Question

An amusement ride consists of a car moving in a vertical circle on the end of a rigid boom. The radius of the circle is 10 m. The combined weight of the car and riders is 5.0 kN. At the top of the circle the car has a speed of 5.0 m/s which is not changing at that instant. What is the force of the boom on the car at the top of the circle?

Answer 1

Force of boom= 1275.5N

Step-by-step explanation:

Combined weight of car and rider=5.0KN= 5000N.

Radius of circleR=10m

EF=ma

Fc= W + FB

FB= (mv^2/r) - W

Fc is centripetal force

Mass= Weight/ g= 5000/9.8

m=510.2Kg

If car's speed is 5m/s

FB= (510.2× 5^2/10) - 5000

FB=( 12755/10) - 5000

FB= 1275.5 - 5000

FB= -3724.5N

Fc= W + FB

Fc= 5000 + (-3724.5)

Fc= 1275.5N