# Question regarding newton third law and weight in a lift

1. May 8, 2006

### sdapin

emm, i m new member here
maybe i m stupid and young enuf to ask these question, but i seriously nid help from u all to explain to me
1st, if we standing on a weighing machine in a lift, when the lift accelerate downward, the reading of W.M. wil decrease?and when accelerate upward the reading wil increase?

2nd, according to newton third law, every action there is an opposite and equal reaction, so y thing in the world can move, said if an object of 5kg move wif acc of 2ms-2, force is 500N, then the reaction force mus be 500N isnt it?so net force eventually become 0N, so y would it move?

sry these question maybe really low standard, but it seems to be hard for me, since i m still young and cant understand wat the book say, nid u all to explain

thanks btw!!!

2. May 8, 2006

### Staff: Mentor

Consider the forces acting on you: your weight (which acts down); the force of the scale (which pushes you up). Then consider Newton's 2nd law, which relates net force and acceleration. Note that the scale reads the force it exerts on you (and you on it), which does not necessarily equal your weight.

Do 3rd law force pairs ("action/reaction" pairs) act on the same object? Equal and opposite forces only add to zero if they act on the same object.

3. May 8, 2006

### sdapin

so can we calculate the force nid to produce the acceleration given?we know the net force. i.e. a guy wif mass 60kg in a lift at acceleration of 2ms-2, is 120N.does tat mean the lift system exert 120N on the lift to accelerate it wif 2ms-2?if not, how much force is nided in order to produce the acceleration of 2ms-2?

4. May 8, 2006

### arildno

Well, it is great that you have the capacity for humility, but you certainly shouldn't wallow in self-contempt.

Your questions are better than that; in fact, they are good novice questions.

5. May 8, 2006

### Hootenanny

Staff Emeritus
Your logic is good, but you would also need to take into account the mass of the lift. Remember the man and lift will be acclerated at 2 m/s/s, not just the man. Also you would need to include any frictional forces present.

~H

6. May 8, 2006

### Staff: Mentor

If you are only concerned with figuring out what the scale reads, all that matters are the forces acting on the man. So, if the acceleration is +2m/s^2, then the net force on the man will be 120N (upward). That net force is due to the combined force of the man's weight (acting downward) and the force that the scale/elevator pushes up on him with (which is what you are trying to find).

Write a mathematical expression for the sum of the forces (be sure to get the signs right) and solve for the force of the scale on the man.

7. May 8, 2006

### nrqed

This is a very common misconception, a *lot* of other people are confused by this.

warning #1 :The answer is that whenever we talk about the action-reaction principle, the two forces involved (the action and the reaction forces) are always acting on different objects . If you push on the wall with a force of 100N, the force will push back on you with a force of the same magnitude and opposite directions. These two forces are party of an action-reaction pair. But notice: one force is on the wall and the other force is on you. Because they are acting on different objects, they don't cancel out.

As for the lift problem, this should be done with a free body diagram. But the end result is this: there are two forces acting on the man: gravity creating a gravitational force down which is what we call the *weight*, given by mg, and a normal force exerted by the floor. If there is a scale under the feet of the man, the scale will be reading the value of the normal force ( warning #2 :the scale does NOT read the weight, but the normal force...only when there is no acceleration, for example when you step on yoru scale in yoru bathroom, does the normal force equal the weight...the problem is that we are so use to that fact that we start believing that a scale always read the weight, which is not true).

Now, Newton's second law says

Normal force minus weight = m times acceleration

where the acceleration will be positive if it is upward and negative if it is downward.

( warning #3 acceleration being upward does not necessarily mean that the lift is *moving* upward or even that the lift is speeding up! This is another common source of confusion)

I hope this helps a bit...

Patrick

8. May 8, 2006

### reincarnated_soul

Weight in a lift

In order to solve problems like this u should first begin by drawing a free body force diagram on a suitable object. In this respect it is the man in the lift bcuz u want to find the reading of scale (the reading of the scale does not mean the man's weight-it is in fact the just the normal contact force or reaction force OF THE MAN ON THE SCALE.......or u might call it apparent weight). U in fact find the reaction force of the scale on the man since u r considerin only the forces acting on the man, but since forces act in action and reaction pairs, the reaction force of the scale on the man is equal to the normal contact force of the man on the scale (a 3rd law pair- remember 3rd law pairs act on different bodies so the forces dont cancel out since when u consider the two objects individually there is only one force on each- one normal force on the man and one normal force on the scale).

