# Why do bodies at rest weight more than zero?

• prober

#### prober

why do bodies at rest weight more than zero?

here is the bull thing...

now I am sitting on my chair and i know Earth exerts a force on me which is called my weight. by definition w=mg

I DO NOT have any acceleration g or anything else. let me prove

a = dv/dt my speed is zero v = 0, coz I am sitting so a = dv/dt = 0

w = ma = m times 0 = zero

which means i have to weight zero. but Earth attracts me so it can't be zero. this looks like a contradiction...can anyone tell me why i weight?

g can be defined as the acceleration due to Earth's gravity or it can be defined as the gravitational field strength ie the gravitational pull on a mass of 1Kg.

here is the bull thing...

now I am sitting on my chair and i know Earth exerts a force on me which is called my weight. by definition w=mg

I DO NOT have any acceleration g or anything else. let me prove

a = dv/dt my speed is zero v = 0, coz I am sitting so a = dv/dt = 0

w = ma = m times 0 = zero

which means i have to weight zero. but Earth attracts me so it can't be zero. this looks like a contradiction...can anyone tell me why i weight?

You have proved nothing because you are using the wrong equation. You just said, in your first line, "w= mg" and then you use "w= ma" instead! The first is true, the second is not.

"g" in w= mg is NOT the acceleration of the body. It is the acceleration a body would[\b] have if it were falling freely toward the earth. g= 9.81 m/s2, approximately, even if the body is not moving.

Welcome to PF!

Hi prober! Welcome to PF! w = ma = m times 0 = zero

which means i have to weight zero. but Earth attracts me so it can't be zero. this looks like a contradiction...can anyone tell me why i weight?

it's the chair! good ol' Newton's second law says that your change in momentum (mass times acceleration) equals the net force on you

the chair is pushing back on you with a force of w, upward, so the net force (total force) on you is w - w = 0 so, w=0 therefore w=mg=0 =>either your mass=0, g=/=0 or g=0 , m=/=0

so, w=0

uhh?

where did that come from? Weight, as used commonly, refers to the weight force. The net force on a body may equal 0, but this is due to the force normal of the surface beneath it cancelling out the force from gravity. Just because gravity provides constant acceleration, does not mean that it is purely an acceleration. It is an actual force, which provides approximately 9.8 m/s^2 of acceleration when it is the only force.

Since there is still a force being applied to body even when the net force = 0, we can feel the force. When you push against a wall, and it doesn't move, you can still feel the force you are applying. The same goes for the weight force.

Besides, each part your body has to apply a force to the rest of the body above it. That is really what is being felt.

As you said, weight w = ma.
Also, you said that since you are not accelerating, a=0 m/s^2.
This is false..The acceleration due to gravity is independent of your mass or position(on the surface)
Therefore, no matter whether you are accelerating or not(on the surface). The value of 'g' is roughly a constant throughout.

i.e, the 'a' in the equation which you have used is not your acceleration, but the acceleration due to the Earth's gravitational field. I think this thing is stated clear by someone here. Hope This would have helped. If not, please kindly listen to the others.. This is all about definitions... From wikipedia: "the weight of an object is the magnitude, W, of the force that must be applied to an object in order to support it (i.e. hold it at rest) in a gravitational field"

This means you need to know how hard the Earth is pulling at you, for this the law of universal gravitation can be used. The force between two bodies due to gravitation is:

F = G * m_earth * m_you / r^2

So, F is the force by which the Earth is pulling at you (as well as the other way around! Newton's third law...) Write this a bit differently:

F = (G * m_earth / r^2 ) m_you = g * m_you

thus, g = (G * m_earth / r^2 ) ~ 9.81 m/s^2

To solve this problem F=ma is not needed, this just tells you the NETTO force on you is zero. This is true because the chair applies an equal but opposite force on you counteracting your weight (again, Newton's third law).

