Newton's 3rd Law Scale

[SaintJude]

I chose the right answer out of 'habit'. Can anyone correct the underlined wording of all the other wrong answer choices?


According to Newton's 3rd Law, which of the following is necessarily true of a person standing on a scale?

A. The scale will exert a force on the person that is equal and opposite to the person's weight.

B. The normal force on the person will be equal and opposite to the person's weight.

C. The Earth will experience force= person's weight.

D. The scale will experience a force from the person that is equal and opposite to the person's weight.

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[SaintJude]

Anybody?

 

[Dasypus]

I think it's a tricksy question. The big thing that the question doesn't mention is if the scale and person are, say, in a moving [accelerating downward] elevator. If they are, and the elevator is moving [accelerating] down, then the normal force, which just is the force by the scale on the person (A & B are two ways of saying the same thing), will be less than the person's weight, and the scale will experience a force less than the person's weight. D is the 'most wrong' since the scale would *never* experience a force equal and opposite to the weight -- unless I suppose someone was coincidentally pushing it down that hard on the person's head).

On the other hand, it doesn't matter what the situation is for C. By definition, the person's weight *is* the force of gravity by the earth on the person, and thus there must be an equal and opposite force of gravity by the person's mass on the earth.

 

[chiddler]

I'm wondering why A is wrong.

 

[MedPR]

I'm wondering why A is wrong.

Yea, and B too.

 

[Dasypus]

Sorry, instead of 'moving elevator' read 'accelerating elevator'. Check out https://docs.google.com/viewer?a=v&q=cache:gvgGAe2BhIsJhysics.usask.ca/~kathryn/phys111/apparent_weight.pdf+necessarily+true+of+a+person+standing+on+a+scale+newtons&hl=en&gl=ca&pid=bl&srcid=ADGEEShxUNxIlw9u4eiqn1SVN72cuVWx9-Mn45PyQL8jaNhGjt01ccMbL0dagq7M0-_3l3arjh_L-04eITOjbLAZ0nJIZlnKe7e43HcgEQcRFjJ1twCMDY_TsNj9wUNmBmo1EZrkRr4S&sig=AHIEtbQFK5drImj6AARs1KiY4qPbAaD0jA for diagrams.

 

[chiddler]

maybe it is which is NOT true!

earth is in free space. it can't really experience a force from a person standing on it, i don't think.

disregarding gravitational attraction.

 

[Dasypus]

maybe it is which is NOT true!

earth is in free space. it can't really experience a force from a person standing on it, i don't think.

disregarding gravitational attraction.

But it's exactly that gravitational attraction that they're asking about.

 

[chiddler]

But it's exactly that gravitational attraction that they're asking about.

that's right...the weight is due to gravitational force.

MILSKI D:!!

 

[milski]

I was actually trying to avoid the question - too much splitting hairs for my taste.

Weight is defined as the force that is exerted via gravity from Earth on the person. If there is any additional force acting on that person, the weight and the normal are not going to be the same. Being in an elevator is one example, being anywhere besides the pole is another. If you remember a question from a few days ago, every time you are not on the pole, you are in a uniform circular motion. That means that you are constantly accelerating and the net force on you is not zero. Since the only forces acting on you are the normal and the weight, you end up with W!=N. In other words, your apparent weight is slightly lower than your real weight. That disqualifies A and B, which would have been fine choices if we were talking about 'apparent weight.' The fact that A & B are virtually the same is another hint that they are the wrong answer - props to {can't remember who, sorry} for this approach.

D does not makes sense - that's the same as the person pulling the scale.

C is just 3rd law - it is always true. There are no exceptions for being in free space (Earth) or being too big (Earth). Of course, the acceleration of Earth due to the force that you are exerting on it via gravity is well, insignificant.

 

[chiddler]

<3