Homework Help: Normal Strenght

1. Apr 20, 2005

Vanity

Hi!

I'm doing my physic 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!
- Alex

2. Apr 20, 2005

whozum

If its a flat surface that the object is sitting on and theres no other forces on it, then that should be good.

3. Apr 20, 2005

Vanity

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... ;)

4. Apr 20, 2005

whozum

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.

5. Apr 20, 2005

Vanity

Yes, I've join a picture of what it looks like (see attachments).

That's what I'm looking for :(

Attached Files:

• physic01.jpg
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6. Apr 20, 2005

Vanity

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

Thanks! :)

7. Apr 20, 2005

whozum

What function relates the hypotenuse (path) with the opposite (force) ?

8. Apr 20, 2005

Vanity

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, wich brings me to find the acceleration with Ftotal = [weight] * [acce.] ;)

Let's hope I'm right. Thanks alot for the tips! :rofl:

9. Apr 20, 2005

whozum

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)