# Force Formula - Question

1. Feb 13, 2012

### -Physician

1. The problem statement, all variables and given/known data
Does the force formula $F=m χ 10\frac{N}{kg}$ which means $F=G$, apply for bodies on rest or motion?

2. Relevant equations
$F=G=m χ 10\frac{N}{kg}$

3. The attempt at a solution
I think it only applies for bodies in rest.

I'm waiting an answer , thanks!

2. Feb 13, 2012

### Staff: Mentor

The formula W = mg gives you the weight of a mass m, regardless of whether it's at rest or in motion.

3. Feb 13, 2012

### -Physician

Weight is G, not W.., anyways I wasn't asking about other formulas, i was asking about my formula $F=mχ10\frac{N}{kg}$

4. Feb 13, 2012

### Staff: Mentor

Where did you get that formula? It sure looks like 'mg' to me, which is weight. Seems like you're using g = 10 m/s^2.

5. Feb 13, 2012

### -Physician

I see you don't know that formula,we use that to find the Pushing force or weight of a body using the mass, which i think that the body is in rest. For example if we have a body with mass 2g, then the weight would be:

$G=mχ10\frac{N}{kg}=\frac{2}{1000}kg χ 10\frac{N}{kg} = \frac{200}{1000}N$(btw. kilogram and kilogram get simplified )

What I wanted to ask is if that formula applies also if the body is in motion

6. Feb 13, 2012

### sankalpmittal

No ! Doc Al got you correct. Your formula is nothing but of weight which is mass times gravity.

F=mg

Newton per kg in your unit gives you gravitational pull..

1 N/kg = 1 m/s2

You are assuming gravity in your formula to be 10 m/s2.

So you state that weight = mg = m*10 N/kg
OR
weight = m*10 m/s2
Of course , I guess everyone here know that kg and kg gets simplified..

Your formula is nothing but simple mg where you substituted for g by 10 metre per second squared.

Edit : Of course this formula applies when body is in motion. Doc Al already stated that in previous posts.

7. Feb 13, 2012

### -Physician

So, the gravity is not always $10\frac{N}{kg}$, then why is the Weight/Force formula
$F=mχ10\frac{N}{kg}$

8. Feb 13, 2012

### Staff: Mentor

That formula applies near the earth's surface.

9. Feb 13, 2012

### -Physician

Thank you very much, I am new on physics, I am only on the first grade so , I need to learn more :) Thanks@!

10. Feb 13, 2012

### -Physician

Anyways, can the body near the earth surface be in motion and in rest aswell, or only in rest or only in motion?
----------------------
Would...:
$1\frac{N}{kg}=1\frac{m}{s^2}$ because :
$1\frac{N}{kg}=1\frac{m}{s^2}=1\frac{kg\frac{m}{s^2}}{kg}=1\frac{m}{s^2}$kg and kg get simplified $=1\frac{m}{s^2}=1\frac{m}{s^2}$

11. Feb 13, 2012

### Staff: Mentor

I don't know what you mean. A rock (for example) can be in motion or at rest. (Not at the same time within the same reference frame, of course.)

Sure, those units are equivalent.