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A car (m=1700 kg) is parked in a road that rises 15 degrees above the horizontal. What are the magnitudes of
a. the normal force
b. the static frictional force that the ground exerts on the tires?

Sagot :

A) You need to work out the forces perpendicular to the road. So we can say:
[tex]Normalforce = (1700\times9.8) \times sin(15) \\ \therefore Normalforce = 4311.93N (2 d.p.)[/tex]
This is the normal force acting against the car.

B) This is very similar to the working out of part A. We can say that since the car is not moving, something is stopping the car from moving forwards. The friction is doing this. We can summarise that the force of the friction is acting parallel to the road. 
We can say:
[tex](1700\times9.8)\times cos(15) = 16092.32N (2d.p.)[/tex]
The is the amount of friction is newtons that is stopping the car from moving forward.]

Hoped this wasn't too confusing, and that it helped!

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