Question

In the arrangement shown in figure above the masses $m$ of the bar and $M$ of the wedge, as well as the wedge angle $\alpha$ are known. The masses of the pulley and the thread are negligible. The friction is absent. Find the acceleration of the wedge $M$.

Answer 1

To analyse the kinematic relations between the bodies, sketch the force diagram of each body as shown in figure below.
On the basis of force diagram, it is obvious that the wedge $M$ will move towards right and the block will move down along the wedge. As the length of the thread is constant, the distance travelled by the block on the wedge must be equal to the distance travelled by the wedge on the floor. Hence $d{s}_{mM}=d{s}_M$. As $\overrightarrow{v_{mM}}$ and $\overrightarrow{v_M}$ do not change their directions and accelerations that's why $\overrightarrow{w_{mM}}\upuparrows \overrightarrow{v_{mM}}$ and $\overrightarrow{w_M}\upuparrows \overrightarrow{v_M}$ and $w_{mM}=w_M=w$ (say) and accordingly the diagram of kinematical dependence is shown in figure below.
As $\overrightarrow{w_m}=\overrightarrow{w_{mM}}+\overrightarrow{w_M}$, so from triangle law of vector addition.
$w_m=\sqrt{w_M^2+w_{mM}^2-2w_{mM}{w}_M\cos \alpha }=w\sqrt{2(1-\cos \alpha )}$ (1)
From $F_x=m{w}_x$, (for the wedge),
$T=T\cos \alpha +N\sin \alpha =Mw$ (2)
For the bar $m$ let us fix (x-y) coordinate system in the frame of ground Newton's law in projection form along x and y axes (figure shown below) gives
$mg\sin \alpha -T=m{w}_{m\left(x\right)}=m[w_{mM\left(x\right)}+w_{M\left(x\right)}]$
$=m[w_{mM}+w_M\cos (\pi -\alpha )]=mw(1-\cos \alpha )$ (3)
$mg\cos \alpha -N=m{w}_{m\left(y\right)}=m[w_{mM\left(y\right)}+w_{M\left(y\right)}]=m[0+w\sin \alpha ]$ (4)
Solving the above equations simultaneously, we get
$w=\frac{mg\sin \alpha }{M+2m(1-\cos \alpha )}$
Note: We can study the motion of the block $m$ in the frame of wedge also, alternatively we may solve this problem using conservation of mechanical energy.