Normal force calculation

In summary, the normal force is defined as the force that acts perpendicularly on an object when it touches a surface. In terms of magnitude, the normal force is equal to the weight force, but in opposite direction. This can be confusing when using scalar equations, but when using vector equations, it is clear that the normal force and weight force are opposite in direction. Additionally, the weight force is exerted by the Earth's gravitational field on the mass, while the normal force is exerted by the surface on the object. Therefore, they are two separate forces and can be added together in equations.
  • #1
ballooza
14
0
if the normal force defined as the force that acts on an object when it touches the surface perpendicularly, wouldn't it be at the opposite direction of the Weight (W) as a vector and with the same absolute value.. so that the formula would be : F(n)= -mg
if this formula is correct (which is surely), so how come it's being calculated like this:
the normal force acting on a 70 kg-person would be
FN = - (70 kg)(-9.8 m/s2) = 686 N
while the weight of the same person is:
F=mg
= 70 * 9.8 = 686 N
1)shouldn't be the weight positive while the normal force is negative?
2) could i be scientifically correct, if i say that the normal force is considered the Re-action force due to the act of the weight of the body (u know... body acts force on surface while surface acts force on body)
while action and reaction forces should be applied to one single point??
background: when we start solving mechanics problems we proof that a body is stable when the net force= zero by formula: net force= F(n)+W+F+F(friction)
so how come we can add F(n)+W together while they don't apply thy forces to the same one point?
 
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  • #2
When talking about weight it is understood the the vector is pointing downward.

Regarding "the normal force" you can avoid confusion by specifying the object upon which the force acts - since there are TWO normal forces when two objects touch - one on each object. Of course, they are equal in magnitude and opposite in direction.
 
  • #3
ballooza said:
if the normal force defined as the force that acts on an object when it touches the surface perpendicularly, wouldn't it be at the opposite direction of the Weight (W) as a vector and with the same absolute value.. so that the formula would be : F(n)= -mg
if this formula is correct (which is surely), so how come it's being calculated like this:
the normal force acting on a 70 kg-person would be
FN = - (70 kg)(-9.8 m/s2) = 686 N
while the weight of the same person is:
F=mg
= 70 * 9.8 = 686 N
1)shouldn't be the weight positive while the normal force is negative?
2) could i be scientifically correct, if i say that the normal force is considered the Re-action force due to the act of the weight of the body (u know... body acts force on surface while surface acts force on body)
while action and reaction forces should be applied to one single point??
background: when we start solving mechanics problems we proof that a body is stable when the net force= zero by formula: net force= F(n)+W+F+F(friction)
so how come we can add F(n)+W together while they don't apply thy forces to the same one point?
1)Well, when you are doing the addition of forces, you need to take note of the direction of the forces. Hence, whether is the force negative depends on which direction you take as postive. Einstein's Relativity.
2)Well, the weight of an object is the gravitational force which the Earth acts on a body and the normal force is also exerted by the surface on the body. Since both forces act on the same body, they can be added together. Its just a misconception on your part that the weight force is exerted by the body.
 
  • #4
ballooza said:
if the normal force defined as the force that acts on an object when it touches the surface perpendicularly, wouldn't it be at the opposite direction of the Weight (W) as a vector and with the same absolute value.. so that the formula would be : F(n)= -mg
if this formula is correct (which is surely), so how come it's being calculated like this:
the normal force acting on a 70 kg-person would be
FN = - (70 kg)(-9.8 m/s2) = 686 N
while the weight of the same person is:
F=mg
= 70 * 9.8 = 686 N
1)shouldn't be the weight positive while the normal force is negative?
You need to distinguish between scalar equations and vector equations. The scalar equation [itex]N=W=mg[/itex] suggests that the magnitude of the normal force is equal to the magnitude of the weight force. In vector form, the equation becomes [itex]\mathbf{N}=-\mathbf{W}=-m\mathbf{g}[/itex]. So the normal force is the same magnitude as the weight force and lies in the opposite direction.
ballooza said:
2) could i be scientifically correct, if i say that the normal force is considered the Re-action force due to the act of the weight of the body (u know... body acts force on surface while surface acts force on body)
while action and reaction forces should be applied to one single point??
No. The gravitational vector field [itex]\mathbf{g}[/itex] assigns a weight force [itex]m\mathbf{g}[/itex] to any mass [itex]m[/itex]. According to Newton's Second Law, the mass [itex]m[/itex] exerts an equal and opposite reaction force [itex]\mathbf{R}=-m\mathbf{g}[/itex] on the Earth's centre of mass, not to be confused with the normal force [itex]\mathbf{N}[/itex] which is applied to surface [itex]m[/itex].
 
