Answers to N3L day questions
Right... one day later than promised, but here's my solution to the two questions I posed earlier.
The first question: If a person exerts a force of 50N on a box and the friction involved is 4N, we know that the box will start moving because it has a resultant force of 46N acting on it. But... we know that from N3L, the box will exert an equal and opposite force which should cancel out with your initial 50N exerted, giving the box a resultant force of 0N. What is the problem here?
My answer: The problem is the result of a misconception with regards to N3L, that the forces are acting on the same body (which is not what N3L states).
N3L states that when body A exerts a force on body B, body B exerts an equal but opposite force back on body B. The main point lies in the fact that both forces act on different bodies, thus they do not cancel out.
In this case, the resultant force cannot be 0N simply because the person exerts 50N on the box and the box exerts 50N on the person (acting on different bodies, no cancelling), thus although the 50N from the box to the person has no effect on the person, the 50N from the person to the box allows the box to move at 46N after overcoming the 4N of friction.
The second question: To correctly identify the pair/pairs of action-reaction forces in a diagram involving a box on a table. Is the weight of the box an action-reaction pair with the contact force of the box on the table?
My answer: The answer is no. This misconception that they are action-reaction pairs arose from doing Physics calculations, which only focuses on the forces acting on a body.
N3L states that when body A exerts a force on body B, body B exerts an equal but opposite force back on body B. The main point lies in the fact that it has to be A -> B and B -> A, then A -> B and B -> A are action-reaction pairs.
Consider the forces that I have mentioned: (let box be A, Earth be B and table be C)
Weight of box => Force of Earth on box (B -> A)
Contact force of table => Force of table on box (C -> A)
We now see that they are not action-reaction pairs. They are just forces acting on the same body which happen to be equal in magnitude and opposite in direction.
So... the actual action-reaction pairs should be:
Weight of box (B -> A) and Force of box on Earth (A -> B)
and
Force of table on box (C -> A) and Force of box on table (A -> C)
The weekend has arrived. It's TGIF! No KTV today though... it has been postponed to next Thursday. I want to go ice-skating... maybe I'll go tomorrow. Heh. I can feel more relaxed that most of the stuff due this and next week are done.
Nothing interesting for the week. I'll be taking a break from the normal evening tuitions though, now that my cousins' exams are over =]
The first question: If a person exerts a force of 50N on a box and the friction involved is 4N, we know that the box will start moving because it has a resultant force of 46N acting on it. But... we know that from N3L, the box will exert an equal and opposite force which should cancel out with your initial 50N exerted, giving the box a resultant force of 0N. What is the problem here?
My answer: The problem is the result of a misconception with regards to N3L, that the forces are acting on the same body (which is not what N3L states).
N3L states that when body A exerts a force on body B, body B exerts an equal but opposite force back on body B. The main point lies in the fact that both forces act on different bodies, thus they do not cancel out.
In this case, the resultant force cannot be 0N simply because the person exerts 50N on the box and the box exerts 50N on the person (acting on different bodies, no cancelling), thus although the 50N from the box to the person has no effect on the person, the 50N from the person to the box allows the box to move at 46N after overcoming the 4N of friction.
The second question: To correctly identify the pair/pairs of action-reaction forces in a diagram involving a box on a table. Is the weight of the box an action-reaction pair with the contact force of the box on the table?
My answer: The answer is no. This misconception that they are action-reaction pairs arose from doing Physics calculations, which only focuses on the forces acting on a body.
N3L states that when body A exerts a force on body B, body B exerts an equal but opposite force back on body B. The main point lies in the fact that it has to be A -> B and B -> A, then A -> B and B -> A are action-reaction pairs.
Consider the forces that I have mentioned: (let box be A, Earth be B and table be C)
Weight of box => Force of Earth on box (B -> A)
Contact force of table => Force of table on box (C -> A)
We now see that they are not action-reaction pairs. They are just forces acting on the same body which happen to be equal in magnitude and opposite in direction.
So... the actual action-reaction pairs should be:
Weight of box (B -> A) and Force of box on Earth (A -> B)
and
Force of table on box (C -> A) and Force of box on table (A -> C)
The weekend has arrived. It's TGIF! No KTV today though... it has been postponed to next Thursday. I want to go ice-skating... maybe I'll go tomorrow. Heh. I can feel more relaxed that most of the stuff due this and next week are done.
Nothing interesting for the week. I'll be taking a break from the normal evening tuitions though, now that my cousins' exams are over =]
posted by notsniw at 3:24 PM
8 Comments:
Now I'm really confused. Ha!
Not clear meh? Tsk
It seems clear enough to me. =)
For the second question, I guess your answer is correct (coz you're not considering a closed system whereby the only two objects are the table and the box. You're considering the 3rd mass (ie. the earth) and the pull of the earth on the box and vice versa. I guess since O level physics don't really consider closed or open systems, it's correct too.
I'm not sure if any students argue with you as to they do constitute an action-reaction pair if you do not take the attraction of the earth's gravity into consideration? If you consider only the box and the table, both masses will exert a pull on each other following Newton's Universal Law of Gravitation:
F = G (m1*m2/r^2)
And since they are stationary, the forces will constitute an action-reaction pair because they are the only 2 objects in the system. That is, the weight of the box is due to the pull of the table and vice versa instead of the Earth.
Eh, so what's your definition of action-reaction pair?
If it is a closed system involving only the table and the box, I guess the only pair of action-reaction pair is:
Force of table on box and Force of box on table.
Force of table on box will be the same as your pull of table on box.
Force of box on table will be the same as your pull of box on table.
The weight of the box (in every general case taken to be due to the Earth's gravitational acceleration because weight = mg), will still not be an action-reaction pair with anything.
Yeah, exactly what you say... if the system only involves the box and table, the so-called 'weight' of the box will be due to the pull of the table rather than the gravitational force, since it's non-existent in this sense.
Though of course it would be hard to explain this to students I suppose.
Hmm, I wouldn't really term the pull of the table on the box as weight, since weight is generically defined in all or most cases to be due to gravitational pull of the Earth, Moon, etc.
Weight of the box in this case would strictly refer to the pull of the Earth on the box, since that is the definition of weight. Otherwise, it would be totally confusing =]
"Weight of box due to table"
>> Force of table on box
Weight of box due to Earth
>> Force of Earth on box
Ah, that's true. I should be more precise on the definition of weight. Basically to me it should be defined as a force due to the attraction of another body, and the norm would be that of the Earth.
Well, I guess your explanation should be sufficient for 'O' level Physics, don't think a lot of them will probe anything deeper than that anyway.
N3L is optional for 'O' levels. Heh
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