Misinterpretation of Newton's Third Law

[Kyoma]

Homework Statement

A man standing on the ground is subjected to 2 forces : weight and normal reaction force. Do they form an action-reaction pair of forces? (That's Newton's Third Law).

I put yes, but I was wrong. I don't get it. I'm new to Newton's Third Law.

 

[tiny-tim]

Hi Kyoma!

An action-reaction pair of forces (in https://www.physicsforums.com/library.php?do=view_item&itemid=373") is a force from A on B and a force from B on A

Your two forces are both from A on B

 

[PeterO]

Homework Statement

A man standing on the ground is subjected to 2 forces : weight and normal reaction force. Do they form an action-reaction pair of forces? (That's Newton's Third Law).

I put yes, but I was wrong. I don't get it. I'm new to Newton's Third Law.

You were distracted by the two forces being equal in magnitude and opposite in direction.

To show that those forces do NOT make a 3rd law couple, all you have to do is put a back pack on the man.
The man himself is no heavier, but the Normal reaction force will now be greater.

Another way is to show they are not a 3rd Law couple is to have the man stand on an incline. Then the normal force won't be in the opposite direction to the weight force [and hopefully a friction force has appeared to stop him slipping over]

Finally, both those forces you mentioned act on the man. For Newtons 3rd law couple the two forces act on different bodies.

The man pushes down on the floor - the floor pushes up on the man.

The Earth pulls the man down - the man pulls the Earth up.

generally the pairs can be described as

Body A acts on Body B - Body B acts on Body A

 

[davidm1732]

Was that the complete question? If so, I'd agree with your answer, and respectfully disagree with the two previous posts...

PeterO...if you put a backpack on the man... he (well the combination of him + backpack) does weigh more... which is why the reaction force increases... If he didn't, and the reaction force increased, you'd have a force imbalance and he'd accelerate upwards...

Tiny-tim... by definition, a reaction force is equal and opposite to the applied force (his weight in this example)...so both forces are not acting in the same direction. Draw a free body diagram of it...again, they have to be opposing, or else you wouldn't have static equilibrium.

PeterO...if you put him on an incline, yes the normal force is now not equal and opposite to the weight, but you have an additional vector component (the frictional force) that when added to the normal reaction force, is equal to the weight...again, if it were not, he'd be accelerating in one direction (down the slope in this case)

PeterO...you also give the exact example as the original question, but argue the opposite answer?...

The man pushes down on the floor - the floor pushes up on the man.

cheers
dsm

 

[Doc Al]

PeterO...if you put a backpack on the man... he (well the combination of him + backpack) does weigh more... which is why the reaction force increases... If he didn't, and the reaction force increased, you'd have a force imbalance and he'd accelerate upwards...

Yes, the normal force and weight will both increase, but they are still not 3rd law pairs.

Tiny-tim... by definition, a reaction force is equal and opposite to the applied force (his weight in this example)...so both forces are not acting in the same direction. Draw a free body diagram of it...again, they have to be opposing, or else you wouldn't have static equilibrium.

Irrelevant.

PeterO...if you put him on an incline, yes the normal force is now not equal and opposite to the weight, but you have an additional vector component (the frictional force) that when added to the normal reaction force, is equal to the weight...again, if it were not, he'd be accelerating in one direction (down the slope in this case)

PeterO...you also give the exact example as the original question, but argue the opposite answer?...

The question is are the weight and the normal force 3rd law pairs, not whether they add to zero. The answer is no. PeterO gave a good example where the weight remains the same but the normal force differs. 3rd law pairs are always equal and opposite.

 

[tiny-tim]

A man standing on the ground is subjected to 2 forces : weight and normal reaction force. Do they form an action-reaction pair of forces? (That's Newton's Third Law).

Tiny-tim... by definition, a reaction force is equal and opposite to the applied force (his weight in this example)...so both forces are not acting in the same direction. Draw a free body diagram of it...again, they have to be opposing, or else you wouldn't have static equilibrium.

Yes, but they're not an action-reaction pair.

There are two action-reaction pairs:

i] the force of the man on the ground, and the force of the ground on the man.

ii] the weight of the man (ie the gravitational force from the Earth), and the gravitational force from the man on the Earth.

 

[PeterO]

Was that the complete question? If so, I'd agree with your answer, and respectfully disagree with the two previous posts...

PeterO...if you put a backpack on the man... he (well the combination of him + backpack) does weigh more... which is why the reaction force increases... If he didn't, and the reaction force increased, you'd have a force imbalance and he'd accelerate upwards...

Tiny-tim... by definition, a reaction force is equal and opposite to the applied force (his weight in this example)...so both forces are not acting in the same direction. Draw a free body diagram of it...again, they have to be opposing, or else you wouldn't have static equilibrium.

PeterO...if you put him on an incline, yes the normal force is now not equal and opposite to the weight, but you have an additional vector component (the frictional force) that when added to the normal reaction force, is equal to the weight...again, if it were not, he'd be accelerating in one direction (down the slope in this case)

PeterO...you also give the exact example as the original question, but argue the opposite answer?...cheers
dsm

dsm: In order of comments:

1.
Putting a back pack on the man does not increase the weight of the man himself. My example was to show that the weight of the man and the reaction force from the floor are not even equal when the man is loaded up with something - like a back pack.
We could never put a limit on Newtons Third law which said "unless he is wearing a back pack", so we should never have thought this was an example of Newtons 3rd law in the first place.

