Question

In the figure shown blocks $A$ and $B$ are kept on a wedge $C.A,B$ and $C$ each have mass $m$. All surfaces are smooth. Find the acceleration (in $m/s^2$) of $C$.

Answer 1

Step1: Draw a free body diagram.

Step2: Find the acceleration of $C$.

Consider free body diagram for mass $A$.

There is a pseudo force ma acting on the
masses $A,B$ if we take reference frame fixed
to $C$.
From $FBD$ of block $A$,
$N_A+m\times a_c\cos {53}^0=mg\cos {37}^0$

$N_A=mg\cos {37}^0-m_{ac}\cos {53}^0\dots \left(i\right)$

Now consider $FBD$ for mass $B$,

$N_B=mg\cos {53}^0+m{a}_c\cos {37}^0\dots \dots \left(ii\right)$

Consider the wedge $C$.

The net horizontal force on the wedge
$=N_A\cos {53}^0-N_B\cos {37}^0$
Substitute the values of $N_A$ and $N_B$, we get
$(mg\cos {37}^{\circ }-m{a}_c\cos {53}^0)\cos {53}^0-(mg\cos {53}^0+m{a}_c\cos {37}^0)\cos {37}^0=m{a}_c$

Therefore,
$m_{a_c}(1+{\cos }^2{53}^0+{\cos }^2{37}^0)$
$=mg(\cos {37}^0\cos {53}^0-\cos {53}^{\circ }\cos {37}^0)$

Thus,
$m{a}_c=0$
Or
$a_c=0$

Final Answer :(0)