Question

A Mercedes-Benz 300SL (m = 1550 kg) is parked on a road that rises 10° above the horizontal. (a) What is the magnitude of the normal force? (b) What is the static frictional force that the ground exerts on the tires?

Could you please show your work as well? Thank you!

Answer 1

Weight of car mg is acting vertically downwards, which can be split into its components mgcos10 acting downwards perpendicular to the road

and mgsin10 acting downwards along the road

Fr is the frictional force that counteracts the pulling force mgsin10 (keeps the car from rolling)

N is the normal reaction

Since the car is parked(at rest)

Consider all Vertical Forces

$\sum F_{y}=0$

$N=mgcos10$

$N=1550kg*9.81m/s*Cos10$

$N=1550*9.81*Cos10$

(a)ANSWER: $N=14974.49N$

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Consider all horizontal forces

$\sum F_{x}=0$

$Fr=mgsin10$

$Fr=1550kg*9.81m/s^{2}*sin10$

$Fr=1550*9.81*sin10$

(b) ANSWER: $Fr=2640.41N$

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