  • #5
jdstokes said:
No. The gravitational vector field assigns a weight force to any mass . According to Newton's Second Law, the mass exerts an equal and opposite reaction force on the Earth's centre of mass, not to be confused with the normal force which is applied to surface .



now i got like two theories.. the first that joyful presented and which i understand that there was really a misconception... but u, jdstokes, i didn't understand urs because it appeared to me like there were three forces, the first that the mass exerts an equal and opposite reaction force on the Earth's centre of mass, the second that the Earth is exerting on body (W)... the third which is the force that's applied to surface (normal force)... so please i want u to come again explaining more to me :smile:
because i surely got u wrong...
 
  • #6
ballooza said:
jdstokes said:
No. The gravitational vector field assigns a weight force to any mass . According to Newton's Second Law, the mass exerts an equal and opposite reaction force on the Earth's centre of mass, not to be confused with the normal force which is applied to surface .



now i got like two theories.. the first that joyful presented and which i understand that there was really a misconception... but u, jdstokes, i didn't understand urs because it appeared to me like there were three forces, the first that the mass exerts an equal and opposite reaction force on the Earth's centre of mass, the second that the Earth is exerting on body (W)... the third which is the force that's applied to surface (normal force)... so please i want u to come again explaining more to me :smile:
because i surely got u wrong...

There are two forces acting on each of two bodies---four forces in total. The gravitational field exerts a force [itex]\mathbf{W}[/itex] on the mass [itex]m[/itex]. Likewise, [itex]m[/itex] exerts an equal and opposite reaction force [itex]-\mathbf{W}[/itex] on the Earth mass [itex]M[/itex]. In addition, the surface of the Earth (or table or whatever) exerts a normal force [itex]\mathbf{N}[/itex] on the surface of [itex]m[/itex], which exerts an equal and opposite force [itex]-\mathbf{N}[/itex] on the surface of the earth. So the nett force on [itex]m[/itex] is [itex]\Sigma\mathbf{F} = \mathbf{N} - \mathbf{W} =0 [/itex] and the nett force on [itex]M[/itex] is [itex]\Sigma\mathbf{F} = \mathbf{W} - \mathbf{N} =0 [/itex].
 
  • #7
Thx jdstokes for helping me to clear up this misconception.
 
  • #8
well, after trying hard to analyze the forces the way u explained to me i think ur mistaken when
u were saying that net force acting on (m) is (N - W) because the force (-W) is acting on earth
mass (M) not (m), and so as the other net force formula... anyway i tried to figure it out with
a sketch which is attached so i hope u take a deep look at it because i had a conclusion written on
the down-right corner of the sketch... NOTE: wherever i mention F(m), it means the net force
acting on (m)...
believe me,,, i really want to know the truth of this whole thing so please help! so just
tell me if it's right or wrong... if wrong, please correct me...
but i understand that u ment by N-W and W-N to clarify the opposite vectors, isn't that right?
 

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  • #9
Finally, you got it right.
 
  • #10
magnitudes vs vectors

I think you are confusing yourself with plus and minus signs and directions.

And your diagram is oddly drawn. Why does it show W and "-W" both acting on m? And N and "-N" both acting on m? That makes no sense.

What are the forces on m? First we have its weight W (actor: the earth) pulling down. (The reaction force is the mass m pulling the Earth up.) Note that W is just the magnitude of the weight: the direction is downward.