Weight is not the force due to gravity, it is the force with which you are being pushed upwards by the surface you are on. Since there is an acceleration due to gravity, as you said you should logically have a positive velocity downwards. But you are not moving. so the force acting downwards ( mg ) must be opposed by an equal and opposite force acting upwards. ( don't confuse this with Newton's third law, the reaction mentioned there acts on the Earth not you. ) So you have mg acting downwards and W acting upwards so the net acceleration is 0.

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Weight is not the force due to gravity, it is the force with which you are being pushed upwards by the surface you are on. Since there is an acceleration due to gravity, as you said you should logically have a positive velocity downwards. But you are not moving. so the force acting downwards ( mg ) must be opposed by an equal and opposite force acting upwards. ( don't confuse this with Newton's third law, the reaction mentioned there acts on the Earth not you. ) So you have mg acting downwards and W acting upwards so the net acceleration is 0.

Hmmm... subtle... I don't think weight is a vector (is it?) To me it's strange to define the force acting on you by the chair should be defined the weight, since if the chair would be pulled from under you, you wouldn't weigh any less (unfortunately...). The way I see it the Earth just pulls at you by a force equal to your weight.

You're right about Newton's third law, it just says that there is a force acting on the chair that is equal but opposite to the force acting on you... But of course I meant that, since you don't move, there is a force on you, by the chair, which is opposite and equal to the force due to gravity. However, this follows from a force balance, not from Newton's third law...

EDIT: I just noticed that indeed the weight is not a vector. In the defenition: 'the weight is the magnitude of the force ... '

here is the bull thing...

a = dv/dt my speed is zero v = 0, coz I am sitting so a = dv/dt = 0

Just a mathematical issue here: if v is zero, it does not mean it's derivative is zero! Take for example:

v = 1 - t^2

v is equal to zero if t = 1, but dv/dt = -2t and thus at t = 1, dv/dt = -2!

whoa

What do we have here. Let me understand something

w= m*g

g--> vector, as in it is resolved into components of i, j,k.

Mass acts as a mere multiplying factor wherein the value of g is magnified by the value of the mass

How does this become scalar, kindly clarify

v = 1 - t^2

v is equal to zero if t = 1, but dv/dt = -2t and thus at t = 1, dv/dt = -2!

Now isn't that actually a discontinuous derivative. I don't remember what this is called.

But i think practically it makes no sense, that's like saying in the absence of velocity we can have acceleration.

On the other sense velocity at t=1 would not exist as in it would become negative for all times greater than 1, also here acceleration would be continuously negative implying retardation, is it actually practically realizable. I doubt so

Now isn't that actually a discontinuous derivative. I don't remember what this is called.

This is just a normal derivative. I don't know any other kinds of derivatives... When a function changes slope at a point, it's derivative becomes discontinuous and doesn't exist in that point. I don't know if that is what you mean?

But i think practically it makes no sense, that's like saying in the absence of velocity we can have acceleration.

This makes perfect sense. When you throw a ball vertically in the air, at a certain point it's velocity changes direction and the ball comes back to earth. At that exact instant it's velocity is zero. However, since the Earth's gravity is still pulling on it, the acceleration is not zero!

On the other sense velocity at t=1 would not exist as in it would become negative for all times greater than 1, also here acceleration would be continuously negative implying retardation, is it actually practically realizable. I doubt so

A negative velocity just means a velocity in the negative direction of your axis. If the z-axis points upward to the sky, then a negative velocity means that the object has a speed towards the earth.

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whoa

What do we have here. Let me understand something

w= m*g

g--> vector, as in it is resolved into components of i, j,k.

Mass acts as a mere multiplying factor wherein the value of g is magnified by the value of the mass

How does this become scalar, kindly clarify

in w=m*g the vector w is not the weight, it is the force by which the Earth pulls at you. The weight is the absolute value of this, so W = |w|, then it is a scalar

Hmmm... subtle... I don't think weight is a vector (is it?) To me it's strange to define the force acting on you by the chair should be defined the weight, since if the chair would be pulled from under you, you wouldn't weigh any less (unfortunately...). The way I see it the Earth just pulls at you by a force equal to your weight.
Oh oh! You are wrong.

