What is the Meaning of Weight for a Freely Falling Body?

[xCross]

Imagine an object is resting upon the floor. The object is pushing the floor with a force mg, which is vertically downward. The floor is also pushing the object back with the same magnitude of force but it is vertically upward.

My question is: which one of these force is called weight?

My answer is the second one. Because we say that a free falling body is weightless. And then normal force doesn't act on the body when it is in free-fall.

Is my answer correct?

 

[Doc Al]

Imagine an object is resting upon the floor. The object is pushing the floor with a force mg, which is vertically downward. The floor is also pushing the object back with the same magnitude of force but it is vertically upward.

My question is: which one of these force is called weight?

Neither. In US physics textbooks, at least, 'weight' refers to the force of gravity acting on an object. The two forces in your example are just the two halves of the normal force interaction.

My answer is the second one. Because we say that a free falling body is weightless. And then normal force doesn't act on the body when it is in free-fall.

Is my answer correct?

The 'weight' in 'weightlessness' refers to the apparent weight, which is the support force holding up an object, not 'true' weight which is the Earth's gravitational force. But some folks (in the UK, I believe) do use weight to mean apparent weight.

 

[Stonebridge]

Neither. In US physics textbooks, at least, 'weight' refers to the force of gravity acting on an object. The two forces in your example are just the two halves of the normal force interaction.


The 'weight' in 'weightlessness' refers to the apparent weight, which is the support force holding up an object, not 'true' weight which is the Earth's gravitational force. But some folks (in the UK, I believe) do use weight to mean apparent weight.

Yes there is some confusion on this. UK books (at least those I am and have been familiar with over the last 40 or so years) talk correctly about "weight" as being the gravitational force of attraction on an object. They also point out that "weightless" is a bit of a misnomer in the case of an astronaut floating in space.
They also talk about apparent or "experienced" weight and "measured" weight as recorded on a spring balance or scale(s).

In the case of the original poster's question, the forces he mentions are numerically equal to the apparent/experienced or measured (with a spring balance) "weight" of an object. This is and always has been the source of confusion.
So it's a bit of a semantic argument too.
Which force is "called" the weight of an object, and which force is "equal to" the weight of the object.
As educators we need to be very clear.
You are right to point this out.

 

[mastoner20]

In the current AP physics B high school textbook in the U.S. it refers to 'weight' as the Force of Gravity vertically down (towards earth/planet you're on). In the first scenario 'weight' is the force of the object pushing straight down on the floor by gravity. (mg, Fg, or rarely termed w) Normal force (N) is the resultant force gainst gravity as part of Newton's second law (every action there's an equal and opposite reaction). In the scenario in which an object is laying on a flat horizontal plane, -(mg)=N.

Similarly to the first scenario, weight is always the force of gravity straight to the center of the earth, so if the object is on an incline, N=-(Mg)cos$$\Theta$$

 

[dgOnPhys]

In the current AP physics B high school textbook in the U.S. it refers to 'weight' as the Force of Gravity vertically down (towards earth/planet you're on). In the first scenario 'weight' is the force of the object pushing straight down on the floor by gravity. (mg, Fg, or rarely termed w) Normal force (N) is the resultant force gainst gravity as part of Newton's second law (every action there's an equal and opposite reaction). In the scenario in which an object is laying on a flat horizontal plane, -(mg)=N.

Similarly to the first scenario, weight is always the force of gravity straight to the center of the earth, so if the object is on an incline, N=-(Mg)cos$$\Theta$$

Hi mastoner20

this is not exactly accurate: the action and reaction pair (third law) for gravitational interaction in the OP example are the object weight and the opposite pull the object exerts on the planet. The other action-reaction pair are the normal reaction of the floor and the opposite force the object applies to the floor. With a single object this just looks like semantics but if you put your object on a table resting on the floor you will see that this is a necessary distinction: the normal reaction from the floor to the table is not equal to the table's weight anymore...

 

[Doc Al]

In the current AP physics B high school textbook in the U.S. it refers to 'weight' as the Force of Gravity vertically down (towards earth/planet you're on).

Correct. Note that weight is a gravitational force acting on the object.

