What is the force of the boom on the car at the top of circle?

[gijungkim]

Homework Statement

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?

a.3.7 kN (Down)

b.1.3 kN (Down)

c.6.3 kN (Up)

d.3.7 kN (Up)

e.5.2 kN (Down)

Homework Equations

Fc = mv^2/r
Fg = mg = 5000N
m = weight / g = 510kg

The Attempt at a Solution

Since the car is at the top of the circle, I thought it would be
Fc = Fb + Fg
Fc = 510 * (5)^2 / 10 = 1275 N
Fb = Fg - 1275 N = 3725 N
So I thought the answer was 3725 N to the down way, but the answer says it is d which is 3.7kN to the up way.
Can you tell me what I did wrong?

 

[haruspex]

Your Fc=Fb+Fg is correct if you are taking down as positive throughout. But how did that equation become Fb = Fg - 1275N?

 

[gijungkim]

Your Fc=Fb+Fg is correct if you are taking down as positive throughout. But how did that equation become Fb = Fg - 1275N?

Oh I'm so dumb haha so since Fb = 1275N - Fg = -3725 N and I picked down as positive, the force is up 3725 N right??

 

[haruspex]

Oh I'm so dumb haha so since Fb = 1275N - Fg = -3725 N and I picked down as positive, the force is up 3725 N right??

Yes.

 

[gijungkim]

Yes.

Thank you so much!