physics question

[TwoJ's]

^^ Exactly. Everyone else is just wrong.

FretBurnr: You were the closest with the momentum. But what we are actually looking for is the change in momentum from the instant of impact until it actually reaches rest. This is a highly complex situation, and is worlds ahead of newton's second law (Net force = m*a). That only works in classical physics, which can only deal with relatively large objects. Newton's second is no good when we look at the molecular level. You will need quantum mechanics to completely understand that.

In this situation, whatever surface you drop it on will bend upon impact, and it will take a very small amount of time to make the object come to rest. You need a function for the force with respect to time. Then and only then can you solve this problem. And if you were to have this function, you can find that P(final)-P(initial) = The integral of F(t)dt from t=0 (initial impact) to t final (object at rest). This is called impulse, and it's what you're looking for.

Bingo. Like I said on the other side, you are asking a problem that you can't get an answer to with the data provided.

[LivingLegend06]

^^ Exactly. Everyone else is just wrong.

FretBurnr: You were the closest with the momentum. But what we are actually looking for is the change in momentum from the instant of impact until it actually reaches rest. This is a highly complex situation, and is worlds ahead of newton's second law (Net force = m*a). That only works in classical physics, which can only deal with relatively large objects. Newton's second is no good when we look at the molecular level. You will need quantum mechanics to completely understand that.

In this situation, whatever surface you drop it on will bend upon impact, and it will take a very small amount of time to make the object come to rest. You need a function for the force with respect to time. Then and only then can you solve this problem. And if you were to have this function, you can find that P(final)-P(initial) = The integral of F(t)dt from t=0 (initial impact) to t final (object at rest). This is called impulse, and it's what you're looking for.

Bingo. Like I said on the other side, you are asking a problem that you can't get an answer to with the data provided.

Because they give the "force of impact" id like to assume the force while it is coming to rest is a constant which make the problem easy enough for a monkey to solve using work energy.

not using gravitational potential energy you can say that the obeject ends up with the same ammount of energy it started with. then its simple force X distance = Force X distance

the first force is from gravity and is going to equal the mass of the object(slugs) X 32.2 = (pounds), the first distance is the distance of the fall 2ft
the second force is the given "force of impact" 500 and the second distance is the distance over which it come to a stop.

simple enough.

[akorcovelos]

lol guys, you all suck at physics.

The object does not simply create a force on impact, it needs to come to rest over some distance over which it will have an acceleration.

If this is some kind of homework problem its a trick question and just say that the object has a mass of (500/32.2) slugs.

not homework, for my job. I need to know how much weight I need to drop to generate the required 500 lb impact force.

[TwoJ's]

So your telling me that I need to drop a 500 lb weight from 2' to generate 500 lbs of impact on an object? Gravity wouldn't increase the force?

And this is what is wrong with the standard measurement system. Pounds are not a measure of mass. For instance, when I step on the scale, it says 220 lbs. My mass is not 220 lbs. My weight is. Weight DOES NOT EQUAL mass. In our measurement system, mass is measured in slugs. So you must multiply slugs times some force (gravity in this case) to get pounds.

[TwoJ's]

Because they give the "force of impact" id like to assume the force while it is coming to rest is a constant which make the problem easy enough for a monkey to solve using work energy.

not using gravitational potential energy you can say that the obeject ends up with the same ammount of energy it started with. then its simple force X distance = Force X distance

the first force is from gravity and is going to equal the mass of the object(slugs) X 32.2 = (pounds), the first distance is the distance of the fall 2ft
the second force is the given "force of impact" 500 and the second distance is the distance over which it come to a stop.

simple enough.

But that's a completely theoretical (and false) circumstance. In a real, inelastic collision, both objects will deform and the relationship between force and time is not linear at all.

Not quite simple enough. If you want to solve it that way, then you will definitely get an answer, but it simply won't mean anything.

The work-energy theorem does not apply to inelastic collisions. Work-energy will give you the velocity when it hits the ground, but that's not what we're looking for.

[fretburnr]

^^ Exactly. Everyone else is just wrong.

FretBurnr: You were the closest with the momentum. But what we are actually looking for is the change in momentum from the instant of impact until it actually reaches rest. This is a highly complex situation, and is worlds ahead of newton's second law (Net force = m*a). That only works in classical physics, which can only deal with relatively large objects. Newton's second is no good when we look at the molecular level. You will need quantum mechanics to completely understand that.

In this situation, whatever surface you drop it on will bend upon impact, and it will take a very small amount of time to make the object come to rest. You need a function for the force with respect to time. Then and only then can you solve this problem. And if you were to have this function, you can find that P(final)-P(initial) = The integral of F(t)dt from t=0 (initial impact) to t final (object at rest). This is called impulse, and it's what you're looking for.

Bingo. Like I said on the other side, you are asking a problem that you can't get an answer to with the data provided.

You make an excellent point, as no collision is purely elastic. In addition, there's no way to account for material flex and energy lost to sound, heat, etc. As it sits, the question is either poorly worded (pounds isn't mass) or is a trick question. If anything beyond that was intended, there's not nearly enough info to model the entire system. I can't continue with a momentum calculation without further knowledge.

[BigFurryWhale]

And this is what is wrong with the standard measurement system. Pounds are not a measure of mass. For instance, when I step on the scale, it says 220 lbs. My mass is not 220 lbs. My weight is. Weight DOES NOT EQUAL mass. In our measurement system, mass is measured in slugs. So you must multiply slugs times some force (gravity in this case) to get pounds.

Man, I'm glad I live in Canada and don't have to deal with the imperial system, metric is so much easier for things like physics.

[LivingLegend06]

not homework, for my job. I need to know how much weight I need to drop to generate the required 500 lb impact force.

what kind of material are you dropping and what kind of material are you landing on? why do you need 500lbs of force?

obviously the most accurate way to do this is to just gently set 500lbs on where ever you need that much force.

[fretburnr]

Man, I'm glad I live in Canada and don't have to deal with the imperial system, metric is so much easier for things like physics.

metric is so much easier for everything, except for talking with an older generation of americans.

[LivingLegend06]

But that's a completely theoretical (and false) circumstance. In a real, inelastic collision, both objects will deform and the relationship between force and time is not linear at all.

Not quite simple enough. If you want to solve it that way, then you will definitely get an answer, but it simply won't mean anything.

The work-energy theorem does not apply to inelastic collisions. Work-energy will give you the velocity when it hits the ground, but that's not what we're looking for.

my answer doesn't have to make sense because the question doesn't make sense! [/thread]

[TwoJ's]

Ha. I guess this is true.

[CorvallisBMW]

metric is so much easier for everything, except for talking with an older generation of americans.

Indeed. We rarely use English measurements for anything in class because it's just so fucking complicated. All physics/chemistry classes are done in metric, as are most of engineering classes.

[LivingLegend06]

Indeed. We rarely use English measurements for anything in class because it's just so fucking complicated. All physics/chemistry classes are done in metric, as are most of engineering classes.

physics is all metric, but not my engineering classes though. Statics, dynamics and strength of materials all use mixed units, as well as a bunch of the general and introductory engineering classes I've had. hopefully next year in pro school it will be all metric. (yes, I go to the same school as you)

[jflip2002]

6 replies and no treadmill?

I caught wind of this late. I was going to insist it really does take off. Its because you see, the wheels aren't pr..... nevermind.

I took chem, not physics.

[akorcovelos]

ok, I need to design a test that will drop a weight, likely steel, 2' onto a steel cone partially embedded into a concrete wall. Maybe this will help.