What is the effect of equal and opposite forces on a body?
I conducted an experiment to find out what is the effect of equal and opposite forces on a body, and discovered they add inertia.
In the picture Pic 1, A and B are horizontally parallel bars placed at a distance from each other at the same height above surface. Equal weights F1 and F2 are tied at two ends of a smooth & light weight string. In the middle of the string is a point object p, and the rope is placed on the bars such that weights F1 and F2 dangle at equal height above the surface and point object p is in the middle of the bars. Weights F1 and F2 act as equal and opposite forces on point object p. F1 and F2 are now referred to as forces and they cause the point p to hang in balance.
Suppose the force required to give point p an acceleration of a, in the direction of B, is f (weight f is placed upon F2 to apply this force).
Now if the magnitude of forces F1 and F2 are doubled, the force required to give point object p , the same acceleration a, in the same direction B, would be 2f.
As we double the opposing forces, inertia m of the point object p is also doubled; thus opposing forces don’t add up to zero, they rather increase inertia or mass of the body.
F1 − F2
= ma − ma, {as F1 = F2 = ma}
= m ( a − a )
= m ( 0a ), mass with zero acceleration.
= m
or F − F = m
This is not correct I know, F − F = 0, but then how do I account for the change in inertial mass of the point object p. And I know mathematical physics has its limitation like; in absence of a second object one cannot tell if one object is rotating or not, it needs a reference point.
Actually both of you are incorrect in your analysis of the situation. Both force and acceleration are vectors, which means that the direction is important.
The formula "a = F/m_total" is only valid if all the masses are moving in the same direction, which they are not in this case.
In the original situation when f is added to F2, f moves down and so does F2 but p moves to the right and F1 is moves upwards. You cannot just add these together as if they were all moving in the same direction!
Also Shahin you have made a basic algebraic mistake in the following:
Quote:
Originally Posted by shahinomar
F1 − F2
= ma − ma, {as F1 = F2 = ma}
= m ( a − a )
= m ( 0a ), mass with zero acceleration.
= m
or F − F = m
m ( 0a ) = 0 not m, zero multiplied by anything equals 0 !
Actually both of you are incorrect in your analysis of the situation. Both force and acceleration are vectors, which means that the direction is important.
The formula "a = F/m_total" is only valid if all the masses are moving in the same direction, which they are not in this case.
Hello Elscarta. Actually I believe that I have consistently treated the forces as directional vectors, and in fact so has Shahinomar. Perhaps it hasn't been terribly clear throughout, but we both used plus and minus signs to indicate directions right and left. In the equation you cited: "a = F/m_total" the F was meant to be the resultant force when you vectorially add F1, F2, and f. So yes, we undserstand that direction is important. I did try to make it this clearer by using vector notation in the later posts - I suppose I should have used this style of notation from the start. Nevertheless, I don't believe there are any errors in my posts. Please let me know if you think otherwise.
Hi ebaines, my point wasn't so much that the forces need to be added as directional vectors but that it is not appropriate to add the masses together and treat them as a single mass being accelerated by a single force.
Another way of seeing this is that each of the separate masses do not experience the same forces and so there is no justification to adding the masses together.
As a further example consider what happens to a driver in a car which hits a wall head on. If the driver has a seatbelt on, then his mass can be added to the mass of the car as they are effectively a rigid body undergoing the same motion and having the same net force acting on them.
But if the driver is not wearing a seatbelt then it is not appropriate to add the driver's mass to that of the car as the driver's motion (and net force acting on him) will differ to that of the car.
Thanks for your time and response!
Sorry! You are incorrect here, and I cannot get into explaining it in this thread. If you like I can do that in a different thread.
One more request; please do not misquote me. In the very first post I have written and accepted that: "or F − F = m This is not correct I know, F − F = 0".
So there is no need to question my algebraic skills really.
So you are suggesting that, if I change the source of the force (from F1 to R1 or weight to rocket), while applying the same force 98 N, the effect will change?
And it will become 'easier' for me to move (or accelerate) point p in the direction of B?
Hi ebaines, my point wasn't so much that the forces need to be added as directional vectors but that it is not appropriate to add the masses together and treat them as a single mass being accelerated by a single force.
Another way of seeing this is that each of the separate masses do not experience the same forces and so there is no justification to adding the masses together.
As a further example consider what happens to a driver in a car which hits a wall head on. If the driver has a seatbelt on, then his mass can be added to the mass of the car as they are effectively a rigid body undergoing the same motion and having the same net force acting on them.
