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.

Have we run into a similar difficulty here?

Regards
Shahin

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.



Not quite. As I understand it, you have a point mass at p of m, two opposing forces F1 and F2 which are equal, and an additional force f caused by placing an additional weight at F2. To analyze this you need to consider the overall mass of the system. The mass of the weights F1 and F2 are F1/g and F2/g, respectively, and the mass of the body that causes force f is f/g. So, applying the basic formula F=ma to the system:

f = (f/g+F1/g+F2/g+m)*a

So the acceleration of the system is

a = f*g/(f+F1+F2+mg)

If you double F1 and F2, the new acceleration is

a2 = f*g/(f+2F1+2F2+mg)

To make a1 = a2, you then need to change f to a new value - let's call it f2, and you get:

a = a2
f*g/(f+F1+F2+mg) = f2*g/(f2+2F1+2F2+mg)

f*(f2+2F1+2F2+mg) = f2*(f+F1+F2+mg)

f*(2F1+2F2+mg) = f2*(F1+F2+mg)

f2/f = (2F1+2F2+mg)/(F1+F2+mg)

So you can see that f2/f is not equal to 2, but is affected by the magnitude of m.



As we double the opposing forces, inertia m of the point object p is also doubled;


I'm not following you here - are you suggesting that the mass m at pont p is somehow doubled? How does this occur?




thus opposing forces don’t add up to zero, they rather increase inertia or mass of the body.


Forces don't increase mass - please clarify what you mean. After you get these points straightened out, repost your new calculations to show us what you get.

Not quite. As I understand it, you have a point mass at p of m, two opposing forces F1 and F2 which are equal, and an additonal force f caused by placing an additional weight at F2. To analyze this you need to consider the overall mass of the system. The mass of the the weights F1 and F2 are F1/g and F2/g, respectively, and the mass of the body that causes force f is f/g. So, applying the basic formula F=ma to the system:

f = (f/g+F1/g+F2/g+m)*a

So the acceleration of the system is

a = f*g/(f+F1+F2+mg)

If you double F1 and F2, the new acceleration is

a2 = f*g/(f+2F1+2F2+mg)

To make a1 = a2, you then need to change f to a new value - let's call it f2, and you get:

a = a2
f*g/(f+F1+F2+mg) = f2*g/(f2+2F1+2F2+mg)

f*(f2+2F1+2F2+mg) = f2*(f+F1+F2+mg)

f*(2F1+2F2+mg) = f2*(F1+F2+mg)

f2/f = (2F1+2F2+mg)/(F1+F2+mg)

So you can see that f2/f is not equal to 2, but is affected by the magnitude of m.



I'm not following you here - are you suggesting that the mass m at pont p is somehow doubled? How does this occur?



Forces don't increase mass - please clarify what you mean. After you get these points straightened out, repost your new calculations to show us what you get.

Thank you for taking time out to answer!

Sorry! Let me make things more clear.

Mass of point p is hypothetically negligible.

F1 is a force that would otherwise give acceleration a to an object of mass m.

F2 is a force equal and opposite to F1.

After the string is placed over the bars as shown in the picture, suppose force required to give point p an acceleration a1 in the direction of B is f. You do need to apply some force to give it acceleration a1.

And after we double the forces F1 and F2 the force required to give the same acceleration a1 to the same point object p would be 2f, nearly.

Why the required force increases? Remember I have used weights as source of forces only because that is easiest for me. I would have liked to create forces from a source that did not have a mass of its own, hypothetical again, and even then the results would be the same.

Let's use real numbers to make the experiment clearer:

Suppose the masses you use to create F1 and F2 are 10 Kg each; let's call these masses m1 and m2. So F1 = F2 = 10 kg*g = 98 Nt. Then you add another mass at F2, let's say that masss is m = 20Kg, which causes force f to be applied to the system. For a mass of m = 20 Kg this f is 20*g = 196 Nt. The sum of forces acting on the system as a whole is 196 Nt, acting on a system of m_total = 10+10+20 Kg = 40 Kg, for an acceleration of a = F/m_total = 196 nt/40kg = 4.9 m/s = 1/2 g.

Now let's double the force F1 and F2 by doubling the masses to 20 Kg each. What do you have to do to force f to keep acceleration of the system at 1/2g? You can calculate:

a = 1/2g = f/m_total = m*g/(m+m1+m2).

1/2 g = mg/(m + 40Kg)

Solve for m and you find m = 40 Kg. So indeed, if you double m1 and m2, you also have to double m to keep acceleration the same.

Hope this helps.

Thanks again, but not quite.

Forgive me if I sound to be playing with words here, trust me I am not.

