How Do You Calculate Normal Force on an Inclined Plane?

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SUMMARY

The discussion focuses on calculating the normal force acting on an object placed on a 5-degree inclined plane. The gravitational force is determined to be 5.47 N based on the object's weight of 0.558 kg. Participants clarify that the normal force (N) can be calculated using the cosine function, specifically N = 5.46 * cos(5), which is equivalent to N = 5.46 * sin(85). The conversation emphasizes the importance of using the correct angle in trigonometric calculations to accurately determine the normal force.

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Hi!

I'm doing my physics lab and I need to find the normal strenght applied on an object. Here's the info I have:

weight: 0,558 kg
gravitational strenght: 0,558 * 9,8 = 5,47 N

Anybody can help me out? Thanks! :smile:
- Alex
 
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If its a flat surface that the object is sitting on and there's no other forces on it, then that should be good.
 
whozum said:
If its a flat surface that the object is sitting on and there's no other forces on it, then that should be good.
the surface is not flat, sorry i forgot to mention that. It's on 5 degrees ~slope~, and it's "slipping" on it. The question says to ignore the friction, so no other forces beside the normal and gravitationnal one... ;)
 
So the object is on a 5 degree incline, do you know which trig function will compensate for this? gravity is working in the vertical direction, and the object is sliding down the 'hypotenuse' of our imaginary triangle.
 
whozum said:
So the object is on a 5 degree incline [...] gravity is working in the vertical direction, and the object is sliding down the 'hypotenuse' of our imaginary triangle.
Yes, I've join a picture of what it looks like (see attachments).

whozum said:
do you know which trig function will compensate for this?
That's what I'm looking for :(
 

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Okay, image has been aproved now.. (sorry for the double post)..
Anybody can help me out from here?

Thanks! :)
 
What function relates the hypotenuse (path) with the opposite (force) ?
 
whozum said:
What function relates the hypotenuse (path) with the opposite (force) ?
Okay, let's see. The object only moves horizontally, so if I add all the forces on Y, it should be 0. So gravitationnal force + normal force (Y) should = 0.

Fg = 0,558 kg * 9,8
= 5,47

N + 5,47Sin(85) = 0
so, N = -5,47Sin(85)

That means the total forces on the object is equal to the force in X, which brings me to find the acceleration with Ftotal = [weight] * [acce.] ;)

Let's hope I'm right. Thanks a lot for the tips! :smile:
 
A few things

Weight means gravitational force.

The decline is at 5 degrees, not 85 degrees. What you'll want to do is stick with the angle of the decline.

Use the property cos(x) = sin(90-x). Basically,

N + 5.46cos(5) = 0
N = 5.46cos(5)

Note cos(5) = Sin(85)
 

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