Question

Solution Let acceleration of ( m ) be ( a_{1} ) (absolute) and that of ( M ) be ( a_{2} ) (absolute). Writing equations of motion only in the directions of ( a_{1} ) or ( a_{2} ) For ( m )
[
m g cos alpha-N=m a_{1}
]
For ( M )
[
N sin alpha=M a_{2}
]
Here, ( N= ) normal reaction between ( m ) and ( M ) As discussed above, constraint equation can be written as,
[
a_{1}=a_{2} sin alpha
]
Solving above three equations, we get
acceleration of rod,
[
a_{1}=frac{m g cos alpha sin alpha}{left(m sin alpha+frac{M}{sin alpha}right)}
]
and acceleration of wedge
[
a_{2}=frac{m g cos alpha}{m sin alpha+frac{M}{sin alpha}}
]

# a A = dg sino Example 4 In the arrangement shown in the figure, the rod of mass m held by two smooth walls, remains always perpendicular to the surface of the wedge of mass M. Assuming all the surfaces to be frictionless, find the acceleration of the rod and that of the wedge. 1 .. 1

Solution