What you have done by using real numbers is show me how much, and proved the same F - F = m. In fact thanks for that.

You give me force F1, then you give me force F2 which is equal and opposite to F1, I put them together and what you get is a system that has mass and no force, no acceleration.

And so equal and opposite forces cause mass.

What you have calculated is

98 Nt - 98 Nt = 20 kg

Please do reply
Shahin

Thanks again, but not quite.

Forgive me if I sound to be playing with words here, trust me I am not.

What you have done by using real numbers is show me how much, and proved the same F - F = m. In fact thanks for that.

Hello Shahin.

I'm sorry - I do not understand what you mean by this statement! I think you have confused force and mass. I did not "prove" that F-F = m.



You give me force F1, then you give me force F2 which is equal and opposite to F1, I put them together and what you get is a system that has mass and no force, no acceleration.


So far so good...



And so equal and opposite forces cause mass.


Sorry - but this is incorect. Forces do not "cause" mass. Equal and opposite forces balance out and cause 0 acceleration; it doesn't matter what the value of the mass is.



What you have calculated is

98 Nt - 98 Nt = 20 kg


Not right. What I have shown is that 98 N + 20 Kg*g - 98 Nt = 20 N * g = 196 N. This is the sum of the forces acting on the system after you apply your extra force f of 196 N. This force will equal the mass of the system (40 Kg in my example) times its acceleration - this comes directly from the principle of F=ma.

Please be sure whenever you write an equation to think through the units of each term in the equation, and make sure they are consistent. In your posting here you are trying to equate force (newtons) to mass (Kg). But they are not the same thing! You must multiply mass by its acceleration to determine the force acting on it. Or conversely, divide a force by the quantity of mass it is acting on to determine the object's acceleration.

Dear Sir,

Thank you yet again, I really appreciate your time and response!

What is creating confusion here is I have used masses m1 and m2 to apply forces F1 and F2, this was just for my convinience.

Let me remove that and use a different mechanism to apply the same forces, let us use two very small size rockets R1 and R2 (I cannot think of something else).

Suppose the rockets weigh only 1 gram each, but when ignited apply a force of 98 Nt each. R1 and R2 apply forces on point object p the same way as F1 and F2 respectively.

Please tell me if you will get a different result? Or tell me what will happen now?

Sir, if you see my very first post I have written F - F = m is not correct. My real question was, does mathematics has its limit and does it fail to explain certain aspects of physics (like you cannot tell rotation without a reference point).

You do agree when I say, "Give me force F1, then give force F2 which is equal and opposite to F1, I put them together and what you get is a system that has mass and no acceleration."

So what is the problem when I write it in short "Equal and opposite forces when put together create a system that has mass and no acceleration."

Or for that matter "Equal and opposite forces create mass."

I do not mean to say forces create matter (matter and mass should not be confused), matter has mass and the later is known to be proportional to the former.

What I mean to explain is E&O forces when put together give a system the same attribute as mass (that matter by default has), and this mass is also proportional to the magnitude of the E&O forces applied.

Kind regards
Shahin

Dear Sir,
.. let us use two very small size rockets R1 and R2 (I cannot think of something else).

Suppose the rockets weigh only 1 gram each, but when ignited apply a force of 98 Nt each. R1 and R2 apply forces on point object p the same way as F1 and F2 respectively.

Please tell me if you will get a different result? Or tell me what will happen now?



With the two rockets of negligible mass applying equal and opposite forces, the system is in equilibrium and there is no movement. But if you then apply that extra force \vec{f} , as you described in your earlier post, then the system will move in the direction of \vec{f} , because now the system is no longer in equilibrium and there is a net force acting on it.



Sir, if you see my very first post I have written F - F = m is not correct. My real question was, does mathematics has its limit and does it fail to explain certain aspects of physics (like you cannot tell rotation without a reference point).

You do agree when I say, "Give me force F1, then give force F2 which is equal and opposite to F1, I put them together and what you get is a system that has mass and no acceleration."


Yes, I agree with that. As Newton taught us: \vec{F}=m\vec{a}, so if the net force \vec{F} is zero, then \vec{a} is zero as well, (regardless of the value of m).



So what is the problem when I write it in short "Equal and opposite forces when put together create a system that has mass and no acceleration."

Or for that matter "Equal and opposite forces create mass."

I do not mean to say forces create matter (matter and mass should not be confused), matter has mass and the later is known to be proportional to the former.


Your statements that "Equal and opposite forces when put together create a system that has mass" and ""Equal and opposite forces create mass" are just plain wrong. Forces do not create mass! No where does the mathematics suggest that it does. You seem to acknowledge this when you say "I do not mean to say forces create matter," so I don't understand why you keep saying that forces create mass??



