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Normal force calculation...

by ballooza
Tags: calculation, force, normal
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ballooza
#1
Sep4-04, 06:08 PM
P: 14
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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Tide
#2
Sep4-04, 06:22 PM
Sci Advisor
HW Helper
P: 3,144
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.
joyful55
#3
Sep4-04, 11:05 PM
P: 22
Quote Quote by ballooza
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.

jdstokes
#4
Sep5-04, 05:12 AM
P: 527
Normal force calculation...

Quote Quote by ballooza
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.
Quote Quote by ballooza
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].
ballooza
#5
Sep6-04, 02:34 AM
P: 14
[QUOTE=jdstokes]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
because i surely got u wrong...
jdstokes
#6
Sep6-04, 04:07 AM
P: 527
[QUOTE=ballooza]
Quote Quote by jdstokes
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
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].
joyful55
#7
Sep7-04, 12:14 AM
P: 22
Thx jdstokes for helping me to clear up this misconception.
ballooza
#8
Sep7-04, 07:58 AM
P: 14
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?
Attached Thumbnails
N force.jpg  
joyful55
#9
Sep7-04, 08:18 AM
P: 22
Finally, you got it right.
Doc Al
#10
Sep7-04, 08:24 AM
Mentor
Doc Al's Avatar
P: 41,475
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].
ballooza
#11
Sep7-04, 10:26 AM
P: 14
WELL, GUYS, I CAN'T BELIEVE I MADE IT........
but i gotta tell ya 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
jdstokes
#12
Sep8-04, 02:11 AM
P: 527
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.
joyful55
#13
Sep10-04, 09:37 AM
P: 22
Just make sure you know which direction you are taking as positive, the negative signs on the formula is trivial.
ballooza
#14
Sep12-04, 03:43 PM
P: 14
thank u people.... i got it, but i'm gonna 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
Sagekeeper
#15
Dec11-10, 09:31 PM
P: 2
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
Studiot
#16
Dec12-10, 03:41 AM
P: 5,462
Please note the date on the original posts.
Sagekeeper
#17
Dec12-10, 10:55 AM
P: 2
I do. but the people viewing it may not


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