is 4000N the right answer to this problem?

[neonerd]

A 1000-kg sports car moving at 20 m/s crosses the rounded top of a hill (radius = 100 m). Determine the normal force on the car.

I got 4000N, just wanted to make sure i was doing this right

 

[chuckywang]

well, 4000N = mr^2/v, but somehow I can't help but think that you have to factor g = 9.8 m/sec^2 in there somehow.

 

[neonerd]

Originally posted by: chuckywang
well, 4000N = mr^2/v

wait, that equals Msmg..

 

[chuckywang]

I think its mg-mr^2/v. The faster you're going, the less the normal force....

 

[neonerd]

mr being mass times radius?

 

[TuxDave]

Originally posted by: chuckywang
well, 4000N = mr^2/v, but somehow I can't help but think that you have to factor g = 9.8 m/sec^2 in there somehow.

I'm thinking along those lines. The question is a little vague too. Is it asking what is the normal force that the car is exerting onto the hill or what.

 

[TechnoKid]

Originally posted by: TuxDave

Originally posted by: chuckywang
well, 4000N = mr^2/v, but somehow I can't help but think that you have to factor g = 9.8 m/sec^2 in there somehow.

I'm thinking along those lines. The question is a little vague too. Is it asking what is the normal force that the car is exerting onto the hill or what.

prob at the crest of the hill

 

[TheLonelyPhoenix]

You're halfway there. 4000 N is the NET force downward. However, the downward force on the car from gravity is 1000*g, which is 98000 N. Thus, the normal force is the difference between the two.

 

[chuckywang]

Originally posted by: TheLonelyPhoenix
You're halfway there. 4000 N is the NET force downward. However, the downward force on the car from gravity is 1000*g, which is 98000 N. Thus, the normal force is the difference between the two.

Yay. I got it right.

<- Haven't taken physics in 2 years.

 

[TheLonelyPhoenix]

Originally posted by: chuckywang

Originally posted by: TheLonelyPhoenix
You're halfway there. 4000 N is the NET force downward. However, the downward force on the car from gravity is 1000*g, which is 98000 N. Thus, the normal force is the difference between the two.

Yay. I got it right.

<- Haven't taken physics in 2 years.

Haven't taken any kinematic physics since high school. Had to pray to a voodoo god to remember as much as I did.

 

[neonerd]

how can the NET force downward though be less than the force from gravity? wuldn't the force from gravity be less than the NET force?

 

[TuxDave]

Originally posted by: neonerd
how can the NET force downward though be less than the force from gravity? wuldn't the force from gravity be less than the NET force?

The net force is 4000N. The difference between that and what gravity is doing is due to the force that the ground is exerting on the car.

 

[chuckywang]

Originally posted by: neonerd
how can the NET force downward though be less than the force from gravity? wuldn't the force from gravity be less than the NET force?

Fgravity - Fnormal = 4000

Think about going round a hill. The faster you go, the more you're be "trying to fly off the hill". If you go fast enough, then you'll fly off the hill at the top, so then N = 0 in that case.

 

[neonerd]

so the answer would be 5800N?

 

[chuckywang]

Originally posted by: neonerd
so the answer would be 5800N?

If I were taking a test, that's the answer I'd put.

 

[neonerd]

Originally posted by: chuckywang

Originally posted by: neonerd
how can the NET force downward though be less than the force from gravity? wuldn't the force from gravity be less than the NET force?

Fgravity - Fnormal = 4000

Think about going round a hill. The faster you go, the more you're be "trying to fly off the hill". If you go fast enough, then you'll fly off the hill at the top, so then N = 0 in that case.

so if the car was going at 31.30 m/s, it would fly off the hill?

 

[EyeMWing]

Originally posted by: neonerd

Originally posted by: chuckywang

Originally posted by: neonerd
how can the NET force downward though be less than the force from gravity? wuldn't the force from gravity be less than the NET force?

Fgravity - Fnormal = 4000

Think about going round a hill. The faster you go, the more you're be "trying to fly off the hill". If you go fast enough, then you'll fly off the hill at the top, so then N = 0 in that case.

so if the car was going at 31.30 m/s, it would fly off the hill?

Yes.

 

[neonerd]

ok, thank you

 

[TheLonelyPhoenix]

Originally posted by: chuckywang

Originally posted by: neonerd
so the answer would be 5800N?

If I were taking a test, that's the answer I'd put.

No no no....... the answer is 94000N

 

[TheLonelyPhoenix]

Force of Gravity = 1000 * 9.8 = 98000 N
Net force = 4000 N

Net Force = Force of Gravity - Normal Force

Do teh math.

 

[neonerd]

just a b it confused on this section and easily get lost among all the equations....especially in this particular section....stuck on another problem, but w/e, I'll get some help from the teacher tommorrow

 

[chuckywang]

Originally posted by: TheLonelyPhoenix
Force of Gravity = 1000 * 9.8 = 98000 N
Net force = 4000 N

Net Force = Force of Gravity - Normal Force

Do teh math.

Enron called. They want you back.

 

[CarlKillerMiller]

hm. I always thought that it had something to do with r(mu)g

 

[Einz]

Originally posted by: chuckywang

Originally posted by: TheLonelyPhoenix
Force of Gravity = 1000 * 9.8 = 98000 N
Net force = 4000 N

Net Force = Force of Gravity - Normal Force

Do teh math.

Enron called. They want you back.

He's got it right, save that extra 0 on the end.
Sum of forces = 0
Fg + Fn + Fc = 0
Fg = -9.8*1000 = -9800N
Fc = m*V^2/r, = 4000N
Fn = 9800-4000 = 5800 N

 

[chuckywang]

Originally posted by: chuckywang

Originally posted by: TheLonelyPhoenix
Force of Gravity = 1000 * 9.8 = 98000 N
Net force = 4000 N

Net Force = Force of Gravity - Normal Force

Do teh math.

Enron called. They want you back.

c'mon! How can nobody find this funny??