What I mean to explain is E&O forces when put together give a system the same attribute as mass (that matter by default has), and this mass is also proportional to the magnitude of the E&O forces applied.


Sorrry - this is incorrect, and I really don't see how you could draw this conclusion. In previous posts you tried to suggest that \vec{F}- \vec{F} = m , but that was a math error on your part.

R1 = 98 N
R2 = -98 N
f = 196 N as before. (F1 = 98 N and F2 = -98 N, and f = 196 N.)

Question 1

Please can you tell me how do you calculate acceleration a now?

Earlier "The sum of forces acting on the system as a whole is 196 Nt, acting on a system of m_total = 10+10+20 Kg = 40 Kg, for an acceleration of a = F/m_total = 196 nt/40kg = 4.9 m/s = 1/2 g."

Now what: The sum of forces acting on the system as a whole is still 196 Nt.
Acting on a system of mass ?
for an acceleration of a = F/m_total = 196 nt / ?kg =?

Question 2

Now let's double the forces R1 and R2 by using some mechanism. Please tell me the required force f to keep the same acceleration a as before.

Thanks for your every reply, always!

R1 = 98 N
R2 = -98 N
f = 196 N as before. (F1 = 98 N and F2 = -98 N, and f = 196 N.)

Question 1

Please can you tell me how do you calculate acceleration a now?

Earlier "The sum of forces acting on the system as a whole is 196 Nt, acting on a system of m_total = 10+10+20 Kg = 40 Kg, for an acceleration of a = F/m_total = 196 nt/40kg = 4.9 m/s = 1/2 g."

Now what: The sum of forces acting on the sytem as a whole is still 196 Nt.
Acting on a system of mass ?
for an acceleration of a = F/m_total = 196 nt / ?kg = ?.


Sum of forces = 196N.

If this force is caused by the addition of a weight of mass 20 Kg, and given that the rockets and string are of negligible mass, the total mass of the system is just 20 Kg, and the sum of forces on the system is \vec {F} = 196 N. So:


\vec {F} = m \vec{a} \\
196N = 20kg * \vec{a} \\
\vec {a} = 9.8 \frac m {s^2} \\


Note that this is the same answer as you would have without the rockets - hence showing that the presence of the rockets is immaterial if their mass is zero and thety act in equal and opposite directions.

However, if this additional force \vec {F} is due to a third rocket of negligible mass, then the total mass of the system is close to zero, and hence the acceleration will be very large. In fact, if the total mass of the system is truly zero, then Newton says the acceleration would be infinite. Of course in reality the rocket is not zero mass, but you see that the acceleration of the system is depenedent on the total mass of the system.



Now let's double the forces R1 and R2 by using some mechanism. Please tell me the required force f to keep the same acceleration a as before.


Given that the rockets are of negligible mass, it does not matter what the values of R1 and R2 are. They could be 0 N, 100 N, 10^6 N, whatever -- as long as they are equal and opposite, they have no effect on the behavior of the system. So the required force is the same as before.

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:


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.

Hope this clarifies my point.

Hi elscarta!

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.

Cheers!
Shahin

Hi ebaines!

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 applied 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 see you get the same answer either way.

Hi ebaines!

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: \vec{F} \ = \ m \vec {a} . If you make m small, then for a given value of \vec{F} the acceleration \vec{a} becomes large.

(Not that I have accepted your answer, and I will come back to you shortly).

But I find you professional, I like the image you have inserted. :)

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.

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.



Correct!



Logical fallout 2
By touching point p I can tell that whether the source of the force 98 N is, F1 or R1.


If what you mean is: by trying to push point p with your finger you can tell if the system has a lot of mass (hence inertia) or not, then yes - this is correct.



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?


Not sure what you mean by "additional signal" - please explain. The tension in the string is the only "signal" that is transmitted. It takes more tension to move the system when it has a lot of mass than if it has a small amount of mass.



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.

Don't forget that the difference between F and R scenarios is that the F scenario adds a lot of mass to the system. You can definitely detect this mass when you try to accelerate it.

What you may not be grassping is the fact that although at rest both systems have 98 N of tension in the string, they have different tensions when you push on p to accelerate the system at 4.9 m/s^2. In the R scenario you push on p with .0147 N of force to the right - this increases the tension to the left of p by 0.0049 N, and decreases the tension to the right side by 0.0049 N, hence making an unbalanced force at R1 and R2 of 0.0049N to the right, which moves both R's to the right at 4.9 m/s^2. But in the F scenario, you have to push with 98.0049 N of force, which increases the tension to the left by 49 N and decreases the tension to the right by 49 N. This in turn causes unbalanced forces at F1 and F2 that are enough to accelerate both obectcts are 4.9 m/s^2.

