# Newton third law

1. May 19, 2012

Newton third law!!!

Hi fellow members!!!
I would like to ask a silly question because my basic concepts are quit weak. Since according to newton third law, with every action, an opposite and Equal force react to resist motion.Since both ACTion and Reaction forces are equal, they will cancel each other thus body should be in equilibrium than how the object moves or accelerated???
Answers will be appreciated!!!
Thnxx

2. May 19, 2012

### lcypf

Re: Newton third law!!!

Actually the action and reaction force are affect on different objects so they could make the objects moves or accelerated. For example ,if I push someone ,I will move backward and he will move forward.(I'm not an american.Maybe there're some grammar error,please correct me.)

3. May 19, 2012

### phinds

Re: Newton third law!!!

If you are facing each other and you push him, you will both move backwards, but yes, this is a good explanation for the OP.

OP, the forces are EQUAL ... that is not the same thing as equilibrium.

4. May 20, 2012

Re: Newton third law!!!

Thnx for rplying phinds & lcpyf but i m also cnsidering here different objects. But the forces r same!
Suppose a ball of mass m is falling under the action of gravity. Consider mass of earth is M. Distance b/w the ball & surface of earth is r at any instant. According to law of gravitation :-
F=GmM/r2
Thus, here gravitation force exerted by the ball on earth=gravitation force exerted by earth on ball= GmM/r2
since, objects are different, but forces are they why gravitation force exerted by earth on ball dominates? Ball should have to be suspended since the forces are equal.

5. May 20, 2012

### Staff: Mentor

Re: Newton third law!!!

OK.
As already pointed out, the equal and opposite forces act on different bodies. Only if the net force on the ball were zero would its acceleration be zero. But that's not the case here.

6. May 20, 2012

### Staff: Mentor

Re: Newton third law!!!

Yes, the force exerted by the ball on the earth is equal to the force exerted by the earth on the ball (although the forces are in opposite directions - the force of the earth on the ball is pulling the ball down, towards the center of the earth, while the force of the ball on the earth is pulling the earth up, towards the center of the ball).

Now let's apply Newton's Second law, $F=ma$ to the earth and the ball:

(force of earth on ball) = (mass of ball) * (acceleration of ball towards earth)
(force of ball on earth) = (mass of earth) * (acceleration of earth towards ball)

The mass of the ball is about $1$ kg.
The mass of the earth is about $6\times 10^{24}$ kg

The two forces are equal. So when you plug in the numbers, you'll see that both the earth and the ball will accelerate towards each other. But because the mass of the earth is so much greater, the earth's acceleration is much less - indeed, it is far too small to measure with even the most sensitive instruments - so we only notice the movement of the ball.

(If we did have sufficiently sensitive instruments, we would be able to see that the earth is moving towards the ball as well as the ball moving towards the earth, that the forces acting on the entire ball+earth system are balanced and the center of gravity of that entire system is not moving. But it's really impossible to see this effect when the masses are so different - $6\times 10^{24}$ is a very big number indeed)

7. May 21, 2012

Re: Newton third law!!!

Thnx nugatory. You saved me. You resolved a big confusion which i was facing these days.
Let me ask another question about x & y component in newton third law.
When the coordinate system of free body is not inclined then
x-component=xcos(theta)
y-component=ysin(theta)
which is understandable but when the coordinate system is inclined to somd angle (theta) then above mentiöned case is inverted.
X-component=xsin(theta)
y-component=ycos(theta)
why this is so?

8. May 21, 2012

### robphy

Re: Newton third law!!!

(in general, a new question should be started in a new thread)

Every vector can be thought of as "the hypotenuse of _some_ right triangle".

In breaking a vector into components,
you FIRST need to choose a coordinate system
and then use a right-triangle who legs are parallel to those coordinate axes.

Remember... in a right-triangle
cosine(angle) goes with the component adjacent-to-that-angle
and sine goes with the component opposite-to-that-angle.

Now you just have to do some geometry to express
some angle in your right-triangle above
in terms of the incline's angle.

Try it out.

9. May 22, 2013