What is the relationship between weight and g-force in determining F?

[Aqeelz]

Hi.

Can you clarify for me the equation for g-force below:

F= ma + mg

I got this equation from my lecturer. He asked the class to determine how was the formula above formulated. But this is not a homework. Is the equation above true for g-force? If it's true, how does the formula come about?

I thought g-force is given by (as stated by Newton's law of universal gravitation):

F = G m1m2/r^2

I really can't see how these two equations are equal. Could you help me out? Thanks!

 

[Aqeelz]

Now that I have had a good look at my question, I thought about the following:

F = ma + mg
F - mg = ma
F - F_g = ma
→ ƩF_net = F - F_g = ma

Actually I will be on a trip to ride a roller coaster. Our class was asked to determine first how was the 'formula for g-force' F = ma + mg formulated.

If an object falls freely downward, it would be subjected to g (acceleration due to gravity). Time this with the object' mass we would obtain its weight/ gravitational force, F_g. However Newton's 3rd law states that for every action there is an equal but opposite reaction, hence this opposite force, denoted as F, which is directed upward (opposite F_g), but has an equal magnitude to that of F_g. The net force, ƩF_net, is then the difference between these two forces (F and F_g) and it must obey Newton's 2nd law, which is

ƩF_net = ma​

Therefore, the F in the equation F = ma + mg (given by my lecturer) is not really a g-force, is it? Because g-force is already given by F_g = mg. In other words, F = ma + mg is not an equation to find the gravitational force, right?

Am I right? If not, please correct me.

Thanks!

 

[AJ Bentley]

Well, mg is the force due to gravity alone - which we call 'weight'.

Is weight a g-force, or do we consider the other 'peculiar' forces (a) experienced on a ride the 'g-force' and leave weight out of it? Or is it all g-force?

It's just semantics, but generally we'd consider weight part of the g-force so it's g+a.
So F = m(g+a)