Normal Force on the top of a Loop-the-Loop


by Earn Success
Tags: force, looptheloop, normal
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#1
Dec13-10, 03:03 PM
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"A roller coaster car does a loop-the-loop. When it is upside down at the very top, which of the following is true?"

The normal force and the weight are in opposite directions.
The normal force and the weight are perpendicular to each other.
The weight is zero.
The normal force and the weight are in the same direction.

My Physics book doesn't really talk about this, and I need to see an example of something before I can dissect it, so I'm not the most analytical person.

All I know is that when there is a surface such as a table, the weight and the N are opposite. If the roller coaster car is upside down, it doesn't even exert a weight on the track at all so where is the normal force...

Sorry, I know this is simple stuff, but I am really struggling in this class simply because sometimes these forces feel imaginary...
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#2
Dec13-10, 03:11 PM
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All I know is that when there is a surface such as a table, the weight and the N are opposite.
The normal force exerted by a surface is always 'out' of the surface towards the object. Since the table is under the weight, the normal force it exerts on the weight must be upward.
If the roller coaster car is upside down, it doesn't even exert a weight on the track at all so where is the normal force...
In the roller coaster case, the normal force is not equal to the object's weight as it is for an object at rest on a table. Regardless, since the 'surface' (the track) is above the car, in what direction must the normal force on the car act?
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#3
Dec13-10, 03:45 PM
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Okay, so the way I'm beginning to see it is that because the FN is just a contact force, and does not depend on the weight of an object, unless that object is at rest. Even if there is an acceleration, the FN is there - as long as there is contact between that object and a surface.

When the object is at rest, the [tex]\Sigma[/tex]F is = 0 because it is at rest, and therefore you can say that the FN is equal and opposite to the weight.

So what if the car was fastened to the top of that loop-the-loop. It would be at rest, so would the [tex]\Sigma[/tex]F still be 0?

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#4
Dec13-10, 04:06 PM
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Normal Force on the top of a Loop-the-Loop


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Dec13-10, 04:50 PM
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Okay, so the way I'm beginning to see it is that because the FN is just a contact force, and does not depend on the weight of an object, unless that object is at rest. Even if there is an acceleration, the FN is there - as long as there is contact between that object and a surface.
OK.
When the object is at rest, the [tex]\Sigma[/tex]F is = 0 because it is at rest, and therefore you can say that the FN is equal and opposite to the weight.
Yes. For an object sitting on a table, the acceleration is zero and thus ΣF = 0. In that case, Fn = mg.

So what if the car was fastened to the top of that loop-the-loop. It would be at rest, so would the [tex]\Sigma[/tex]F still be 0?
If the car were fastened to the top of the loop and not accelerating, then sure ΣF = 0. Whatever is fastening it to the track would provide the upward force to counter gravity.

But in these loop-the-loop problems they usually want you to pretend that the car isn't fastened to the loop, so the only force acting on it besides gravity would be the normal force.

(Don't bump threads after 20 minutes! )
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#6
Dec14-10, 03:53 PM
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Ok thanks for all the help. Big test tommorrow, if I get a 90+ all my classes will be above 90 :P


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