In the first scenario 'weight' is the force of the object pushing straight down on the floor by gravity.

Incorrect. The force that the object exerts on the floor is the normal force, not the weight. (In many circumstances, the normal force happens to have a value equal to the weight, but they are different forces.)

(mg, Fg, or rarely termed w) Normal force (N) is the resultant force gainst gravity as part of Newton's second law (every action there's an equal and opposite reaction).

No, as dgOnPhys points out, the normal force is not an 'equal and opposite reaction' to the weight. They are not 3rd law pairs.

 

[nikhil khatri]

acc. to Newton law ,every to object in universe attract each other,similarly Earth attract us.so the force with which Earth attract us is our weight,not the force of normal reaction.

my qestion is that,if some one ask our weight then what we should reply either 60kg or 588N.

 

[Doc Al]

my qestion is that,if some one ask our weight then what we should reply either 60kg or 588N.

In everyday life, it doesn't matter. In a physics class, weight is a force so 588N is correct.

 

[Sankalp Sethi]

Because we say that a free falling body is weightless. And then normal force doesn't act on the body when it is in free-fall.

A free falling body is assumed to be weightless as the apparatus measuring the weight too is assumed to be moving with the same velocity and acceleration. For example , a spring balance attached to a metallic sphere shows 0 weight only when both are dropped (are in free fall) . There is no such relation between weight and normal reaction.

 

[jtbell]

The 'weight' in 'weightlessness' refers to the apparent weight, which is the support force holding up an object, not 'true' weight which is the Earth's gravitational force. But some folks (in the UK, I believe) do use weight to mean apparent weight.

When I taught a "descriptive physics" course about twenty years ago, Paul Hewitt's book Conceptual Physics used "weight" to mean the support force. I don't know if the current edition still does this.

I personally think this makes more sense than the usual physics definition of "weight." After all, the physiological effects of your "apparent weight" in an upward-accelerating elevator or rocket are the same as those of your "real weight" when standing on the ground. But I'm not going to argue with my textbook in an intro-level class.

 

[Doc Al]

When I taught a "descriptive physics" course about twenty years ago, Paul Hewitt's book Conceptual Physics used "weight" to mean the support force. I don't know if the current edition still does this.

I also taught a similar course using Hewitt's book years ago (more than twenty, I'm afraid ). I'll have to dig up my copy. (I think I have a newer one as well somewhere--I'll check it out tonight.)

I personally think this makes more sense than the usual physics definition of "weight." After all, the physiological effects of your "apparent weight" in an upward-accelerating elevator or rocket are the same as those of your "real weight" when standing on the ground. But I'm not going to argue with my textbook in an intro-level class.

I hear you and largely agree. I tend to use the convention that was beaten into me during my Halliday & Resnick days, which is still standard in most US intro texts (I think). Nonetheless, whenever I think of it, I try to use 'gravitational force' instead of 'weight' to avoid adding to the confusion.

 

[Saw]

In US physics textbooks, at least, 'weight' refers to the force of gravity acting on an object. The two forces in your example are just the two halves of the normal force interaction.

So... are there four forces involved? Gravitational force exerted by the Earth on the object and gravitational force exerted by the object on the Earth -on the one hand- plus normal force exerted by the Earth on the object and by the object on the Earth -on the other hand-?

 

[Doc Al]

So... are there four forces involved? Gravitational force exerted by the Earth on the object and gravitational force exerted by the object on the Earth -on the one hand- plus normal force exerted by the Earth on the object and by the object on the Earth -on the other hand-?

Absolutely!

 

[jtbell]

whenever I think of it, I try to use 'gravitational force' instead of 'weight' to avoid adding to the confusion.

Same here!

 

[jtbell]

Gravitational force exerted by the Earth on the object and gravitational force exerted by the object on the Earth

plus normal force exerted by the Earth on the object and by the object on the Earth

...which form two distinct action-reaction (Third Law) pairs.

 

[Gallin]

So when is the normal force different than the gravitational one?

 

[D H]

So when is the normal force different than the gravitational one?

When you are not at the north or south pole.

