Actual Working of Weighing Machine (Normal vs gravitational force)

[Steve4Physics]

When a person is standing on it Normal force equal to m(person)*g acts downwards along with m(person)g .
So , f(downwards)= N+m(person)g= 2m(person)g.

At the risk of being one the excessive number of cooks...

Gravitational attraction acts between 2 masses, A and B. Each attracts the other in accordance with Newton's 3rd law.

If you put a 3rd object, C, in-between A and B, the gravitational force between A and B simply 'passes through' C without affecting C. So C is 'transparent' to this force. There is no gravitational force on C due to the gravitational attraction between A and B.

(There will be additional forces of gravitational attraction between A and C and between B and C, but that's a different issue.)

If A is a person, B is the earth and C is a weighing machine, then the weighing machine is 'transparent' to the forces between A and B. The gravitational attraction force of B on A 'passes through' C; it does not exert a force on C.

 

[jbriggs444]

You seem to have misinterpreted my question.
When a person is standing on it Normal force equal to m(person)*g acts downwards along with m(person)g .
So , f(downwards)= N+m(person)g= 2m(person)g.

A rope is being tugged by two teams with force $\vec{F_1}$ rightward by the one team and a equal but opposite force $\vec{F_2}$ to the left by the other team. The magnitude of both forces is $F$.

What is the the tension in the rope?

a) $2F$
b) $F$
c) $0$
d) Something else
e) Cannot be determined from the given information

The same rope is being tugged by team left with the same force $F$ as before. Team right ties their end to a handy wall and goes away to eat lunch.

Now what is the tension in the rope?

a) $2F$
b) $F$
c) $0$
d) Something else
e) Cannot be determined from the given information

What is the net force on the rope in the two scenarios above? Is this different from the tension in the rope?

 

[Orodruin]

Answer 3: The same as the weight of the person

Again, I do not think this is an accurate interpretation of the OP given the attempted free body diagram in #5. The OP (wrongly) believes the gravity on the person acts on the scale. The gravity on the scale was never considered by the OP.

 

[kuruman]

The gravity on the scale was never considered by the OP.

That is not how I would interpret this OP statement in post #1 (color emphasis mine)

But when a body is standing on machine 2 forces act simultaneously (namely the normal force and gravity[/color]) .

(Even though I said I would not post any more here, I had to reply.)

 

[Lnewqban]

Was that person inside the original square?

 

[Orodruin]

That is not how I would interpret this OP statement in post #1 (color emphasis mine)

(Even though I said I would not post any more here, I had to reply.)

That is an interpretation, but the OP also says “when a person stands on”. If you look at the attempted FBD in #5, it certainly seems like the mg force is related to the person rather than the scale.

I am not saying I cannot see your interpretation. I am saying it was a bit over the top to call out everyone ignoring the mass of the scale when their interpretation of the OP was equally valid (and the one OP intended).

 

[kuruman]

I am saying it was a bit over the top to call out everyone ignoring the mass of the scale when their interpretation of the OP was equally valid (and the one OP intended).

It certainly was not my intention to "call out everyone" and, if my post seemed to indicate that, I apologize to all. My point of view is that it makes no sense to draw a free body diagram for a system with zero (or negligible) mass. FBDs serve as guides for writing Newton's second law which requires a non-zero mass to make sense. I will say no more.