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f=ma

seraine
Posts: 734
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1/17/2012 6:51:07 PM
Posted: 4 years ago
Acceleration due to gravity is 9.8 m/s/s, which means that its acceleration is 9.8, correct? And since the equation for force is f=ma, that should mean that the height something falls from doesn't change its force, but I know that's false. What am I missing?
M.Torres
Posts: 3,626
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1/17/2012 6:54:30 PM
Posted: 4 years ago
At 1/17/2012 6:51:07 PM, seraine wrote:
Acceleration due to gravity is 9.8 m/s/s, which means that its acceleration is 9.8, correct? And since the equation for force is f=ma, that should mean that the height something falls from doesn't change its force, but I know that's false. What am I missing?

... the fact that "mass" is a variable. lol
: At 11/28/2011 1:28:24 PM, BlackVoid wrote:
: M. Torres said it, so it must be right.

I'm an Apatheistic Ignostic. ... problem? ;D

I believe in the heart of the cards. .:DDO Duelist:.
RoyLatham
Posts: 4,488
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1/18/2012 7:04:18 PM
Posted: 4 years ago
At 1/17/2012 6:51:07 PM, seraine wrote:
Acceleration due to gravity is 9.8 m/s/s, which means that its acceleration is 9.8, correct? And since the equation for force is f=ma, that should mean that the height something falls from doesn't change its force, but I know that's false. What am I missing?

The 9.8 is at the surface of the earth. The force drops as the square of the distance from the center of the earth. The mass of the object doesn't change; it's weight changes with the force of gravity.

The earth has a radius of about 4000 miles, so the force of gravity changes by only a small amount near the surface. Even 100 miles up, it isn't much less. Satellites stay up because gravity is balanced by centrifugal force. there's plenty of gravity up there.
DetectableNinja
Posts: 6,043
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1/18/2012 7:14:37 PM
Posted: 4 years ago
Lol. 9.8 is for noobs. In the IB world, we use 10.
Think'st thou heaven is such a glorious thing?
I tell thee, 'tis not half so fair as thou
Or any man that breathes on earth.

- Christopher Marlowe, Doctor Faustus
Ren
Posts: 7,102
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1/18/2012 7:19:23 PM
Posted: 4 years ago
At 1/17/2012 6:51:07 PM, seraine wrote:
Acceleration due to gravity is 9.8 m/s/s, which means that its acceleration is 9.8, correct? And since the equation for force is f=ma, that should mean that the height something falls from doesn't change its force, but I know that's false. What am I missing?

No, it's not.

You're misinterpreting force, methinks, perhaps for energy or work.

Force is something that causes a change in velocity, direction, or position. Therefore, the force acting on an object remains a constant despite whatever height from which it falls, because the acceleration of gravity is indeed a constant.

Work, on the other hand, is the product of force over time, or the accumulation of energy that can be applied to something (transferred). So, the amount of work, or transferable energy that an object generates, does increase with the height from which it drops.
seraine
Posts: 734
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1/18/2012 7:46:42 PM
Posted: 4 years ago
At 1/18/2012 7:19:23 PM, Ren wrote:
At 1/17/2012 6:51:07 PM, seraine wrote:
Acceleration due to gravity is 9.8 m/s/s, which means that its acceleration is 9.8, correct? And since the equation for force is f=ma, that should mean that the height something falls from doesn't change its force, but I know that's false. What am I missing?

No, it's not.

You're misinterpreting force, methinks, perhaps for energy or work.

Force is something that causes a change in velocity, direction, or position. Therefore, the force acting on an object remains a constant despite whatever height from which it falls, because the acceleration of gravity is indeed a constant.

Work, on the other hand, is the product of force over time, or the accumulation of energy that can be applied to something (transferred). So, the amount of work, or transferable energy that an object generates, does increase with the height from which it drops.

How would I determine the work of an object?
seraine
Posts: 734
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1/18/2012 7:47:00 PM
Posted: 4 years ago
At 1/17/2012 6:54:30 PM, M.Torres wrote:
At 1/17/2012 6:51:07 PM, seraine wrote:
Acceleration due to gravity is 9.8 m/s/s, which means that its acceleration is 9.8, correct? And since the equation for force is f=ma, that should mean that the height something falls from doesn't change its force, but I know that's false. What am I missing?