Another force on m is the normal force N pushing m up (actor: the surface). (The reaction force is the mass pushing down on the surface.) Again note that N is the magnitude of the normal force: the direction of the force on m is upward.

Treating W and N as positive numbers (they are just magnitudes of those forces), and choosing up to be the positive direction, the net force on m is: [itex]F_{net} = N - W[/itex].

Of course, if you represented the normal force on m with the vector [itex]\vec{N}[/itex], and the weight of m with the vector [itex]\vec{W}[/itex], then you could say that [itex]\vec{F_{net}} = \vec{N} + \vec{W}[/itex].
 
  • #11
WELL, GUYS, I CAN'T BELIEVE I MADE IT...
but i got to tell you doc that's how i studied it in books... i believe in that now that they r so odd but how should it be?
in other words, please can u show me how could i draw them?
after thati can rest in peace
 
  • #12
You're quite correct ballooza, I stuffed up the signs. The vector equations should be [itex]\Sigma\mathbf{F} = \mathbf{N} + \mathbf{W}[/itex] for [itex]m[/itex] and [itex]\Sigma\mathbf{F} = -(\mathbf{N} + \mathbf{W})[/itex] for [itex]M[/itex]. Your diagram should have [itex]\mathbf{N}[/itex] and [itex]-\mathbf{N}[/itex] drawn in opposite directions with their tails centred at the interface of the block and the Earth. [itex]\mathbf{W}[/itex] should originate from the block's centre of mass and point downward, whereas [itex]-\mathbf{W}[/itex] should point up, originating from the Earth's centre of mass. I agree the signs are confusing. If you understand why this drawing is so, then you have got it.
 
  • #13
Just make sure you know which direction you are taking as positive, the negative signs on the formula is trivial.
 
  • #14
thank u people... i got it, but I'm going to make another sketch to correct those vectors, but i'll consider that vectors of forces start from the point of the centre mass of the body, if u know what i mean.. but i might be late because I'm very busy these days... so don't wait much... but i'll come again this week..
thanx again
 
  • #15
It's not safe to assume that normal force always equals the negative of gravity force. normal force isn't alway opposite from the ground! so you cannot claim that FN=-ma. don't post false equations please
 
  • #16
Please note the date on the original posts.
 
  • #17
I do. but the people viewing it may not
 

What is the Normal Force?

The normal force is a contact force that acts perpendicular to the surface of an object in contact with another surface. It arises from the electromagnetic interactions between the atoms and molecules of the two surfaces and prevents objects from passing through each other. In essence, it is the force exerted by a surface to support the weight of an object placed on it.

How is the Normal Force Calculated?

The normal force can be calculated using the following formula:

\[N = mg\cos(\theta)\]

  • \(N\) is the normal force.
  • \(m\) is the mass of the object.
  • \(g\) is the acceleration due to gravity (approximately 9.81 m/s² on the surface of the Earth).
  • \(\theta\) is the angle between the object's weight vector (vertical) and the surface it rests on. If the object is on a horizontal surface, \(\theta = 0^\circ\) and \(\cos(\theta) = 1\).

When the object is on a horizontal surface, the formula simplifies to \(N = mg\).

How Does the Normal Force Vary in Different Situations?

The normal force can vary in different situations:

1. On a Horizontal Surface:

When an object rests on a horizontal surface, the normal force is equal in magnitude but opposite in direction to the object's weight. It cancels out the gravitational force, resulting in a net force of zero in the vertical direction.

2. On an Inclined Surface:

If the object is on an inclined surface, the normal force is still perpendicular to the surface but is less than the object's weight. It can be calculated using the formula \(N = mg\cos(\theta)\), where \(\theta\) is the angle of inclination.

3. In Free Fall:

During free fall or when an object is in mid-air, the normal force is typically zero because there is no supporting surface. In such cases, the only force acting on the object is gravity.

Why is Calculating the Normal Force Important?

Calculating the normal force is important in physics and engineering to understand how objects interact with surfaces. It plays a crucial role in determining whether an object remains at rest or in equilibrium on a surface and helps in analyzing forces and motion in various physical scenarios, including mechanics, structural engineering, and materials science.

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