Inactive Tutor answered 10/15/23
The gravitational force on the truck is mg sin(θ).
To stop the parked truck from sliding, the static frictional force must be equal in magnitude.
Plug in m = 5000 kg, g = 9.8 m/s2 and θ = 7° to get
(5000)(9.8) sin(7°) = 5971.7 N
Answer (rounded to 2 significant figures)
The frictional force on the truck is 6.0 × 103 N