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is 4000N the right answer to this problem?

N

neonerd

Diamond Member
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
 
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.
Click to expand...

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.
 
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.
Click to expand...

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.
Click to expand...

prob at the crest of the hill
 
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.
 
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.
Click to expand...

Yay. I got it right. 🙂

<- Haven't taken physics in 2 years.
 
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.
Click to expand...

Yay. I got it right. 🙂

<- Haven't taken physics in 2 years.
Click to expand...

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

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.
 
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?
Click to expand...

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.
 
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?
Click to expand...

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.
Click to expand...

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

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.
Click to expand...

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

Yes.
 
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 🙂
 
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.
Click to expand...

Enron called. They want you back.
Click to expand...

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
 
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