How Do You Calculate Normal Force on an Inclined Plane?

  • Thread starter Thread starter Vanity
  • Start date Start date
  • Tags Tags
    Normal Strength
Click For Summary

Homework Help Overview

The discussion revolves around calculating the normal force acting on an object placed on a 5-degree inclined plane. The original poster provides the weight of the object and its gravitational force, seeking assistance in determining the normal force in this context.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between gravitational force and normal force on an incline, questioning which trigonometric functions apply. There is a focus on the forces acting on the object and how they relate to the incline's angle.

Discussion Status

Some participants have provided insights into the trigonometric relationships relevant to the problem, while others have pointed out potential misunderstandings regarding the angles involved. The discussion is ongoing, with various interpretations being explored.

Contextual Notes

The original poster is working under the assumption that friction can be ignored, and there is a need to clarify the correct angle to use in calculations. The presence of a diagram has been noted, which may aid in visualizing the problem.

Vanity
Messages
14
Reaction score
0
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
 
Physics news on Phys.org
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 :(
 

Attachments

  • physic01.jpg
    physic01.jpg
    16.2 KB · Views: 521
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)
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
660
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
8
Views
3K
  • · Replies 20 ·
Replies
20
Views
7K
  • · Replies 12 ·
Replies
12
Views
5K