#1
So F=ma
does F=mg(small g,or in other words the constant for gravity,that equals 9.81)
and does F=mgv?

Vectors are all the same!
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Last edited by Azrael_031 at Jan 15, 2009,
#3
if the object is falling than the accelaration=9.81

also


F=time x velocity

or

F=(time x displacement)/time
Last edited by slashrules!!! at Jan 15, 2009,
#4
I'm not to sure about that last one.
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#6
Quote by slashrules!!!
if the object is falling than the accelaration=9.81


O rly ^^

That was kinda obvious,and I already stated that i know that g has a vector.

I'm calculating the force of an object falling down on it's own with more weight(5 times more tbh) and how much velocity would it take to compensate the same amount of force on the other object

Edit:to the guy above me,what's with stating the obvious?!I know that g only acts in free fall!
Hodam,a stojim...
Last edited by Azrael_031 at Jan 15, 2009,
#7
F=mg is simply the weight of the object with the mass m(g=9.81 btw, at least on earth)
And F doesn't equal mgv.
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#9
Quote by nico42
F=mg is simply the weight of the object with the mass m(g=9.81 btw, at least on earth)
And F doesn't equal mgv.


Could you please then give me the formula,I need one so i can calculate something!
Hodam,a stojim...
#10
Quote by Azrael_031
So F=ma
does F=mg(small g,or in other words the constant for gravity,that equals 9.18)
and does F=mgv?

Vectors are all the same!


It depends on a variety of things, not to mention the constant for gravity is 9.81, not 9.18

And yes to the first if it is being dropped, and no to the second unless v = 0

And gravity always works in free fall.

For example, if the mass of an object is 8 lbs, the velocity is 0 meters per second because it was not thrown, it was only dropped, and the location is sea level, making gravity 9.81 meters per second squared, F = ma, and since a = g in this because gravity is the acceleration, F = (8)(9.81), 78.41 N.
Last edited by punkforlife93 at Jan 15, 2009,
#11
Quote by SGPlayer08
What's the problem at hand you're trying to solve?


Told you already.

I have one object that is 5kg,affected only by g.
The other one,1,1 kg(let's say 1,0!) that's affected by both g and v

wait could f=mv^2?
Hodam,a stojim...
#12
Quote by punkforlife93
It depends on a variety of things, not to mention the constant for gravity is 9.81, not 9.18

And yes to the first if it is being dropped, and no to the second unless v = 0


Sorry,typo,I'll correct that

Edit:right'o,thanks,now i get it!

I need to calculate the velocity of both objects,then force!

EDIT2:damn it,idk s nor t!
Hodam,a stojim...
Last edited by Azrael_031 at Jan 15, 2009,
#13
Quote by Azrael_031
So F=ma
does F=mg(small g,or in other words the constant for gravity,that equals 9.81)


Gravitational acceleration is still acceleration, so yeah.
#14
Dear me, you are confusing the second law of kinematics with formulae of kinematic energy and stationary energy.

Here:

This is the main principle:
E(p) - E(k) = 0
[E(p) is stationary energy and E(k) is kinematic]

In basic physics, there can exist only two kinds of energy in an object. If an object is moving on a table, it has only kinematic energy (if examined from the point of the table). But if you lift the object one meter up in the air, it has the stationary (also known as potential) energy of E(p) = m * g * h, where h = 1.0 m, m is the mass of the object and g is, of course, the famous constant (9.807 ... m / s^2).

So, if an object is dropped from the height of 8 meters, it has only stationary energy in the beginning, but while it drops, it gains kinematic energy until it hits the ground transforming the energy in different forms to, for example, shaping the object.

E(k) = 1 / 2 * m * v^2, thus

E(p) - E(k) = 0
m * g * h - 1 / 2 * m * v^2 = 0

and with basic calculus you should be able to solve your problem.

Sorry my scientific English.


EDIT: I just had to check my grammar. It appears that there is no such thing as "kinematic energy". Replace it with "kinetic energy". And "stationary energy" with "potential energy". They aren't synonyms. My apologies.
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Last edited by nyrhinen at Jan 15, 2009,
#16
Quote by nyrhinen
Dear me, you are confusing the second law of kinematics with formulae of kinematic energy and stationary energy.

Here:

This is the main principle:
E(p) - E(k) = 0
[E(p) is stationary energy and E(k) is kinematic]

In basic physics, there can exist only two kinds of energy in an object. If an object is moving on a table, it has only kinematic energy (if examined from the point of the table). But if you lift the object one meter up in the air, it has the stationary (also known as potential) energy of E(p) = m * g * h, where h = 1.0 m, m is the mass of the object and g is, of course, the famous constant (9.807 ... m / s^2).

So, if an object is dropped from the height of 8 meters, it has only stationary energy in the beginning, but while it drops, it gains kinematic energy until it hits the ground transforming the energy in different forms to, for example, shaping the object.

E(k) = 1 / 2 * m * v^2, thus

E(p) - E(k) = 0
m * g * h - 1 / 2 * m * v^2 = 0

and with basic calculus you should be able to solve your problem.

Sorry my scientific English.


Dear ME!We never studied that,I'm just pulling physics out of my ass as much as we have studied

But thanks for the help much appreciated
Hodam,a stojim...
#17
Quote by punkforlife93
By the way, I edited my post and gave an example if you need any more help. And yes, F = mv^2 as well.

Wait, wait! How come F = m * v^2? One newton (the unit) can be disassembled to 1 kg * m / s^2. You are saying, according to the units, that one newton equals kg * (m / s)^2 = kg * m^2 / s^2. Incorrect.
The important thing is to never stop questioning. - Albert Einstein
#18
Quote by Azrael_031
So F=ma
does F=mg(small g,or in other words the constant for gravity,that equals 9.18)
and does F=mgv?

Vectors are all the same!


F=ma

If an object is falling down and only gravity is acting upon it, then F=mg. (substitute gravity acceleration for the "acceleration).

If an object is falling down and you want to counteract it so it's in a neutral state, then:

F= m*g - m*(d/dt(velocity))

Where d/dt(velocity) is the derivative of velocity in terms of time.

Is this what you want to know? To find the velocity, you're gonna have to do the integral of it, that's the way I see it. I could be wrong, but it looks right to me.

derivative of velocity is acceleration by the way.
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#19
Quote by nyrhinen
Wait, wait! How come F = m * v^2? One newton (the unit) can be disassembled to 1 kg * m / s^2. You are saying, according to the units, that one newton equals kg * (m / s)^2 = kg * m^2 / s^2. Incorrect.


Oh, I see what I got wrong, actually F = 1/2mv^2. That's a simple mistake, same thing with KE.
#20
Quote by punkforlife93
Oh, I see what I got wrong, actually F = 1/2mv^2. That's a simple mistake, same thing with KE.


No. Just No. Kinetic energy, T = (1/2) mv^2, but F =/= (1/2)mv^2. It just doesn't work. F = ma = mg = E/s but force can never have units of energy.

Quote by Azrael_031
Told you already.

I have one object that is 5kg,affected only by g.
The other one,1,1 kg(let's say 1,0!) that's affected by both g and v

wait could f=mv^2?


What? g is the acceleration cause by gravity. You can't put velocities into newton II. If you need to solve situations with accelerations and initial velocities, without using calculus, you need to use the 3 kinematic equations (we call them Suvat equations over here, because of the 5 variables involved, dunno if you have another name for them?) but without a more detailed description (either diagramatical or mathematical) we can't really help you.