How Do You Calculate Normal Force on an Inclined Plane?

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To calculate the normal force on an object on a 5-degree incline, the gravitational force acting on the object is 5.47 N, derived from its weight of 0.558 kg. Since friction is ignored, the normal force and the component of gravitational force perpendicular to the incline must balance out. The correct trigonometric function to use is cosine, leading to the equation N = 5.47 * cos(5). This reflects the relationship between the normal force and the gravitational force acting on the object as it slides down the incline. Understanding these principles is essential for accurately determining the normal force in inclined plane scenarios.
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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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