Almost Physics II - Normal Force

Post Reply
momo
Golden Member
Golden Member
Posts: 0
Joined: Mon Nov 04, 2013 7:10 pm

My posts are covering basic physics laws, I hope readers are aware that I am still newbie and that what you read in my post is far away to replace the Marshals, Explorers and Rescue team teaching advises through the organized trips. I discourage readers to keep the content of this posts as a standard when driving as the desert may be an hazardous terrain, the beautiful straight dune that you might see on a drawing maybe completely different in real. So please, what you read is just there for general culture.


Normal Force

It is important to know what normal force is. We often ignore it when we think about the gravitational force, but it remains there whenever there is an object sitting on a surface. If this force doesn't exists, any object will sink into that surface.

Definition: The normal force FN on the object is equal but in opposite direction to the gravitational force Fg applied on the object. In other words, it is the force that the ground applies on your car to resist to the gravitation force so that your car doesn't sink into the ground.
[ATTACH=CONFIG]1890[/ATTACH]

Normal force (FN) = mass (m) * gravitational acceleration (g) (9.81 m/s2)

So if we consider the gravitation force positive, the normal force will be negative and reversal.
This is only true for a horizontal ground. Why? What happens gradually between horizontal and nearly vertical uphill?

[ATTACH=CONFIG]1891[/ATTACH][ATTACH=CONFIG]1892[/ATTACH]

Well a third force F1 appears, it is the result of the difference of two force vectors (FN – Fg). Well that is the major force which drags you down a dune. We will discuss the effect of this force in the next thread. But it is good to know that at horizontal ground this force F1 is null but near vertical hill this force is at its maximum value and equal almost the gravitational force, the normal force will be at this moment close to zero.

For math enthusiast, normal force on a dune calculated following: N = m * g * cosine θ
θ equal the dune approach angle in degree unit.

I will be explaining more in the next thread with schema the gravitational, normal and auxiliary forces that play a major role in a car roll-over.

To memorize: The Normal Force decreases when the dune approach angle is becoming high, letting the creation of a new forces.

Regards,
Mohanned

Any correction or clarification are most welcome.
Links:
http://en.wikipedia.org/wiki/Normal_force
http://programmedlessons.org/VectorLessons/vch07/vch07_5.html
Alok708
Newbie
Newbie
Posts: 0
Joined: Mon Jun 04, 2012 11:24 pm
Location: Dubai

Hey Momo,

Great learning's. You are virtually putting me back into my physics classes back in college (not that I have great memories of physics sessions!!).

Feeling hungry for more, keep sharing.

Cheers!
momo
Golden Member
Golden Member
Posts: 0
Joined: Mon Nov 04, 2013 7:10 pm

Alok708;35191 wrote:Hey Momo,

Great learning's. You are virtually putting me back into my physics classes back in college (not that I have great memories of physics sessions!!).

Feeling hungry for more, keep sharing.

Cheers!
Alok708, it is a pleasure to read your feedback. I wish some members who dislike my posts could write to me a PM or post the reason, then we can improve the articles and make them a good learning base for all members.

Cheers
Mark B
Marshal
Marshal
Posts: 1
Joined: Sun Mar 23, 2014 10:50 am

Mono. Don't stop with the technical explanations. We all whiteness what happens out there and some have a feel for it, but don't discount the value of a technical explanation. Personally I find the posts useful. Thanks.
Mark B
Marshal
Marshal
Posts: 1
Joined: Sun Mar 23, 2014 10:50 am

Momo. Sorry for the misspelling of your handle.
momo
Golden Member
Golden Member
Posts: 0
Joined: Mon Nov 04, 2013 7:10 pm

Mark B;35424 wrote:Momo. Sorry for the misspelling of your handle.


I have missed your two last posts, thanks anyway.

Regards
Post Reply