Why are there two different equations for normal force on an angle?

AI Thread Summary
The discussion revolves around the two different equations for normal force on an inclined plane: Fn = mg cos(θ) and Fn = mg/cos(θ). The first equation applies when an object slides down a slope under the influence of gravity alone, where the normal force is the y-component. The second equation is relevant in scenarios where additional forces act on the object, such as a car navigating a banked curve, where the normal force acts as the hypotenuse of a triangle formed with weight and centripetal force components. The confusion arises from the change in axis orientation and the context of the forces involved. Understanding the situation and visualizing the forces with a diagram is crucial for clarity.
student34
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Homework Statement



I feel like I am being told that there two different kinds of normal force.

1st. In grade 11 we were always told that normal force on an angle equals m*g*cosθ, where Fn equals the y-component, and θ is the angle between Fn and mg.

2nd. Now in university we are told that the normal force can also equal (m*g)/cosθ, where θ is still the angle between Fn and mg, but mg is the y-component and Fn is the hypontuse.

What is going on? The only difference that I see is changes to what the x and y axises are. Why does changing the x and y axises change the ratio of Fn and mg?
 
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Normal force is a force of constraint, it keeps the object moving on the constrained path. It depends on the situation. When an object slides along a slope, with only gravity acting on it, the normal force is Fn=mgcos(θ) where θ is the angle of inclination (also the angle of the normal with respect to the vertical).

What was the other problem? If there are other forces, the normal force can be different. For example, if the object on the slop is pressed against the slope by some force F, and it is stationary, the normal force is mg/cos(θ)

ehild
 
ehild said:
Normal force is a force of constraint, it keeps the object moving on the constrained path. It depends on the situation. When an object slides along a slope, with only gravity acting on it, the normal force is Fn=mgcos(θ) where θ is the angle of inclination (also the angle of the normal with respect to the vertical).

What was the other problem? If there are other forces, the normal force can be different. For example, if the object on the slop is pressed against the slope by some force F, and it is stationary, the normal force is mg/cos(θ)

ehild

Yes, the 2nd normal force where normal force is mg/cos(θ) is about a car taking a banked corner. I am really confused. Why does this problem make Fn the hypotnuse and mg the vertical component? And why do we change x and y axises to do it?
 
A drawing always helps.The car drives along a horizontal circle. That means the net force acting on it is horizontal, and equal to the centripetal force, Fcp. Gravity is vertical so the centripetal force is provided by the horizontal component of the normal force while the vertical component balances the weight of the car. If you add the vectors W and N you get a horizontal vector. See the yellow triangle: N is the hypotenuse.

ehild
 

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ehild said:
A drawing always helps.The car drives along a horizontal circle. That means the net force acting on it is horizontal, and equal to the centripetal force, Fcp. Gravity is vertical so the centripetal force is provided by the horizontal component of the normal force while the vertical component balances the weight of the car. If you add the vectors W and N you get a horizontal vector. See the yellow triangle: N is the hypotenuse.

ehild

Ah, I get it, thanks.
 
You are welcome. Do not forget to draw next time:-p

ehild
 
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