A $4000$kg truck is parked on a $15°$slope. How big is the friction force on the truck? The coefficient of static friction between the tires and the road is $0.90$.

The friction force is equal to the product of the coefficient of friction and the normal force. The force that stops one solid object from sliding or rolling over another is known as friction. Frictional forces, such as the traction needed to walk without slipping, can be beneficial, but they can cause a lot of resistance to movement.

Let the horizontal weight force is $W_y=W\cos \left(\theta \right)$and vertical weight force is $W_x=W\sin \left(\theta \right)$.

According to Newton's second law of motion, the friction force can be written as $f_r=\mu N$

Here we are given

localid="1648206782774" $$\begin{gathered} \mu =0.9 \\ m=4000kg \\ g=9.8m/s^2 \end{gathered}$$

Substitute the vales , here,$f_r$is the friction force,$\mu$is the coefficient of static between friction force,$N$is the normal force

$$\begin{gathered} f_r=\mu N \\ {f}_r=\mu .W\sin \left(\theta \right) \\ {f}_r=\mu .mg\sin \left(\theta \right) \\ {f}_r=0.9\times 4000\times 9.8\times \sin 15° \\ {f}_r=9,140.45N \end{gathered}$$