Let's say your standing in a lift (or elevator). If the lift accelerates downwards, you actually weigh less! This may sound strange, but it really isn't. You still have the same mass, it's just that the floor exerts less force on you. If you've ever been in a lift you know this from experience. Another example where you can even become weightless, is in a parabolic flight. The plane accelerates down so quickly, that you can just 'fly' in mid air.

The confusion comes from the everyday usage of the word 'weight'. Then it usually means what we call 'mass' in science.

in w=m*g the vector w is not the weight, it is the force by which the Earth pulls at you. The weight is the absolute value of this, so W = |w|, then it is a scalar

Obviously that is not classifiable as it automatically becomes a scalar once u take the magnitude, what i thought u were saying is that weight by itself is a scalar

This is just a normal derivative. I don't know any other kinds of derivatives... When a function changes slope at a point, it's derivative becomes discontinuous and doesn't exist in that point. I don't know if that is what you mean?

I was wrong on this the function is a minima at t=+/-1.

This makes perfect sense. When you throw a ball vertically in the air, at a certain point it's velocity changes direction and the ball comes back to earth. At that exact instant it's velocity is zero. However, since the Earth's gravity is still pulling on it, the acceleration is not zero!

Raises an interesting question in my opinion. Actually till this point the g acts as a retardation on the ball and the moment the ball reaches this point the retardation equals the acceleration thereby canceling each other out. So again classically it does not hold that it has acceleration.

Raises an interesting question in my opinion. Actually till this point the g acts as a retardation on the ball and the moment the ball reaches this point the retardation equals the acceleration thereby canceling each other out. So again classically it does not hold that it has acceleration.
Having a velocity of zero on doesn't mean there can't be any acceleration...?
v=0
a= dv/dt = 9,8 m/s²
F=ma, not F=mv!

And what do you mean by retardation cancels acceleration out? Remember that acceleration is a vector.

Fg = mg. If there's a gravitational force F and the ball has a mass m there is always a acceleration vector which points in the same direction as the Fg (to the centre of the Earth ). Even when you throw the ball up, or when it's velocity changes direction, the Fg points to the centre of the Earth and so does the acceleration.

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You may see it more clearly raknath If you sketch a velocity time graph the acceleration being the gradient of the graph.For objects that move with simple harmonic motion the acceleration is a maximum when the velocity is zero.

And what do you mean by retardation cancels acceleration out? Remember that acceleration is a vector.

Fg = mg. If there's a gravitational force F and the ball has a mass m there is always a acceleration vector which points in the same direction as the Fg (to the centre of the Earth ). Even when you throw the ball up, or when it's velocity changes direction, the Fg points to the centre of the Earth and so does the acceleration.

I have a point to make here, I don't deny that theoretically it makes perfect sense to say that acceleration exists in the absence of velocity, in fact at zero velocity most equations would have negative acceleration. My point about the ball going up is simple, the Earth continues to retard the ball even as it moves up and here the g acts as retardation.

When the retardation overcomes the acceleration of the ball, the ball stops and thereby you would have the ball accelerating in the opposite direction.

Hence my point remains that at zero velocity, the acceleration of the ball would be equal to the retardation and the ball would stop, so acceleration here is discontinuous.

I have a point to make here, I don't deny that theoretically it makes perfect sense to say that acceleration exists in the absence of velocity, in fact at zero velocity most equations would have negative acceleration. My point about the ball going up is simple, the Earth continues to retard the ball even as it moves up and here the g acts as retardation.

When the retardation overcomes the acceleration of the ball, the ball stops and thereby you would have the ball accelerating in the opposite direction.

