Newton’s Second Law of Motion states, “In an inertial frame of reference, the net force on a body is equal to the product of the body’s mass and its acceleration”, i.e. $$F=ma \tag{1}$$ $$\begin{align*} & \text{where,} \\ & m=\text{ mass of the body}\end{align*}$$ Net force due to multiple forces is equal to the vector sum of all of them, i.e. $$\vec{F_{net}}= \vec{F_{1}}+\vec{F_{2}}+…+\vec{F_{n}}\tag{2}$$ In the system for the experiment,

1. Equations of forces on the vehicle:
   $$T- F_f=Ma$$ $$F_N=F_{g1}$$ $$\begin{align*} &\text{where},\\&T= \text{Tension on the string}\\ &F_f= \text{Frictional force on the vehicle}\\ &M= \text{Mass of the vehicle}\\ &a= \text{Acceleration of the system}\\ &FN= \text{Normal force on the vehicle}\\ &F_{g1}=Mg= \text{Weight of the vehicle} \end{align*}$$
2. Equations of forces on the hanger: $$F_{g2}-T=ma,\\ \begin{align*} &\text{where,}\\ &F_{g2}= \text{Weight of the hanger and discs}\\ &m= \text{Mass of the hanger and discs}\\ \end{align*} $$ On solving for a, by eliminating T, we get $$ a=\frac{m- µM}{m+Mg} $$