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Cheers.

- Thread starter Monster007
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Cheers.

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If you are travelling fast enough the normal force may be larger than your weight at that point.

- #3

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- #4

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The normal force cannot be greater than gravity on a hill because it's a reaction to gravity.

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- #6

A.T.

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The normal ground reaction force adjusts to whatever is necessary to prevent the car from sinking into the ground. On the hill some of the weight force goes into accelerating the car downwards, so the ground reaction force doesn't have to balance that part.

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In a word "acceleration". If the car is not accelerating, then the normal force is the weight. But, if the car is accelerating, which happens on a vertically curved road, the normal force will vary according to the direction of the acceleration.

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Ok, can you potentially explain that further? when the car is exactly on top of the hill, if it were stationery then the normal reaction force would be equal to the weight force, yet when moving it is less than the weight force. Doesn't all the weight force go into accelerating the car down?The normal ground reaction force adjusts to whatever is necessary to prevent the car from sinking into the ground. On the hill some of the weight force goes into accelerating the car downwards, so the ground reaction force doesn't have to balance that part.

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You're thinking: If a car is "sitting" on a flat road, then the normal force must be the weight. The velocity and acceleration of the car are irrelevant.Ok, can you potentially explain that further? when the car is exactly on top of the hill, if it were stationery then the normal reaction force would be equal to the weight force, yet when moving it is less than the weight force. Doesn't all the weight force go into accelerating the car down?

To see the problem with that argument consider two cars "sitting" instantaneously on the road. One has, however, just fallen from a height and has hit the road. It is decelerating rapidly and has a much higher normal force at that instant.

- #10

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Yep that makes sense to me, the car that has fallen from a height is exerting a greater force on the road and therefore the force the road exerts back is higher. But in the case of the car on the hill are you suggesting that the car's acceleration in the vertical direction is decreasing? and therefore the normal is less?You're thinking: If a car is "sitting" on a flat road, then the normal force must be the weight. The velocity and acceleration of the car are irrelevant.

To see the problem with that argument consider two cars "sitting" instantaneously on the road. One has, however, just fallen from a height and has hit the road. It is decelerating rapidly and has a much higher normal force at that instant.

- #11

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Here's a way to look at it. Imagine the car reaching the top of a hill with different speeds.Yep that makes sense to me, the car that has fallen from a height is exerting a greater force on the road and therefore the force the road exerts back is higher. But in the case of the car on the hill are you suggesting that the car's acceleration in the vertical direction is decreasing? and therefore the normal is less?

At high speed, the car leaves the surface and the normal force is zero.

Get the speed just right and the car stays on the surface but with instantaneously 0 normal force.

Design the track just the right shape - parabola - and the car will stay in contact with the track but exert zero normal force. The track matches the natural parabolic path of the car under gravity at that velocity.

Slow the car down and it will try to take a different parabolic path. The car would like to go below the road, but the solid surface prevents it. A normal force is created, keeping the car in a forced path.

The slower the car travels, the greater the difference between the natural curved path it wants to take and the path along the track it is forced to take. Hence the greater the normal force to keep the car on the track.

- #12

A.T.

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No, because the car doesn't accelerate down at 1g, but less than that, depending on it's speed and curvature of the hill.when the car is exactly on top of the hill ...Doesn't all the weight force go into accelerating the car down?

- #13

sophiecentaur

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I can't find a comment in this thread about the (very relevant) sign of the curvature of the track. When the car is going over a 'hill' the curvature is negative so the centripetal force will be of opposite sign . In a valley, thr curvature is positive so the centripetal force will be positive. The net normal force on the car can be zero, on a hump, at which point. the car will no longer follow the track and 'fly off' on a parabolic trajectory. (The design of a real ride would never allow that - with a pretty wide margin, I'd bet)

- #14

A.T.

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Did you mean "negative too" rather than "opposite sign"?When the car is going over a 'hill' the curvature is negative so the centripetal force will be of opposite sign.

- #15

sophiecentaur

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I just meant that the sign of the force is diferent for the valley and for the top of the hill. It can add or subtract from the 'g force' that's experienced.Did you mean "negative too" rather than "opposite sign"?

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Here's a way to look at it. Imagine the car reaching the top of a hill with different speeds.

At high speed, the car leaves the surface and the normal force is zero.

Get the speed just right and the car stays on the surface but with instantaneously 0 normal force.

Design the track just the right shape - parabola - and the car will stay in contact with the track but exert zero normal force. The track matches the natural parabolic path of the car under gravity at that velocity.

Slow the car down and it will try to take a different parabolic path. The car would like to go below the road, but the solid surface prevents it. A normal force is created, keeping the car in a forced path.

The slower the car travels, the greater the difference between the natural curved path it wants to take and the path along the track it is forced to take. Hence the greater the normal force to keep the car on the track.

Yes, this all makes sense conceptually (really well explained) and I am more than happy that the maths agrees nicely, my question is what is it that causes the normal force to be less than the weight force at the top of the hill? I understand that if the car is travelling slowly more normal force will exist because the car wants to head into the surface, but how would i finish this sentence? When the car is travelling quicker the normal reaction force gets smaller because .....

Thanks heaps for all the help. I'm having real trouble with normal reaction forces in situations like this.

- #17

A.T.

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Read up on constraint forces, which adjust to whatever is needed to enforce certain kinematics. On the atomic level the adjustment is varying amount of EM repulsion, depending on the distance between atoms.my question is what is it that causes the normal force to be less than the weight force at the top of the hill?

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... At that point I'd be tempted to trust the mathematics. This situation at the top of the hill exists only instantaneously and even then only in an idealised fashion. It's because of the centripetal acceleration and the curvature of the slope. More of gravity is deployed making the car change direction and less is left to push it into the track.Yes, this all makes sense conceptually (really well explained) and I am more than happy that the maths agrees nicely, my question is what is it that causes the normal force to be less than the weight force at the top of the hill? I understand that if the car is travelling slowly more normal force will exist because the car wants to head into the surface, but how would i finish this sentence? When the car is travelling quicker the normal reaction force gets smaller because .....

.

An easier situation is your weight on an elevator accelerating down. As long as the elevator accelerates the normal force is reduced by precisely the elevator's acceleration times your mass.

Centripetal acceleration is also acceleration. If that's isn't easy to grasp, then trust the mathematics.

- #19

A.T.

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Well, he agrees that it makes sense conceptually and mathematically. So I don't really know what the remaining issue is.At that point I'd be tempted to trust the mathematics.

- #20

pbuk

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... the downwards acceleration is greater so the net downwards force must be greater; the net downwards force is the weight (which is constant) less the normal reaction force so in order for the net downwards force to increase the normal reaction force must decrease.When the car is travelling quicker the normal reaction force gets smaller because .....

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