But if the driver is not wearing a seatbelt then it is not appropriate to add the driver's mass to that of the car as the driver's motion (and net force acting on him) will differ to that of the car.
Hope this clarifies my point.
Hello Elscarta. You are incorrect saying that it is wrong to add the masses and treat them as a single system. In the problem posed by the OP, the masses are tied together via a string - consquently they MUST accelerate as one, and the magnitude of displacement of one mass must equals the magnitude of displacement of the other mass. If you can think of a mechanism where this isn't case please let us know. This means that the total mass of the system times the acceleration of the system equals the net force aplied to the system. The only trick is to recognize that the net force is the diffrerence between F1 and F2 (since they act in opposite directions on the system).
In your example of the car and driver with seatblet - you are correct that the masses can be added since the driver is tied to the car seat. You is precisely analogous to the situation as described by the OP.
Below is an example of a similar problem that will make it clear. If m1 = 1 Kg and m2 = 2 Kg, you can determine the total acceleration of the system as follows:
F1 = 1 KG * g
F2 = 2 Kg * g
F2 - F1 = (m1+m2) * a
a = (2Kg * g - 1 Kg * g)/(2 Kg + 1 Kg) = 1/3 g
Or, you could go through a more detailed exercise of calculating the tension in the string - which turns out to be 4/3 kg*g - and then calculate the acceleration of m1 and m2:
a1 = (T-m1g)/m1 = (4/3g - 1g)/1 = 1/3 g
a2 = (T-m2g)/m2 = (4/3g-2g)/2 = -1/3 g
So you are suggesting that, if I change the source of the force (from F1 to R1 or weight to rocket), while applying the same force 98 N, the effect will change?
And it will become 'easier' for me to move (or accelerate) point p in the direction of B?
Yes, because you have posited that the rockets have small mass. Therefore they present little inertia to overcome as the system accelerates. So the system accelerates faster than it would jhave if you use weights instead.
Remember the main principle: . If you make m small, then for a given value of the acceleration becomes large.
Hi ebaines.Yes you are correct. It has been a while since I have done physics.
I knew that equal and opposing forces do not increase the inertial mass of a point object, but confused myself it trying to figure out why the initial situation suggested that it did.
From your postings it makes sense now. The increase in inertial mass is not at the point object P, but rather the whole system which is tied together.
In the example regarding the rockets. If you doubled the thrust of each rocket, without changing the rockets themselves, there would be no difference in the acceleration as there is no change in inertial mass of the whole system.
Mass of point p is negligible, say 1 gram
Mass of rocket R1 is negligible, say 1 gram
Mass of rocket R2 is negligible, say 1 gram
Mass of m1 is 10 kg
Mass of m2 is 10 kg
Force applied by R1 is 98 N, in the direction of A
Force applied by R2 is 98 N, opposite to A
Force applied by suspended mass m1, F1 is 98 N, in the direction of A
Force applied by suspended mass m2, F2 is 98 N, opposite to A
(lets ignore the mass of string and other frictions).
Scenario 1: No E&O forces applied on point p.
To give point p an acceleration of 4.9 m/s2 in the direction of B, force required is very small. f = 1/1000 * 4.9 N = 0.0049 N (Ignoring all friction and that point p in this scenario is placed on a surface rather than held between bars as shown in the picture, obviously.)
I apply this force with bare fingers only. And we have no disputes here!
Scenario 2: Forces applied on p are F1 and F2 as shown in the first picture.
Masses m1 and m2 are used to apply forces F1 and F2.
To give point p an acceleration of 4.9 m/s2 in the direction of B, force required is:
0.0049 N + 4.9 * 10 N + 4.9 * 10N = 98.0049 N
I apply this force with bare fingers only
Both of us agree to this that the force required to give p the same acceleration will be higher in this scenario as calculated.
Scenario 3: Forces applied on p are R1 and R2, R1 replaces F1 and R2 replaces F2.
Here you are saying that the force require to give p the same acceleration 4.9 m/s2, in the direction of B, will be 0.0147 N and not 98.0049 N.
Logical fallout 1
In scenario 3 when I try to move point p (with my bare fingers) in the direction of B it will be much easier for me to do that, compared to scenario 2.
Logical fallout 2
By touching point p (as in trying to move it) I can tell whether the source of the force 98 N is, F1 or R1.
Logical fallout 3
98 N force apart from increasing tension in the string also sends "additional signal" through the string telling point p about its mechanism?
I cannot agree with you on this. F1 or R1, only thing they do is apply force (increase tension in the string), and so no reason why the effect on point p should change if the force is the same 98 N. And moving point p would take the same effort in both cases.
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