1. Feb 10, 2006

hi

1-I'm not shure that it is possible for two objects A and B to exert force on eachother, with A exerting 5 N and B exerting 9 N at the same time. If it is, then due to newton's third law A will also exert reactionary force -9 N back on B and B will exert reactionary force -5 N back on A.

So will resultant force on object A be 9 N and on B also 9 N?

2-Two objects with

m1 = 2000000 kg
m2 = 4 kg

are connected with rope. We pull object m2 with force F = 2000000 N.

F = ( m1 + m2 )*a - > a = F/( m1 + m2 ) = 1 m/S^2

If we threat m1 and m2 as separate systems, then F1 + F2 = F .
Net force on object m2 is F2 = F - F1 = 4N and net force on object m1 is F1.

But even if net force on m2 is 4N, we still pulled on m2 with force F. Considering that, does m2 experience all the destructive effects of being pulled with 2*10^6 N, or would damage on m2 be no greater than if it was pulled on pulled by F2 ( and m1 was wasn't connected to it with a rope )?

3-The following always confused me. If two forces(each 10^6 N) are exerted on a person, one from left and other from right, then resultant force is zero. But even if net force is zero, the two forces would cause a lot of damage to that person.

Well how is that different from air pressure where air puts pressure of 10^5 N/m^2 on your skin and there is equal pressure exerted on your skin from opposite direction?

Why doesn't skin get damaged from being pressed by such two large forces? How are the two examples different?

thank you

2. Feb 10, 2006

### Astronuc

Staff Emeritus
Our bodies are adapted to 1 atm (0.101325 MPa) and perhaps slight variations about the nominal value. Higher pressure (force) can be damaging.

106N if exerted on 1 m2 would be equivalent to ~10 atm, which would be excessive pressure.

106N is the force exerted by 102040 kg's. That's a lot of mass, which produces a lot of weight.

Ref - http://en.wikipedia.org/wiki/Newton
http://en.wikipedia.org/wiki/Kilogram-force

3. Feb 10, 2006

### arildno

Sticking to Newton's 3.law as valid, the (total) force from A on B is NECESSARILY equal in magnitude to the (total) force from B on A.

If you treat the two masses as separate systems,, then the net force upon m2 is F-T, where T is the rope tension. The force on m1 is T.

1. Even if you are pushed into pulp by those two forces, your CENTER OF MASS will experience no acceleratio.
2. DO NOT CONFUSE pressure with force!!
On a skin piece of 1cm^2, the air pressure given would give rise to a force:
$$F=10^{5}\frac{N}{m^{2}}*1*(10^{-2}m)^{2}=10N$$
The skin piece might well be strong enough to sustain that force without being damaged.

Consider however a force of 10N be applied to an area of only 1mm^2.

The local pressure there would be 100 times bigger than on the 1cm^2 patch, and the small area might well rupture.

This is why you easily can puncture your skin with a needle, whereas it is difficult to do so with a television screen.

4. Feb 10, 2006

### Staff: Mentor

I'm not sure I understand the setup. Newton's 3rd law tells us that is A exerts a force F on B, then B exerts a force -F on A. It's impossible for A and B to exert different forces on each other.

(The resultant force on any object is due to all the forces acting on it.)

As you seem to realize, just because the net force on something is small does not mean that the large individual forces have no effect on the object! Set a 2 million kg block of steel on top of you. You'll be crushed, but the net force on you will still be zero (assuming the floor doesn't collapse).

Similarly, imagine a man holding two ropes, being pulled in two directions by other men. (A double tug of war!) Even if the net force on the man is small, his arms still must exert a force equal to the tensions in the ropes. If they pull hard enough, he can be torn apart.

The net force determines the acceleration of an object, but the individual forces will determine the stress on the object.

Right!

Your skin is pressed from the outside by the air, and from the inside by whatever's just below your skin. Apparently, our skin is able to deal with that stress. (But don't go swimming at the bottom of the ocean!)

