Confused about Newton's 3rd law

[dyn]

Hi.
I am confused about the use of Newton's 3rd law in the following situation. If someone steps horizontally out of a skyscraper with zero vertical velocity. Then obviously they fall downwards. But if i apply Newton's 3rd law ; persons weight act downwards on the air so doesn't the air apply an equal force on the person acting upwards resulting in the person basically remaining stationary in the vertical direction ; ie. just "floating " there. Where am i going wrong ?
Thanks

 

[Nugatory]

There is a force pair between the faller and the air. The upwards force of the air on the faller is equal to and opposite to the force of the faller on the air.

There is a force pair from gravity between the faller and the earth. The downwards force of gravity on the faller is equal to and opposite to the upwards force of gravity on the earth.

Until the faller reaches terminal velocity the downwards force of gravity is greater than the upwards force of air resistance so the faller accelerates downwards. Eventually the faller reaches terminal velocity and the two forces balance - but acceleration is changing speed so no further acceleration means that the faller continues to move down at a constant speed.

 

[Dale]

persons weight act downwards on the air

The person’s weight acts downwards on the person. The weight is a gravitational interaction with the earth, the air is not involved at all. The third law pair for the weight is an equal and opposite gravitational force from the person pulling up on the earth.

The interaction with the air is the air resistance or drag. The air exerts drag upwards on the person and the person exerts an equal and opposite force on the air.

 

[dyn]

There is a force pair between the faller and the air. The upwards force of the air on the faller is equal to and opposite to the force of the faller on the air.

There is a force pair from gravity between the faller and the earth. The downwards force of gravity on the faller is equal to and opposite to the upwards force of gravity on the earth.

Until the faller reaches terminal velocity the downwards force of gravity is greater than the upwards force of air resistance so the faller accelerates downwards. Eventually the faller reaches terminal velocity and the two forces balance - but acceleration is changing speed so no further acceleration means that the faller continues to move down at a constant speed.

Is the force of the faller on the air the same as the downwards force of gravity on the faller ?

 

[Dale]

Is the force of the faller on the air the same as the downwards force of gravity on the faller ?

Only at terminal velocity. However, even then they do not form a 3rd law pair.

 

[dyn]

If I consider a free body diagram of the person. At t=0 when the person has no vertical velocity does the diagram just have bodyweight acting downwards as the only force ? As the person falls does the diagram have weight acting downwards and air resistance acting upwards ? But at no point are these 2 forces a 3rd law pair ?
Have I got this all correct ?
Thanks

 

[Dale]

Yes, that is all correct!

 

[dyn]

Is the force of the faller on the air the same as the downwards force of gravity on the faller ?

Only at terminal velocity. However, even then they do not form a 3rd law pair.

Thanks for replying. Before terminal velocity what is the force of the faller on the air if it is not the same as the person's weight ?

 

[Nugatory]

Before terminal velocity what is the force of the faller on the air if it is not the same as the person's weight ?

It is some fairly complicated function of the speed. Often something of the form $F=Kv^a$ with $K$ and $a$ constants determined by wind tunnel measurements will be a pretty good approximation.

 

[dyn]

Thanks to both of you for your help

 

[sophiecentaur]

It is important not to confuse the notion of Third Law Pairs with Equilibrium. For a rocket in space, there is a third law pair in the force of ship on the mass of ejected material and an equal force of the material on the ship. In that situation, there is no equilibrium and the ship accelerates (without limit). The same situation in air will have two pairs and an equilibrium situation can arise at terminal velocity.

 

[tech99]

It is important not to confuse the notion of Third Law Pairs with Equilibrium. For a rocket in space, there is a third law pair in the force of ship on the mass of ejected material and an equal force of the material on the ship. In that situation, there is no equilibrium and the ship accelerates (without limit). The same situation in air will have two pairs and an equilibrium situation can arise at terminal velocity.

I do understand the confusion people get into over the Third Law, where it seems impossible for an object to move. But it appears to me that an object moving at any speed in a fluid exerts a force on its molecules, and the molecules exert an equal force, or reaction, on the object? This seems to be Third Law, just like lab experiments using dynamics trolleys.

 

[Nugatory]

But it appears to me that an object moving at any speed in a fluid exerts a force on its molecules, and the molecules exert an equal force, or reaction, on the object?

There is; those forces form a third-law pair. That's been stated several times in posts further up in this thread.

However, there is no particular reason why the force of the air on the object should be equal and opposite to the force of gravity on the object - those aren't a third law pair, clearly because they're both acting on the same thing. All we can say is that if those forces do happen to be the same, the net force on the object will be zero so the object will not change its speed.

 

[tech99]

There is; those forces form a third-law pair. That's been stated several times in posts further up in this thread.

However, there is no particular reason why the force of the air on the object should be equal and opposite to the force of gravity on the object - those aren't a third law pair, clearly because they're both acting on the same thing. All we can say is that if those forces do happen to be the same, the net force on the object will be zero so the object will not change its speed.

Thank you for that clarification.

 

[sophiecentaur]

I do understand the confusion people get into over the Third Law, where it seems impossible for an object to move. But it appears to me that an object moving at any speed in a fluid exerts a force on its molecules, and the molecules exert an equal force, or reaction, on the object? This seems to be Third Law, just like lab experiments using dynamics trolleys.

For an object moving through a fluid (or over a friction surface) there is just a single third law pair. This force pair is due to friction and can transfer energy both to the fluid / surface and to the object. This is not an equilibrium situation but N3 (present in all situations) has nothing to do with that.

 

[Mister T]

persons weight act downwards on the air so doesn't the air apply an equal force on the person acting upwards resulting in the person basically remaining stationary in the vertical direction

A net force of zero doesn't imply a velocity of zero. It implies an acceleration of zero.

 

[dyn]

A net force of zero doesn't imply a velocity of zero. It implies an acceleration of zero.

I did state that the person started off with no velocity in the vertical direction which would mean no net force would remain at a velocity of zero

 

[Nugatory]

I did state that the person started off with no velocity in the vertical direction which would mean no net force would remain at a velocity of zero

But there is a net force when the person start outs: the upwards force of air resistance is zero but the downwards force of gravity is not.

Thus the person accelerates downwards while the speed and air resistance both increase. Only when the speed has increased enough that the air resistance is as strong as the gravitational force is there no net force.

 

[zanick]

don't forget buoyancy… that's an upward force acting on the body falling equal to the weight of the air it displaced.

 

[cjl]

True, but that's quite small relative to weight in most cases.

 

[zanick]

yes, in most cases.. but still 1.2kg per cubic meter certainly not 0. :)

 

[sophiecentaur]

yes, in most cases.. but still 1.2kg per cubic meter certainly not 0. :)

The problem with introducing more and more fine details is that we can lose the basic picture. Before moving into 'practical' situations we need to consider the very simplest situations.

 

[Mister T]

yes, in most cases.. but still 1.2kg per cubic meter certainly not 0. :)

But we're talking about this case. The gravitational force exerted on a human is large compared to the buoyant force. If you want to worry about such details take the effect of Earth's spin into account, that effect is still small enough to ignore but larger than the effect of buoyancy.