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

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.