5. Feb 10, 2006

### Orefa

Hmmm, this was just answered, but since I typed it up already...

Question 1: your assumption is where the problem is. Forces exerted betwen two bodies are always equal and opposite. It's the law.

Question 2: your calculation of 'a' should yield approximately 0.999998 instead of 1. If you examine each mass as a separate system then the force on each one is given by F = ma, so F1 will be approximately 2,000,000 * 0.999998 = 1,999,996 N and F2 will be approximately 4 * 0.999998 = 3.999992 N.

The F force applied on one side of M2 is partly transferred to M1 on the other side, resulting in F1 (the portion transferred to mass M1) which is subtracted from F to give the resulting force F2. But this transfer assumes that there is no elasticity or deformation along the way. If mass M2 is strong as a paper bag then of course the shear will tear it apart and your system will not maintain its integrity, which will render the calculations worthless.

Question 3: air pressure from the atmosphere is countered and exactly balanced by internal pressure within every organ of your own body, including skin. You have air in your lungs, you have gas pressure within your veins and organs. Jump into a vacuum and you will swell up and die from this internal pressure. Same thing for deep water fish: the extreme pressure of the deep sea is exactly countered by the same pressure within every cell of the fish.

6. Feb 10, 2006

May I ask just one more question?

If objects A and B are positioned next to each other (touching) in isolated system and I apply force to A for just a brief instant, then same force will be applied to B and B will exert same force back to A. B will start moving and if force I exerted on A lasted just for that brief moment when B was touching A, then A will stay in equillibrium?

7. Feb 10, 2006

### Staff: Mentor

No.

Let's sort this out carefully. If you exert a force ($F_1$) on A, then A will exert an equal force on you. A and B will also exert equal (and opposite) forces ($F_2$) on each other. In general $F_2 < F_1$. Also, there will be slight delay from the time that $F_1$ is applied and the time that A hits B. No reason to think A will remain in equilibrium.

8. Feb 10, 2006

### lightgrav

True, A won't move very far while F_1 is applied , but A might still accelerate meanwhile (due to F_1 - F_2), and end up with non-zero velocity.

But after F_1 is removed, F_2 will also become zero, (A and B no longer touch)
and they will continue to move with whatever velocity they've acquired
(zero acceleration afterward, so returned to equilibrium)

9. Feb 11, 2006

Why would there be a slight delay? If forces ( that includes both F1 and F2 ) happen simultaneously and if object A is next to B, then A won't travel any distance before hitting B, so the moment A feels force F1, B should feel force F2.

So why would F2 be smaller than F1-where did that difference between two forces ( F1 and F2 ) go ?

If ( for the sake of argument ) forces F1 and F2 were equal, then object A would indeed stay in equilibrium if force F1 was applied just for that brief instant when A and B were at rest and touching?

10. Feb 11, 2006

### arildno

1. The effect of the force upon A is transmitted THROUGH A with a finite velocity.
Thus, the region of A directly in contact with B will not respond to that initial force until a small time interval has passed, and hence, the reaction force from B on A will not be simultaneous with the initial force.

11. Feb 11, 2006

### Orefa

something_about: It seems that you are correct in the generalities but Doc Al is correct in the details. Consider http://bednorzmuller87.phys.cmu.edu/demonstrations/mechanics/collisions/demo1404.html" [Broken]. All balls are touching so swinging one ball at one end results in one ball moving at the other end without any of the others budging. So apparently you are correct. But are you really? If you use more precise observations you see that the middle balls actually move a little and the last ball does not go as far as the first one. Some of the kinetic energy of the initial swing was dissipated by the interposed balls as they did move a little, changed shape a little through their own elasticity (which took a fraction of a second), warmed up from the collision, produced sound waves... The fact is that these collisions do not happen instantly, just very fast.

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12. Feb 12, 2006