Hence my point remains that at zero velocity, the acceleration of the ball would be equal to the retardation and the ball would stop, so acceleration here is discontinuous.
It seems you are assuming there is some kind of force that accelerates as the ball goes up and another force (gravity) which retards the velocity. Is that correct?

This is not the case though. As soon as the ball leaves your hand no force other than gravity pulls the ball. So saying that retardation overcomes acceleration makes no sense.

I think what you're not getting, is that the acceleration vector can point in the opposite direction of the velocity. Think about that. The ball is being accelerated downwards even if it is still going upwards.

It seems you are assuming there is some kind of force that accelerates as the ball goes up and another force (gravity) which retards the velocity. Is that correct?

Ok let me place my case very clearly here,
1. We have a force F which is exerted on the ball as the ball leaves the person's hand, hence the acceleration a on it and F=m*a, where m becomes the mass of the ball
2. There is a gravitational force f which is acting counter to this ball moving upward

Hope this makes my case clear

This is not the case though. As soon as the ball leaves your hand no force other than gravity pulls the ball. So saying that retardation overcomes acceleration makes no sense.

I think what you're not getting, is that the acceleration vector can point in the opposite direction of the velocity. Think about that. The ball is being accelerated downwards even if it is still going upwards.

Nope if fully understand this, its a vector response

Ok let me place my case very clearly here,
1. We have a force F which is exerted on the ball as the ball leaves the person's hand, hence the acceleration a on it and F=m*a, where m becomes the mass of the ball
2. There is a gravitational force f which is acting counter to this ball moving upward

Hope this makes my case clear

Nope if fully understand this, its a vector response
So acceleration is continuous :)

Hence my point remains that at zero velocity, the acceleration of the ball would be equal to the retardation and the ball would stop, so acceleration here is discontinuous.
Let's bring this to a head right here, right now. raknath could you please answer the following two questions:
1. Once the ball has left the person's hand, what is the acceleration of the ball?
2. Is the function a(t) = constant a continuous function?

I have a point to make here, I don't deny that theoretically it makes perfect sense to say that acceleration exists in the absence of velocity,
in fact at zero velocity most equations would have negative acceleration.
Acceleration describes how velocity changes. You can certainly have zero instantaneous velocity and non-zero acceleration.

My point about the ball going up is simple, the Earth continues to retard the ball even as it moves up and here the g acts as retardation.
OK, you can think of it that way. Gravity exerts a downward force on the ball, which produces a downward acceleration. It "retards" the ball on the way up, but that same force speeds the ball up on the way down.

When the retardation overcomes the acceleration of the ball, The "retardation" is the acceleration of the ball. Perhaps you mean that the retardation/acceleration overcomes the velocity. (Sloppy terminology, but OK.)
the ball stops and thereby you would have the ball accelerating in the opposite direction.
After reaching its highest point, the ball starts to move in the opposite direction. The acceleration, 9.8 m/s^2 downward, hasn't changed.

Hence my point remains that at zero velocity, the acceleration of the ball would be equal to the retardation and the ball would stop, so acceleration here is discontinuous.
Again, this is meaningless: What you call the "retardation" is the acceleration, which is downward. (Realize that "acceleration" in physics doesn't just mean "speeding up"--it refers to changing velocity, whether speeding up or slowing down.)

Ok let me place my case very clearly here,
1. We have a force F which is exerted on the ball as the ball leaves the person's hand, hence the acceleration a on it and F=m*a, where m becomes the mass of the ball
When the ball loses contact with the hand, the hand no longer exerts its upward force on the ball.
2. There is a gravitational force f which is acting counter to this ball moving upward
Gravity is always acting downward. Once the force of the hand stops acting, the only force on the ball is the downward force of gravity. F= mg = ma, thus a = g (downward).

Right got the point thanks guys :)

But i think practically it makes no sense, that's like saying in the absence of velocity we can have acceleration.

Assuming we CAN'T have an acceleration in the absence of velocity is like stating "whatever doesn't move, won't move for ever".