A 1400-kg sports car (including the driver) crosses the rounded top of a hill (radius = 87.0 m ) at 18.0 m/s .
a) Determine the normal force exerted by the road on the car.
b)Determine the normal force exerted by the car on the 65.0- kg driver.
c)Determine the car speed at which the normal force on the driver equals zero.

Respuesta :

(a) The normal force exerted by the road on the car is 8,506.2 N.

(b) The normal force exerted by the car on the driver is 394.9 N.

(c) The speed of the car at which the normal force on driver is zero is 29.2 m/s.

Normal force exerted by the road on the car

The normal force exerted by the road on the car is calculated as follows;

Normal force = weight of the car - centripetal force of car

Weight of the car = (1400 x 9.8) = 13,720 N

Centripetal force of the car = (1400 x 18²)/87 = 5,213.8 N

Normal force = 13,720 N -  5,213.8 N

Normal force = 8,506.2 N

Normal force exerted on the driver

Normal force = weight of driver - centripetal force of driver

Weight of driver = (65 x 9.8) = 637 N

Centripetal force of driver = (65² x 18²)/(87) = 242.07 N

Normal force =  637 N - 242.07 N = 394.9 N

Speed at which normal force on driver is zero

N = mg - mv²/r

0 = mg - mv²/r

mv²/r = mg

v²/r = g

v² = rg

v = √rg

v = √(87 x 9.8)

v = 29.2 m/s

Thus, the normal force exerted by the road on the car is 8,506.2 N.

The normal force exerted by the car on the driver is 394.9 N.

The speed of the car at which the normal force on driver is zero is 29.2 m/s.

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