Normal Force of Car Hitting A Bump

[Hypnos_16]

Homework Statement

A car moving at 13.5 m/s encounters a bump that has a circular cross-section with a radius of 56.8 m. What is the normal force exerted by the seat of the car on a 70.8 kg passenger when the car is at the top of the bump?

v = 13.5
r = 56.8
m = 70.8kg

Homework Equations

So i know the forces acting on a object on the inside of a Loop, i know it being on the outside and hitting a bump is a factor but to me it seems like it should just be

Fn = m(v^2 / r) + mg

The Attempt at a Solution

in using that formula I've come up with
Fn = m(v^2 / r) + mg
Fn = 70.8(13.5^2 / 56.8) + (70.8 x 9.81)
Fn = 70.8(3.21) + 695
Fn = 227 + 695
Fn = 922N

but that's not the right answer,
can someone help me out?

 

[PhanthomJay]

Homework Statement

A car moving at 13.5 m/s encounters a bump that has a circular cross-section with a radius of 56.8 m. What is the normal force exerted by the seat of the car on a 70.8 kg passenger when the car is at the top of the bump?

v = 13.5
r = 56.8
m = 70.8kg

Homework Equations

So i know the forces acting on a object on the inside of a Loop, i know it being on the outside and hitting a bump is a factor but to me it seems like it should just be

Fn = m(v^2 / r) + mg

[

You have a signage error. Draw a free body diagram of the passenger and identify the forces acting on the passenger including the gravity and normal forces. In which direction do these forces act? What is the direction of the net force and acceleration at the top of the bump?

 

[Hypnos_16]

In a circle the acceleration is towards the middle, in the case that the man is on top of a bump wouldn't the force of gravity also be towards the middle?
meaning that that all the forces are being applied downwards?

 

[PhanthomJay]

In a circle the acceleration is towards the middle

yes, toward the center of the circle

in the case that the man is on top of a bump wouldn't the force of gravity also be towards the middle?

yes. again this is correct

meaning that that all the forces are being applied downwards?

Incorrect, you are missing a force. Acceleration is not a force. In addition to the gravity force, there is a contact force between the seat and the passenger, the Normal force. What is the direction of the normal force? (hint: calculate the gravity force (mg) and centripetal force (mv²/r). The centripetal force is the net force acting on the passenger).

 

[Hypnos_16]

So then it's the Centripetal Force, which is the net force acting on the car, minus the gravity, since Normal Force opposes Gravity.

 

[PhanthomJay]

You've got your plus and minus signs mixed up.
Per Newton 2, F_net =ma, and since a =v²/r, then F_net = mv²/r. Normal contact forces push toward an object. In this case , the normal force of the seat on the passenger points up (it pushes toward the passenger). Gravity forces act down. The net force, which is the algebraic sum of those 2 forces, acts down, since centripetal net forces always act in toward the center of the circle. So try that equation again. You actually have the correct magnitude of the Normal force, but you should understand in what direction it acts.