\text{A}$ with respect of $\text{B}?$

• A. $4.9{\text{ms}}^{-2}$ in vertical direction
• B. $9.8{\text{ms}}^{-2}$ in vertical direction
• C. zero
• D. $19.6{\text{ms}}^{-2}$ in vertical direction

Answer 1

The correct option is A $4.9{\text{ms}}^{-2}$ in vertical direction

FBD of blocks $\text{A}$ and $\text{B}$:


From the FBD of blocks $\text{A}$ and $\text{B}$, the acceleration of blocks $\text{A}$ and $\text{B}$ along the inclined plane is,

$a_1=\frac{m_{A}g\sin {60}^{\circ }}{m_A}=g\sin {60}^{\circ };$

$a_2=\frac{m_{B}g\sin {30}^{\circ }}{m_B}=g\sin {30}^{\circ }$


The relative acceleration of $\text{A}$ w.r.t. $\text{B}$ in vertical direction is given by

${\left(a_{rel}\right)}_y=a_1\cos {30}^{\circ }-a_2\cos {60}^{\circ }$

$\Rightarrow {\left(a_{rel}\right)}_y=\frac{3g}{4}-\frac{g}{4}=\frac{2g}{4}$

$\Rightarrow {\left(a_{rel}\right)}_y=\frac{g}{2}=4.9{\text{m/s}}^2$

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