Question

Find the accelerations of rod $A$ and wedge $B$ in the arrangement shown in figure above if the ratio of the mass of the wedge to that of the rod equals $\eta$, and the friction between all contact surfaces is negligible.

• A. $w_A=\frac{g}{(1+\eta \cot \alpha )}$, $w_B=\frac{1+g}{(\tan \alpha +\eta \cot \alpha )}$
• B. $w_A=\frac{g}{\left(2\eta {\cot }^2\alpha \right)}$, $w_B=\frac{1+g}{(\tan \alpha +\eta \cot \alpha )}$
• C. $w_A=\frac{2g}{\left(\eta {\cot }^2\alpha \right)}$, $w_B=\frac{g}{(1+\tan \alpha +\eta \cot \alpha )}$
• D. $w_A=\frac{g}{(1+\eta {\cot }^2\alpha )}$, $w_B=\frac{g}{(\tan \alpha +\eta \cot \alpha )}$

Answer 1

Answer: (D)

$w_A=\frac{g}{(1+\eta {\cot }^2\alpha )}$, $w_B=\frac{g}{(\tan \alpha +\eta \cot \alpha )}$

Let us draw free body diagram of each body, i.e. of rod $A$ and of wedge $B$ and also draw the kinematical diagram for accelerations, after analysing the directions of motion of $A$ and $B$. Kinematical relationship of accelerations is
$\tan \alpha =\frac{w_A}{w_B}$ (1)
Let us write Newton's second law for both bodies in terms of projections having taken positive directions of $y$ and $x$ axes as shown in figure below.
$m_{A}g-N\cos \alpha =m_A{w}_A$ (2)
and $N\sin \alpha =m_B{w}_B$ (3)
Simultaneous solution of (1), (2) and (3) yields
$w_A=\frac{m_{A}g\sin \alpha }{m_A\sin \alpha +m_B\cot \alpha \cos \alpha }=\frac{g}{(1+\eta {\cot }^2\alpha )}$ and
$w_B=\frac{w_A}{\tan \alpha }=\frac{g}{(\tan \alpha +\eta \cot \alpha )}$
Note: We may also solve this problem using conservation of mechanical energy instead of Newton's second law.