# A 4000 kg truck is parked on a 15 degrees slope. How big is the friction force on the truck? [Friction Numerical]

Friction Numerical Question) A 4000 kg truck is parked on a 15 degrees slope. How big is the friction force on the truck?

**Solution**

The forces which act on the truck are shown in the Figure above.

With m = 4000 kg and θ =15 degrees, the downward force of gravity mg has been decomposed into components along and perpendicular to the slope as we’ve done before. There is a normal force from the road with magnitude N.

The truck has no acceleration so there is no net force and so there must be a force going up the slope, which of course is from static friction, denoted by F_{stat}.

For the forces along the slope to cancel, we must have

F_{stat}= mg sin θ = (4000 kg)(9.80 m/s^{2}) sin 15° = 1.01 × 10^{4}N

**So the magnitude of the friction force is 1.01 × 10 ^{4} N.**