Inclined planes, Banked highways, and Normal force

  • Thread starter trogdor5
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  • #1
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Alright, I don't have a specific homework problem here, just a general question. I've attached two pages that I will be referencing.

Figure 1. Inclined Plane
2Oi7y.jpg


As can be seen here, the FN = mgcos(∅)

Figure 2. Banked Curve
YFoXe.jpg


It seems as if the opposite is true here. FNcos(∅) = mg

The same holds true for the horizontal components. My question is, what is happening here? Why is there different notation all of a sudden. I completely understand the proof for the banked curve I just don't follow the inclined plane problem. It seems as if the two should be identical. What am I missing here?

Any effort would be extremely appreciated! Honestly, the work the guys who answer questions here is amazing and I really plan on contributing to the community when I feel I'm qualified to help!
 

Answers and Replies

  • #2
haruspex
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Look at the third 'force' in each case.
For the inclined plane, the third force is perpendicular to the normal force. You can think of it as assisting the normal force in opposing gravity. To eliminate the third force you resolve parallel to the normal force.
For the banked curve, the third force (the resultant in this case) is the one required to keep the car going around a vertical axis, so is horizontal. Some of the normal force goes into providing that, and only what's left over gets to oppose gravity. To eliminate the third force you resolve vertically.
 
  • #3
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Oh wow, thank you very much! I get it now
 

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