Hope this helps.

You have only repeated yourself so this does not help. Can you give me any fresh evidence or any other example to support your explanation?

What you have said is not rational, try to analyze this for yourself: let me put you (or any other person) in place of the point object p. And as in scenario 2, you first have 980 N force applied on your right and left hand sides (100 kg weight tied with each your hands the similar way as shown in the picture).

You agree in this case that it will be difficult for you/him to move in either direction.

However if the same force 980 N is applied using 'the rockets', you say the experience for the person will be different and it will be easier for him to move either side as if there was no force applied on him.

Why he should be able to move around easily in the later case, when the forces applied on him are always the same?

You are able to answer this question because you have been told about the source of the force in advance. If you could only see the tension in the string which remains the same in both case, and not know the mechanism behind the force, you would not be able to answer this above question.

Cheers!

Hi ebains,

I am new to this site and saw this thread quite intersteing

I couldn't agree with you on this. I think F1 and R1 will have the same effect , can you explain further please

Hi ebains/shahin,


Good Thread...

Need to find answer and will let you know shortly :)

Cheers

You have only repeated yourself so this does not help. Can you give me any fresh evidence or any other example to support your explanation?


Hello Shahinomar. I will try - see below.



What you have said is not rational, try to analyze this for yourself: let me put you (or any other person) in place of the point object p. And as in scenario 2, you first have 980 N force applied on your right and left hand sides (100 kg weight tied with each your hands the similar way as shown in the picture).

You agree in this case that it will be difficult for you/him to move in either direction.


Difficult because for me to move I have to move those weights as well.



However if the same force 980 N is applied using 'the rockets', you say the experience for the person will be different and it will be easier for him to move either side as if there was no force applied on him.

Why he should be able to move around easily in the later case, when the forces applied on him are alway the same?


Because of the basic principal that equal and opposite forces cancel out and hence have no bearing on the issue. This seems to be the main principal which you are not quite grasping. So here's an example of how this works: suppose I was to arrange it so that there was a force of 50,000 N pressing on you from the front and an equal and opposite force of 50,000 N pressing on you from the rear. Do you believe it would be more difficult for you to move than if there was no such set of balancing forces? Your initial answer might be "yes," perhaps because in your every day life you don't typically experience a scenario of precisely equal and opposite forces, so you may naturally think that whenever forces are applied to an abject it makes it more difficult to move that object. But consider - as you sit there right now at your computer this is approximately the magnitude of force that you are experencing due to the air pressure on your body! And you don't even feel it. You have no problem moving left or right, front or back. Why? Because the air pressure all around you is constant, and consequently the air pressure forces cancel out and have no effect on your ability to move. This is precisely the principal that applies in your example of rockets and weights. Let's continue the example. Suppose now you are sitting on a chair with rollers, and you push back away from the table, and you measure how hard you had to push to get you and chair rolling at, sy 1 m/s^2. Now repeat, but this time you have a 200 Kg mass in your lap. You agree it would now be more difficult to push back and accelerate at the same rate as before? In the first case you have equal and opposite forces of 50000 N pressing on you and you had relatively little difficulty pushing back. In the second case you have equal and opposite forces of 50000 N of forces pressing on you and 200 extra Kg of mass in your lap, and you find it more difficult to move. Do you agree?

This is analogous to being pulled by equal and opposite forces of 100 N and trying to move either 0.003 Kg of mass or 20.001 Kg of mass. Moving 0.003 Kg of mass is definitely easier than moving 20.001 Kg, regardless of the magnitude of equal and opposite forces.



You are able to answer this question because you have been told about the source of the force in advance.



Not so. I agree that if I just stand there without trying to mnove and with 100 N of force being applied to the left and right I won't have any way of telling whether the force is caused by rockets or weights. But if I try to accelerate at 4.9m/s^2 in one direction or the other, it will be easier for me to do this if I don't have to drag the heavy masses of the two big weights along with me. Since the rockets weigh less, they don't slow me down as much as the weights. So it takes less effort on my part to move left or right for the case with the rockets than the weights.




If you could only see the tension in the string which remains the same in both case, and not know the mechanism behind the force, you would not be able to answer this above question.

Cheers!

ahh.. but the tension in the string in the two cases is not the same once you try to accelerate point p at 4.9 m/s^2! I showed this to you earlier in my previous post. Please go back and review that!