Answer
a) The force exerted by the ground on the front wheels is $1.57\times10^4N$
b) The force exerted by the ground on the rear wheels is $5.74\times10^4N$
Work Step by Step
The forces that act on the truck is
- The weight of the truck and its contents $W=(7460kg)\times(9.8m/s^2)=7.31\times10^4N$, pointing downward.
- Normal force exerted on front wheels $N_f$, pointing upward.
- Normal force exerted on rear wheels $N_r$, pointing upward.
Choose the rotation axis to be at the rear wheels, so that $N_r$ creates no torque.
- For $W$, the lever arm is $0.63m$, and $\tau_W\lt0$ because the torque created is clockwise.
- For $N_f$, the lever arm is $0.63+2.3=2.93m$, and $\tau_f\gt0$ because the torque created is counterclockwise.
Because the truck is in equilibrium,
1) The net torque is zero: $$\tau_f-\tau_W=0$$ $$2.93N_f-0.63W=0$$ $$N_f=\frac{0.63W}{2.93}=1.57\times10^4N$$
2) The net force is zero: $$N_f+N_r-W=0$$ $$N_r=W-N_f=5.74\times10^4N$$