An inertial observer sees a person apparently standing at rest on the surface of the Earth as undergoing uniform circular motion. That person is standing still relative to the rotating Earth. The individual forces acting on that person's body must sum to yield the force centripetal force that corresponds to this uniform circular motion. The normal force and gravitational force are be equal in magnitude but opposite direction except at the poles.

 

[Saw]

...which form two distinct action-reaction (Third Law) pairs.

Ok, I see. I paraphrase for your confirmation.

When an object is standing on the Earth, there are two interactions: one attractive (gravitational) and another repulsive (electromagnetic = normal).

Each one has its own pair of 3rd Law forces. The 3rd Law pairs (acting on different objects) do not counteract each other.

Hence it is the mixture between two forces of different kind and opposite sign (which are not 3rd Law pairs and do act on the same object) what causes the equilibrium, isn’t it?

For example:

Gravitational Force by the Earth on the object (pushing the object downwards) is canceled by Normal Force also by the Earth on the object (pushing the object upwards).

Gravitational Force by the object on Earth (pushing the Earth upwards) is canceled by Normal Force also by the object on the Earth (pushing the Earth downwards).

The fact that the forces that cancel against each other have equal magnitudes is not due to one being a reaction against the other, as noted, but to… the fact that we have assumed that they do cancel against each other. In fact, when a ball bounces off the floor after being dropped from some height, they do not cancel: in that case, normal force prevails at collision time and hence the bouncing.

Some language problems:

Usually the gravitational force is referred to as weight. But in common language one tends to think that weight is that sensation that arises when weight combines with an opposite force that makes it “apparent”. So people think that whenever that sensation is absent, like in free fall, the body is weightless, which is not technically true: it is only “supportless”. Hence people should stop saying that or… physics could change its nomenclature and reserve the word “weight” for “apparent weight”. As the latter is too ambitious at this stage, your suggestion is at least try to avoid the word “weight”, to avoid discussions... Good.

Usually Newton’s third Law is explained saying that for every action, there is a “reaction”: a force by A on B triggers a “reaction force” by B on A of the same nature, same magnitude, same direction and opposite sense. But in common language we use the expression “reaction” also to refer to situations where a force by A on B triggers and opposite force also by A on B. This is our case, where gravitational force on the object exerted by the Earth triggers, when the object touches the ground, a reaction through another force, normal force, exerted also by the Earth itself on the same object. Who should rectify here? To me, physics should not use the word “reaction” in the context of 3rd Law. The two sides of a 3rd Law are really the two sides of the same thing. It is an interaction, a single phenomenon, having effects on both sides… What would you advise?

 

[Doc Al]

So when is the normal force different than the gravitational one?

Whenever you are accelerating. For example, jump off of a bench and the normal force goes to zero yet the gravitational force hasn't changed. Or simply stand in an elevator while it accelerates upward--the normal force is now greater than the gravitational force.

 

[Doc Al]

Each one has its own pair of 3rd Law forces. The 3rd Law pairs (acting on different objects) do not counteract each other.

Hence it is the mixture between two forces of different kind and opposite sign (which are not 3rd Law pairs and do act on the same object) what causes the equilibrium, isn’t it?

Exactly. 3rd law pairs can never produce equilibrium on their own, since they act on different bodies.

To me, physics should not use the word “reaction” in the context of 3rd Law. The two sides of a 3rd Law are really the two sides of the same thing. It is an interaction, a single phenomenon, having effects on both sides… What would you advise?

The 'action/reaction' terminology is old-fashioned and misleading, for exactly the reasons you point out. Avoid it! Much better to refer to them as 3rd law pairs.

 

[Saw]

The 'action/reaction' terminology is old-fashioned and misleading, for exactly the reasons you point out. Avoid it! Much better to refer to them as 3rd law pairs.

Thanks a lot. I corrected my notes accordingly and now everything fits better in my head.

 

[ashishsinghal]

Freely falling body is not weightless. It just doesn't feel any weight because one feels weight cause of the normal force acting on it

 

[D H]

Freely falling body is not weightless. It just doesn't feel any weight because one feels weight cause of the normal force acting on it

That depends on what you mean by "weight". "Weight" is one of those semi-scientific terms that has multiple meanings. With one of those meanings, also called scale weight or apparent weight, a freely-falling body is weightless.