Now consider the forces on the man:
1. Since the man and the scale r in contact, the man experiences a normal contact force or reaction force from the scale. i.e., the push of the scale on the man and this acts upwards on the man. Call that force R.

2. Since the man has mass in a gravitational field, he experiences a pull of gravity and this acts downwards on the man.Call that force W.

Now when the lift is accelerating upwards, it means the man is also experiencing the same acceleration upwards since he is a part of that system. And since the acceleration is upwards it means the man experiences a net force in the upward direction which gives rise to the acceleration.

Now from the forces acting on the man the upward force is R and this must be greater than the W since the man is havin a net upward acceleration. If 'a' is the acceleration and 'm' is the mass of the man then from Newton's second law,

Net force=mass x acceleration
Therefore, R - W = ma

And by putting the values of 'W' and 'ma', R can be found which would be the reading of the scale.

Conversely for a lift accelerating downwards, W must be greater than R so there is a net force downwards and hence
W - R = ma.
***B***

9. May 9, 2006

### sdapin

thx for all answer to my QN. really help a lot!
ok now regarding third law, i saw a situation in my reference material
lets say there is 2 ball., one is stationary, one is moving the stationary, after they collide, the moving one wil stop and the stationary one wil move wif the same velocity of the moving one jus now
my qn is, y the moving one before collision wil stop after collision?since when it collide wif the stationary one, an equal force is produce in opposite direction, then the force wil not push the ball backward?
it is a closed system btw, no external force, and 2 balls having the same mass

Last edited: May 9, 2006
10. May 9, 2006

### Staff: Mentor

The force of the second ball does push the first ball back! So much so, that the first ball stops moving.

This phenomenon, that one ball stops moving and the other ball starts moving, is a consequence of conservation of momentum and energy. And it can be understood using Newton's 3rd and 2nd laws.

When the balls collide they exert equal and opposite forces on each other. (Newton's 3rd law.) The first ball, already moving, is slowed down by the force from the second ball; the second ball is sped up by the force from the first ball. Since they have the same mass and the same force, the balls experience the same acceleration (but in opposite directions)--Newton's 2nd law.

11. May 10, 2006

### sdapin

if a smaller mass ball collide wif a bigger mass ball or vice versa, wat wil happen and y?'
sry so many question, but i need to totally understand the whole topic, else i will stuck there...thanks

12. May 10, 2006

### sdapin

the force exerted back is always enuf to stop the 1st ball immediately no matter how fas it move toward the stationery one?can u somehow give some example of calculation in oder to make me more understand?what i wonder is, y the 1st ball wont bounce back like what we seen in real world, if throw a ball to a wall, the ball will bounce back?

Last edited: May 10, 2006
13. May 10, 2006

### Staff: Mentor

In this special case: A moving ball made a direct collision into an identical stationary ball, conserving mechanical energy.
It's a combination of conservation of momentum:
$$mv = mv_1 + mv_2$$
And conservation of energy:
$$1/2 m v^2 = 1/2mv_1^2 + 1/2mv_2^2$$

If you solve these, you'll find that v_1 = 0 & v_2 = v is the only physically possible solution.
The two balls colliding and exchanging momenta is commonly seen in the real world! You need to spend more time hanging out around pool tables.

As far as bouncing a ball off a wall... that's quite different. The wall doesn't move!

Go back to the example of one ball hitting the other. If you make the second ball heavier and heavier, the collision will begin to look more and more like the collision of a ball with a wall. The incoming ball will bounce back from the much heavier target ball.

14. May 21, 2006

### sdapin

well thx for info
i hav another QN, for the lift, y we cant add the net force into the weight?since add up it oso produce a same net force issit?they r acting on same object oso, net force downward and it reaction force on the man, weight downward and it reaction on the man too?am i get somthing wrong?

the ball qn, y the 1st ball will bounce back if collide wif bigger obj?

really really sry for so many question , i jus cant get it...

15. May 21, 2006

### Hootenanny

Staff Emeritus
RE: The lift

You can consider the net force on the man, that is what reincarnated_soul was doing; There are only two forces acting on the man, weight and the normal reaction force. Do you agree? Using Newton's second law;

$$\sum \vec{F} = m\vec{a}$$

Weight(mg) is acting down and the normal reaction force is acting upwards, therefore, $\sum \vec{F} = R - mg$ and thus;

$$R - mg = m\vec{a}$$

This can also be written as;

$$R = m\vec{a} + mg$$

Using this equation it is easy to see how the reaction force (the reading on the scale) with vary with acceleration.