2.
The weight of a man is the force of the Earth attracting a the man, it is NOT the force of the man on the ground. Take the ground away and the man still has weight - and as a result will accelerate down at a rapid rate until he meets a new ground - painfully probably
Newtons 3rd law couples can never give equilibrium - they act on different bodies.
Now a whole bunch of 3rd law couples might collectively hold one or more bodies in equilibrium

3.
OK so on the incline you have an extra force [same when you put the back pack on - the man was subjected to 2 downward forces; his weight and the backpack pushing down]
The fact that there are now need 2 forces to balance his weight means firstly, nether one of those extra 2 forces can constitute a 3rd law couple with his weight, and also that neither one of them is even in the opposite direction to the weight force.

4.
You need you read more carefully
The original question posed a question about the weight of the man [a force acting on the man] and the Normal Reaction force of the floor on the man [another force acting on the man.

My examples were 3rd Law couples

The Earth pulls the man down - the man pulls the Earth up
The first force act on the man, the second force acts on the Earth

The man pushes on the floor - the floor pushes on the man
The first force acts on the floor, the second force acts on the man.

Note that the Earth and the floor are not even the same body. One is possibly made of wood and the other is mostly molten rock with a crisp outer crust.

2 of those 4 forces act on the man - one from the first couple and one from the second couple.
The question asked was do these two constitute a 3rd law couple [were they an example of Newtons Third Law]. The answer is NO.

EDIT: David let's agree that at 12:00 Grenwich Mean time on Septemebr 6th - you will push South on a tree in a park with a Force of exactly 100N. I will be pushing with a force of 100N North on a tree in a Park near me at that time. Both those forces will be acting on the Earth [due to the trees growing in/on the Earth]. They will be equal in magnitude and opposite in direction. Will they constitute an action-reaction pair? I don't really think either of us actually have to got to the Park on the 6th.

 

[davidm1732]

So I was incorrect... and hence the original answer stands...this isn't an example of Newtons third law. A quick reference link to a NASA site is included below which gives another example of this, and specifically explains why this force pair does not fall unders Newtons third...

http://www-istp.gsfc.nasa.gov/stargaze/SNewton3.htm

I think the wording is key in this problem (illustrated by PeterO's fourth point...)


However... getting tangential now... if in free fall, does the man still have weight? He has mass, and he's subject to a force that'll accelerate him due to gravity, but how do we measure weight... by measuring the reaction force, which there is none! Perhaps that's best left for another thread...

As far as the park... I think that would constitute a pair, (you have to be more specific now about which force pair!). My pushing on you, and you pushing back on me would be a pair, wouldn't it? Not sure that the tree between us affects things.

Thanks
interesting discussion
dsm

 

[PeterO]

So I was incorrect... and hence the original answer stands...this isn't an example of Newtons third law. A quick reference link to a NASA site is included below which gives another example of this, and specifically explains why this force pair does not fall unders Newtons third...

http://www-istp.gsfc.nasa.gov/stargaze/SNewton3.htm

I think the wording is key in this problem (illustrated by PeterO's fourth point...)However... getting tangential now... if in free fall, does the man still have weight? He has mass, and he's subject to a force that'll accelerate him due to gravity, but how do we measure weight... by measuring the reaction force, which there is none! Perhaps that's best left for another thread...

As far as the park... I think that would constitute a pair, (you have to be more specific now about which force pair!). My pushing on you, and you pushing back on me would be a pair, wouldn't it? Not sure that the tree between us affects things.

Thanks
interesting discussion
dsm

last point first - I was very specific,; I said action-reaction pair, so unless you can come to my park in Australia and we leave out the tree it will not be an example.

Now for free fall.
The person in free fall certainly has weight, but they might not experience that weight in the usual way.
We usually notice our weight by the reaction forces from our surrounds.
You probably don't really feel the Earth pulling you down, you are more likely to notice the chair pushing up.
When you get into a lift to go up, the floor pushes on you with a force greater than your weight - so you think "wow, I feel heavy now!", but just fro a little while.
When you are coming back down, you "feel" lighter.
But you don't actually lose weight by riding lifts, you have to exercise and have a better diet.
Apparent weight is the property we have that can change. The apparent weight of an astronaut in orbit is zero. That does NOT mean there is no gravity up there though. Typically g = 8.6 - 8.9 depending on how high the orbit is.

Oh, and I didn't need NASA to explain it to me.

Here is a series of questions to hopefully cement the Newtons 3rd law in your mind.

They relate to the example of a person doing a standing jump onto a chair. They are all True False - be careful, use Newton's third Law and read carefully.

In order to jump off the floor, the person has to push harder on the floor than the floor pushes on him?

Once in the air, the force of the air on the man balances his weight?

When the man lands on the chair, the chair pushes harder on the man than the man pushes on the chair, in order to stop him?

Actually seat part of the chair was only made of cane, so could not support him, and his feet broke through the chair.

He broke the chair because he pushed harder on the chair that the chair pushed on him?Four questions on a common theme - and they have something else in common too. what is it?