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

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ebaines

Correct!

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.

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.

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.

shahinomar

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 alway 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!

kiranvirus

hi ebains,

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

I couldnt agree with you on this. I think F1 and R1 will have the same effect , can you explain further please

rameshg123

hi ebains/shahin,


Good Thread...

Need to find answer and will let you know shortly

Cheers

ebaines

Hello Shahinomar. I will try - see below.

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

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.

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.

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!