... the fact that "mass" is a variable. lol

I'm assuming that mass is a constant.
DetectableNinja
Posts: 6,043
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1/18/2012 8:11:04 PM
Posted: 4 years ago
At 1/18/2012 7:46:42 PM, seraine wrote:
At 1/18/2012 7:19:23 PM, Ren wrote:
At 1/17/2012 6:51:07 PM, seraine wrote:
Acceleration due to gravity is 9.8 m/s/s, which means that its acceleration is 9.8, correct? And since the equation for force is f=ma, that should mean that the height something falls from doesn't change its force, but I know that's false. What am I missing?

No, it's not.

You're misinterpreting force, methinks, perhaps for energy or work.

Force is something that causes a change in velocity, direction, or position. Therefore, the force acting on an object remains a constant despite whatever height from which it falls, because the acceleration of gravity is indeed a constant.

Work, on the other hand, is the product of force over time, or the accumulation of energy that can be applied to something (transferred). So, the amount of work, or transferable energy that an object generates, does increase with the height from which it drops.

How would I determine the work of an object?

That depends on the situation.

The most simple work equation is W=Fd cos angle of displacement.

For work by a spring you have W=-1/2kx^2

Et cetera, et cetera, et cetera.
Think'st thou heaven is such a glorious thing?
I tell thee, 'tis not half so fair as thou
Or any man that breathes on earth.

- Christopher Marlowe, Doctor Faustus
Ren
Posts: 7,102
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1/18/2012 8:54:50 PM
Posted: 4 years ago
At 1/18/2012 7:46:42 PM, seraine wrote:
At 1/18/2012 7:19:23 PM, Ren wrote:
At 1/17/2012 6:51:07 PM, seraine wrote:
Acceleration due to gravity is 9.8 m/s/s, which means that its acceleration is 9.8, correct? And since the equation for force is f=ma, that should mean that the height something falls from doesn't change its force, but I know that's false. What am I missing?

No, it's not.

You're misinterpreting force, methinks, perhaps for energy or work.

Force is something that causes a change in velocity, direction, or position. Therefore, the force acting on an object remains a constant despite whatever height from which it falls, because the acceleration of gravity is indeed a constant.

Work, on the other hand, is the product of force over time, or the accumulation of energy that can be applied to something (transferred). So, the amount of work, or transferable energy that an object generates, does increase with the height from which it drops.

How would I determine the work of an object?

Force times displacement times cosine[theta].

So, let's say that you dropped a 7 kg bowling ball from 8 meters in the air.

F = 68,600N

The angle (cosine[theta]) measurement regards the direction of the force in relation to the direction of the displacement. Those 68.6kN are acting on the bowling ball in the same direction in which the bowling ball is moving, meaning that the angle of displacement is at 0 degrees. Therefore, cosine[theta] would be cos[0], which equal 1.

Therefore,

W = 548,800J

Of course, if we were to increase this distance to 10 meters, then,

W = 686,000J

Which would be that increase you were talking about.

Of course, this all leads to Power, which is the rate at which Work is applied.

To determine time, we would need to determine velocity based on acceleration. We can determine it based on the equation,

v^2 = v0^2 + 2a(x - x0),

where v is velocity, v0 is the starting velocity, a is acceleration, x is the ending position, and x0 is the starting position.

So, the bowling ball wasn't moving and did not have a predetermine trajectory, so both v0 and x0 are zero. This means we're left with:

v^2 = 2(9.8)(8)

v^2 = 78.4

v = 8.85m/s by the time it hit the ground.

In other words, it accelerated to 8.85 m/s, which means that the distance it traveled is

8.85/9.8 = 0.9 seconds.

So, the amount of power generated from dropping a bowling ball from 8 meters in the air is 609.77kJ/s.
RoyLatham
Posts: 4,488
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1/18/2012 10:50:49 PM
Posted: 4 years ago
The acceleration is approximately constant, so the velocity is a*t and the distance is d = (a*t^2)/2. t is time. This ignores air resistance.

Using a = 10 approximately,

So if something is dropped 20 meters, it will take 20 = (10*t^2)/2; t = 2 sec.

