rruck shown in figure below:

For ${?}^{in}$ e enter of mass is located at the middle of the distance between front and rear tires. $4.$ nustion of motion for translational and rotational motion will be as follows: feach tire equal to $30\frac{N}{m}.$

a$)$

$M{x}^{}+65x+35=0$

$I{}^{}+35x+180=0$

$M\stackrel{}{x}+90x+45=0$

$\frac{1}{b}ar()+45x+1275=0$

c$)$

$M\stackrel{}{x}+60x+30=0$

$I{}^{}+30x+85=0$

d$)$

$M{x}^{}+155x=0$

$l{}^{}+95=0$ Solve and draw fbd