# Why is W=m*g relevant if I am standing still and not moving ?

1. Aug 1, 2013

### urbano

1. The problem statement, all variables and given/known data

Why is gravity still 9.81m/s/s when we are standing still ? and why do I have a weight when I am standing still ?

2. Relevant equations
F=m*a
W=m * g

a= (final velocity - initial velocity)/ time

3. The attempt at a solution

So it requires more of a theory explanation and I'm a touch confused...

If acceleration is based on a final velocity minus the initial velocity and then divided by time, how can I be accelerating at 9.81 m/s/s if I'm standing still ?

If I jump in the air the earth pulls me back towards its center at 9.81 m/s/s. That much makes sense. ..but standing still on the footpath I don't quite get how I can be accelerating towards the center of the earth given I am having no change in velocity, and a change in velocity is demonstrated through acceleration.

Obviously I'm missing something fundamental here
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 1, 2013

### SteamKing

Staff Emeritus
You are confusing F = ma with F = GMm/r^2.

3. Aug 1, 2013

### rude man

Standing still, you're not accelerating. The gravitational force Mg pulling you down is balanced by the Earth pushing you up.

If standing on the Equator, you are subject to a small acceleration = M(w^2)R where w = Earth rotation in rad/sec and R = radius of Earth. Even when standing still. The reason is that acceleration is not just a change in speed but also direction. As the Earth turns you are constantly changing the acceleration vector's direction since it always points towards the Earth's center. At the North or South pole this acceleration is entirely absent. In between it's multiplied by [cos(L)]^2 where L = latitude.

The above ignores the acceleration of the Earth's rotation about the Sun. The Moon also plays a tiny part.

4. Aug 1, 2013

### Pythagorean

It's just a matter of interpretation. You're right, g doesn't describe your acceleration. It describes the acceleration field generated by Earth's mass.

5. Aug 1, 2013

### urbano

Thanks for the help with this.....much appreciated.

Still trying to get my head around the F=m*a is equivalent to W= m*g statement. So this is not true then (in the case where I am standing still on the footpath)

If I am not accelerating to the center of the earth due to the force of the footpath etc pushing against me but am accelerating as I am changing position due to the earth rotating, is there some equation I should be paying attention to that involves my mass and the rate of rotation (is it 30 km/h per second ?)

So in terms of basic physics (i.e that of a noob to physics), in which situation might you use W= m*g ?

6. Aug 2, 2013

### ehild

We call "weight" the force, a body pushes a horizontal support with, when in rest with respect to the support. On the surface of the Earth, the object experiences the force of gravity: -mg (pointing downward) and the normal force from the ground : N (upward). Being in rest, the sum of forces acting on the object is zero. ƩFi=N-mg=0. The normal force is equal to mg: N=mg. But the normal force exerted by the ground has the same magnitude as the force exerted by the body to the ground, that is, the weight of the body.
In the space station, the astronaut do not feel force from his seat, the normal force is zero, he feel weightless.

It is usual to speak about weight of bodies in general, and W=mg is meant about it. So you can ask what is the weight of an 1kg body at 100 km above the Earth. Then use the local g: GM/r2, where r is the distance from the centre of Earth.

ehild

7. Aug 2, 2013

### rude man

I gave you that equation. It's a = (w^2)R. w is the rate of rotation and R is the radius of the earth. w = 2pi/24 hrs = 7.3e-5 rad/sec. R is about 8000 miles ~ 1.3e7 m.

So in terms of basic physics (i.e that of a noob to physics), in which situation might you use W= m*g ?[/QUOTE]

When you're stepping on a scale, for one thing. The scale measures the amount of force (expressed in kg which is very confusing) the Earth pushes on you. So if you're at the Equator you weigh less than you do at the North or South pole (ignoring differences in R). The difference is m(w^2)R with m = your mass. The difference is only (w^2)R/g or about 0.7%.