At the end of 2 second, it will be traveling 10*2 = 20 m/sec.
RoyLatham
Posts: 4,488
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1/18/2012 11:06:06 PM
Posted: 4 years ago
Oh, I see. the concern is the force of impact, not the force acting on the mass. The force acting on the mass is constant. The force of the mass hitting something depends on how fast it decelerates.

Suppose it hits the dirt and comes to rest in 1 cm. The velocity was 20 m/sec, so

v = 20 = a*t ==> t = 20/a

It takes 1 cm, so 0.01 = (a*t^2)/2

0.02 = 400/a

a = 400/.02 = 20,000 m/sec^2 = 200 g's

One of the tests for the ruggedness of a military computer is to mount the computer to an anvil, then strike it with a 400 lb hammer swung through a six foot arc. The method yields lots of force.
Ren
Posts: 7,102
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1/19/2012 12:46:55 AM
Posted: 4 years ago
At 1/18/2012 8:54:50 PM, Ren wrote:
v^2 = 2(9.8)(8)

v^2 = 78.4

v = 8.85m/s by the time it hit the ground.

In other words, it accelerated to 8.85 m/s, which means that the distance it traveled is

8.85/9.8 = 0.9 seconds.

So, the amount of power generated from dropping a bowling ball from 8 meters in the air is 609.77kJ/s.

Dude, I totally effed this up.

v^2 = 156.8

v = 12.52 m/s by the time the bowling ball hits the ground.

12.52/9.8 = 1.28 seconds to hit the ground.

P = 788.39kJ/s
seraine
Posts: 734
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1/19/2012 4:54:15 PM
Posted: 4 years ago
At 1/18/2012 11:06:06 PM, RoyLatham wrote:
Oh, I see. the concern is the force of impact, not the force acting on the mass. The force acting on the mass is constant. The force of the mass hitting something depends on how fast it decelerates.

Suppose it hits the dirt and comes to rest in 1 cm. The velocity was 20 m/sec, so

v = 20 = a*t ==> t = 20/a

It takes 1 cm, so 0.01 = (a*t^2)/2

0.02 = 400/a

a = 400/.02 = 20,000 m/sec^2 = 200 g's

One of the tests for the ruggedness of a military computer is to mount the computer to an anvil, then strike it with a 400 lb hammer swung through a six foot arc. The method yields lots of force.

For my project, I have to determine the work that a certain object will have before it hits the rubber bands, and then use that to determine how much the rubber bands will stretch. How would I determine it's force right before it begins to stretch the rubber bands?

If I knew the distance and the time which it will take to fall using the equation time= the square root of (2*distance/gravity)
,could I then use that to find the acceleration of a falling object and then the force which it will have?
seraine
Posts: 734
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1/19/2012 5:02:33 PM
Posted: 4 years ago
At 1/18/2012 8:54:50 PM, Ren wrote:
Force times displacement times cosine[theta].

So, let's say that you dropped a 7 kg bowling ball from 8 meters in the air.

F = 68,600N

How did you determine this? Wouldn't it be 7*9.8=68.6?

The angle (cosine[theta]) measurement regards the direction of the force in relation to the direction of the displacement. Those 68.6kN are acting on the bowling ball in the same direction in which the bowling ball is moving, meaning that the angle of displacement is at 0 degrees. Therefore, cosine[theta] would be cos[0], which equal 1.

Therefore,

W = 548,800J

Of course, if we were to increase this distance to 10 meters, then,

W = 686,000J

My project is to determine how far a rubber band will stretch with a certain amount of force/work applied. If I knew that a rubberband stretched about 1/2 cm per Newton, could I somehow convert the joules into newtons to see how far the rubber band will stretch?

Which would be that increase you were talking about.

Of course, this all leads to Power, which is the rate at which Work is applied.

To determine time, we would need to determine velocity based on acceleration. We can determine it based on the equation,

Would the equation
time=the square root of (2*distance)/gravity))
also work?

v^2 = v0^2 + 2a(x - x0),

where v is velocity, v0 is the starting velocity, a is acceleration, x is the ending position, and x0 is the starting position.

So, the bowling ball wasn't moving and did not have a predetermine trajectory, so both v0 and x0 are zero. This means we're left with:

v^2 = 2(9.8)(8)

v^2 = 78.4

v = 8.85m/s by the time it hit the ground.

In other words, it accelerated to 8.85 m/s, which means that the distance it traveled is

8.85/9.8 = 0.9 seconds.

