A car is on a driveway that is inclined 10 degrees to the horizontal. A force of 490 lb is required to keep the car from rolling down the driveway.

a) Find the weight of the car.

b) Find the force the car exerts on the driveway.

a) weight of the car = 2816,1 lbs

b) 2773 lbs

Step-by-step explanation:

The equilibrium force is 490 lbs. That force keep the car at rest, then

∑ Fy = 0 and ∑Fx = 0

Forces acting on the car:

The external force 490 lbs

weight of the car uknown

Normal force

sin∠10° = 0,174

cos∠10° = 0.985

∑Fx = 0 mg*sin10° - 490 = 0 ∑Fy = 0 mg*cos10° - N = 0

mg*0,174 = 490

mg = 490 / 0,174

mg = 2816,1 lbs

weight of the car = 2816,1 lbs

The Normal force

mg*cos10° - N = 0 2816,1 * 0,985 = N

N = 2773 lbs

Then equal force in magnitude and in opposite direction will car exets on the driveway