Solution:
$$
\begin{aligned}
&\text { Here, }[a]=L T^{-2}=\left(a \beta^{-2}\right) \\
&{[v]=L T^{-1}=a \beta^{-1}} \\
&\therefore \quad a=L, \beta=T \\
&{[L]=a \gamma} \\
&\therefore \quad \gamma=\frac{[L]}{\alpha}=\frac{L}{L}=1
\end{aligned}
$$
Coefficient of friction,
\(\mu=\frac{F}{R}=M^{0} L^{0} T^{0}\) i.e. dimensionless
Now, \(\alpha^{0} \beta^{0} \gamma^{-1}=L^{0} T^{0}(1)^{-1}=1\),
which is dimensionless.