So, the amount of power generated from dropping a bowling ball from 8 meters in the air is 609.77kJ/s.
Ren
Posts: 7,102
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1/19/2012 5:50:45 PM
Posted: 4 years ago
At 1/19/2012 5:02:33 PM, seraine wrote:
At 1/18/2012 8:54:50 PM, Ren wrote:
Force times displacement times cosine[theta].

So, let's say that you dropped a 7 kg bowling ball from 8 meters in the air.

F = 68,600N

How did you determine this? Wouldn't it be 7*9.8=68.6?

Yes, but if you're dealing with Newtons and meters, then you need to convert kg to g.

So, 68.6kN becomes 68,600N.

Which would be that increase you were talking about.

Of course, this all leads to Power, which is the rate at which Work is applied.

To determine time, we would need to determine velocity based on acceleration. We can determine it based on the equation,

Would the equation
time=the square root of (2*distance)/gravity))
also work?

Absolutely.

My project is to determine how far a rubber band will stretch with a certain amount of force/work applied. If I knew that a rubberband stretched about 1/2 cm per Newton, could I somehow convert the joules into newtons to see how far the rubber band will stretch?

Sure.

As it turns out, the guy who posted right before me gave you the answer to that question -- the amount of work to compress a string, or anything else elastic:

(1/2)kx^2

It's a derivative of the basic work equation, where you apply the exertion of the elastic material (k) and compound the distance (x^2) due to elasticity.

Determining k, or the elastic material's constant, has already been done for you -- it's 1/2cm/N, or .005m/N.

If you did not have that, you would determine it based on:

F = -mg; then k = F/x.

Since you have that constant, you can move right on to:

x = -F/k (these are all derivatives of each other, as you can see)

Where x is the displacement (how much it's stretched), F is the force, and k is the elastic material's constant.

Why would you need to convert it into Joules?

Right now, if you have the number for F (Newtons), then all you have to do is plug in:

x = (the inverse of your number) times .005

x is the displacement, or the answer to your question.
Ren
Posts: 7,102
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1/19/2012 6:00:14 PM
Posted: 4 years ago
At 1/19/2012 5:50:45 PM, Ren wrote:
At 1/19/2012 5:02:33 PM, seraine wrote:
At 1/18/2012 8:54:50 PM, Ren wrote:
Force times displacement times cosine[theta].

So, let's say that you dropped a 7 kg bowling ball from 8 meters in the air.

F = 68,600N

How did you determine this? Wouldn't it be 7*9.8=68.6?

Yes, but if you're dealing with Newtons and meters, then you need to convert kg to g.

So, 68.6kN becomes 68,600N.

Which would be that increase you were talking about.

Of course, this all leads to Power, which is the rate at which Work is applied.

To determine time, we would need to determine velocity based on acceleration. We can determine it based on the equation,

Would the equation
time=the square root of (2*distance)/gravity))
also work?

Absolutely.

My project is to determine how far a rubber band will stretch with a certain amount of force/work applied. If I knew that a rubberband stretched about 1/2 cm per Newton, could I somehow convert the joules into newtons to see how far the rubber band will stretch?

Sure.

As it turns out, the guy who posted right before me gave you the answer to that question -- the amount of work to compress a string, or anything else elastic:

(1/2)kx^2

It's a derivative of the basic work equation, where you apply the exertion of the elastic material (k) and compound the distance (x^2) due to elasticity.

Determining k, or the elastic material's constant, has already been done for you -- it's 1/2cm/N, or .005m/N.

If you did not have that, you would determine it based on:

F = -mg; then k = F/x.

Since you have that constant, you can move right on to:

x = -F/k (these are all derivatives of each other, as you can see)

Where x is the displacement (how much it's stretched), F is the force, and k is the elastic material's constant.

Why would you need to convert it into Joules?

Right now, if you have the number for F (Newtons), then all you have to do is plug in:

x = (the inverse of your number) times .005

x is the displacement, or the answer to your question.

I forgot to mention:

If you're dealing with Joules instead of Newtons (giving you a reason for the conversion), then you can apply a derivative of the initial equation instead, to go from:

(1/2)kx^2 = W

To:

x = sqrt[W/(0.0025)]

Where 0.0025 = (1/2)k