g-force equation??? (1 Viewer)

[Teh Duke]

Hey guys,

My textbook (Jacaranda) says that the equation for g-force is (g+a)/9.8 , but i think it should be (g-a)/g.

For example:
Take g to be -9.8 ms^-2 (Therefore up is defined as positive)
If a rocket is taking off, the net force on the rocket itself will be directed upwards (+ve direction), lets say for arguments sake that the acceleration acting on the rocket is 15ms^-2, using the equation (T-mg)/m.
When the rocket is taking off, we know that the occupants feel a g-force greater than 1.

So, using the old equation:
g= -9.8
a= +15

g-force = (-9.8+15)/9.8 = 0.531 g (which is wrong)

Using (g-a)/g:
g= -9.8
a= +15

g-force = (-9.8-15)/-9.8 = 2.531 g (which is right)


In the worked solutions in the book they seem to ignore the fact that a and g are in opposite directions which is why they still get the same solution. I just think it's a bit weird that they ignore the directions, because if you tried to use their equation to work out the g-force acting on something that is accelerating downwards (same direction as g), you would get the wrong answer if you ignored the directions of a and g.

 

[Aaron.Judd]

I'ver always known it as (g+a)/9.8

 

[Aaron.Judd]

Use g as 9.8 and it works fine.

 

[Forbidden.]

回复: Re: g-force equation???

Use g as 9.8 and it works fine.

It's on the HSC Physics data sheet.

 

[I-Love-Jesus]

Yeah, usually g is taken as 9.8, not -9.8.
'g' is the magnitude of the acceleration due to gravity, 9.8m.s^-2, but if we are considering the direction of this acceleration, it is written as '-g', which equals -9.8m.s^-2.

For example, in the formula sheet, the formula for vertical displacement is given as:
∆ y = u_yt + a_yt²
If we are using gravity as the acceleration in the formula, we substitute 'a' as '-g', which is -9.8. Hence:
∆ y = u_yt - gt²

Since -g = -9.8, it follows that g = 9.8.

I hope that helped

 

[Forbidden.]

回复: Re: g-force equation???

Yeah, usually g is taken as 9.8, not -9.8.
'g' is the magnitude of the acceleration due to gravity, 9.8m.s^-2, but if we are considering the direction of this acceleration, it is written as '-g', which equals -9.8m.s^-2.

For example, in the formula sheet, the formula for vertical displacement is given as:
∆ y = u_yt + a_yt²
If we are using gravity as the acceleration in the formula, we substitute 'a' as '-g', which is -9.8. Hence:
∆ y = u_yt - gt²

Since -g = -9.8, it follows that g = 9.8.

I hope that helped

Although the actual value is right there, as you pointed out the concept behind acceleration and gravity is left to the student.

 

[Teh Duke]

Now i know why they do it the way they do.
It's because when they derive the equation, they used the reaction force (Reaction = mg) which is equal and opposite to the weight force. Thats why both the acceleration and the weight force are taken in the same direction.