It's one of those weeks...
It's one of those weeks that are like the worst weeks of the year...
It's now tuition everyday for six out of seven days...
And there's still the stuff that's due on Monday.
What else could get worse? Ha.
All in all, a really hectic week for me.
Oh yar... 5th October will be officially known as the N3L day. What's N3L? It's Newton's 3rd Law :P
Why N3L day? It's because my closer NIE friends and I discussed quite a bit on the misconceptions that students have for N3L. Questions courtesy of yours truly :P
Anyway, the first question is something like this: 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?
The second would be: 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?
It's boring, but it's important that we teach the correct concepts to students... and not teach them what we think is correct but is actually wrong.
Weekend lo... but no time to rest :/
It's now tuition everyday for six out of seven days...
And there's still the stuff that's due on Monday.
What else could get worse? Ha.
All in all, a really hectic week for me.
Oh yar... 5th October will be officially known as the N3L day. What's N3L? It's Newton's 3rd Law :P
Why N3L day? It's because my closer NIE friends and I discussed quite a bit on the misconceptions that students have for N3L. Questions courtesy of yours truly :P
Anyway, the first question is something like this: 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?
The second would be: 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?
It's boring, but it's important that we teach the correct concepts to students... and not teach them what we think is correct but is actually wrong.
Weekend lo... but no time to rest :/
posted by notsniw at 2:07 PM
7 Comments:
Let's see how un-rusty my physics is...
If I remember correctly the best possible way is to teach them the concept of force using the conservation of momentum (p = mv):
F = d(mv)/dt = m(dv/dt) + v(dm/dt)
Force is essentially the rate of change of momentum.
For the 1st qn, using Newton's 1st law, we know that the box would move because:
An object at rest will remain at rest unless acted upon by an external and unbalanced force . An object in motion will remain in motion unless acted upon by an external and unbalanced force.
The net force on an object is the vector sum of all the forces acting on the object. Since the resultant vector force isn't zero (50-4=46), the box would start moving.
For the resultant 46N, we now apply the 3rd law. There is an equal and opposite force acting against the box as it moves due to the resultant 46N: air resistance (this is because we have taken the frictional force into consideration already). Although the force opposing the box is equal and opposite (46N in the opposite direction), the accelerations are not. The object with less mass will have a higher acceleration - in this case, the box, as its mass is smaller relative to the mass of the air resistance acting on it.
For the second question, I would say the answer would be a yes - the box pushes down on the table as much as the table pushes the box up. This is because the box remains stationary on the table - thus the resultant forces are the same. This is on the assumption that the system consists only of the 2 objects (closed system).
My 2 cents from memory. Did I make any mistakes?
Hmmm, both answers not quite what I expected. Ha... We'll wait to see if there are any other replies =]
I know nuts about Physick. Yay~ kekeke
i agree w toukarin on the use of the mathematical form of F = d(mv)/dt = m(dv/dt) + v(dm/dt) to explain Q1.
Newton's 1st law states that "An object that is not moving will not move until a force acts upon it.
An object that is in motion will not change velocity until a force acts upon it."
Essentially this is also the law of inertia which indicates that the net force on an object is the vector sum of all the forces acting on the object. If this sum is zero, the state of motion of the object does not change. But in the case of Q1, the sum is not zero.
As a result of Newton's 1st law, we arrive at F = ma where it is defined in physics that the mass of a body is "reciprocally proportional to the bodies".
Derivation as follows:
F = k d(mv)/dt, where k = constant of proportionality, F and v in vector form
When mass is known to be constant,
F = kma, where F and a in vector form
Using SI units where k = 1 or unity,
F = ma, where F and a in vector form
Therefore F=ma (Newton's 2nd law) becomes a quantitative restatement of Newton's 1st law. Now mass is a measurement of inertia.
Then comes Newton's 3rd law. Even in the case of the 2 forces being equal in magnitude but opposite in direction, the accelerations are not. Based on Newton's 2nd law, the less massive object will have a greater acceleration. Example: Compare the force on the ground hit by the ball and the Earth's force on the ball. The magnitude is the same and opposite but due to Newton's 2nd law, the mass of the ball is smaller, thus a greater acceleration of the ball as compared to the Earth.
Q2 - I also think the answer is yes.
Newton's 1st law: An object that is not moving will not move until a force acts upon it. No other force acting on either body. As mentioned by toukarin, I also interpret a closed system.
Newton's 2nd law:
box: F=ma, where a = constant
table: F=ma, where a = constant
Formula verfies what Newton's 1st law suggested, neither box nor table moves.
Newton's 3rd law: Constant acceleration in both bodies resulting in no motion imply mutual actions of 2 bodies upon each other. The changes made by these actions are equal, thus changes of the velocities made toward contrary parts are reciprocally proportional to the bodies, fulfilling F = k d(mv)/dt, in turn fulfilling F = d(mv)/dt = m(dv/dt) + v(dm/dt), conservation of momentum preserved.
Just add on my humble opinion.
Pls correct me if i'm wrong.
madchemist > yea... you and your chemistry
eostkh > hmmm, not quite what I had in mind also :P It's for Sec level, so bringing cheem formulae in won't help in explaining I guess. Impressed with the long answers though. Heh
Hmmm, basically I only require Newton's 3rd law in the explanation I guess. I think it would be sufficient to explain.
Will reveal my answers (well, hopefully correct ones) on Wednesday. Heh
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