A4000 kg truck is parked on a 15 degree slope. how big is the friction force on the truck? the coefficient of static friction between the tires and the road is 0.90.

fr = 10145 N

Explanation:

To solve this exercise let's use Newton's second law

     F = ma

Let's write a reference system, where the x-axis is parallel to the slope and the y-axis is perpendicular

Let's decompose the weight that is not in the direction of the axes

     cos θ = Wy / W

      sin θ = Wx / W

      Wy = W cos θ

      Wx = W sin θ

We write Newton's second law on each axis

Case 1. Truck stopped on the upward slope

X axis

       fr - Wx = 0

       fr = Wx

YAxis

      N- Wy = 0

      N = Wy

The friction force has the equation

      fr = μ N

Let's calculate the force of friction

      fr = W sin θ

      fr = mg sin θ

      fr = 4000 9.8 without 15

      fr = 10145 N

Case2 Truck stopped on the descending slope

X axis

      Wx -fr = 0

      fr = Wx

      fr = W sin θ = m g sin 15

      fr = 10145N

We can see that the value of the friction force is the same regardless of the direction of the slope

because if the tool that you are using is not balanced it there would be false mass.