Do we have any physics gurus in here?

[CWO]

#treadmill

[ELVIS]

dude....we can find an old stolen bbq pit no problem. i bet you are pushing it for physics help/

god bless.

[jluv]

Physics? I got nothing. How about a poem?

There once was a Texas Rat Snake.
Who had a tree house he needed to make.
He hired a Pit Bull with cropped ears.
And they had a few beers.
And that turned out to be a mistake.

See, the Pit Bull was quite a good actor.
And when the Rat Snake was driving the tractor,
The Pit Bull said "Fuck it“,
Jumped out of the bucket,
And he claimed that the beer was a factor.

He sued for eleventy million dollars.
Because bitches love dogs with gold collars.
It turns out he was fine,
So he settled for about nine.
Now he's one of the dfwmustangs' ballers.

[soap]

[sc281]

Physics? I got nothing. How about a poem?

There once was a Texas Rat Snake.
Who had a tree house he needed to make.
He hired a Pit Bull with cropped ears.
And they had a few beers.
And that turned out to be a mistake.

See, the Pit Bull was quite a good actor.
And when the Rat Snake was driving the tractor,
The Pit Bull said "Fuck it“,
Jumped out of the bucket,
And he claimed that the beer was a factor.

He sued for eleventy million dollars.
Because bitches love dogs with gold collars.
It turns out he was fine,
So he settled for about nine.
Now he's one of the dfwmustangs' ballers.

Nice.

[mstng86]

I like turtles

[Baron Von Crowder]

Physics? I got nothing. How about a poem?

There once was a Texas Rat Snake.
Who had a tree house he needed to make.
He hired a Pit Bull with cropped ears.
And they had a few beers.
And that turned out to be a mistake.

See, the Pit Bull was quite a good actor.
And when the Rat Snake was driving the tractor,
The Pit Bull said "Fuck it“,
Jumped out of the bucket,
And he claimed that the beer was a factor.

He sued for eleventy million dollars.
Because bitches love dogs with gold collars.
It turns out he was fine,
So he settled for about nine.
Now he's one of the dfwmustangs' ballers.

[racrguy]

Physics? I got nothing. How about a poem?

There once was a Texas Rat Snake.
Who had a tree house he needed to make.
He hired a Pit Bull with cropped ears.
And they had a few beers.
And that turned out to be a mistake.

See, the Pit Bull was quite a good actor.
And when the Rat Snake was driving the tractor,
The Pit Bull said "Fuck it“,
Jumped out of the bucket,
And he claimed that the beer was a factor.

He sued for eleventy million dollars.
Because bitches love dogs with gold collars.
It turns out he was fine,
So he settled for about nine.
Now he's one of the dfwmustangs' ballers.

Well done, as usual!

[Nash B.]

I like to work in metric units for stuff like this, so 210lb. = 95.2 kg, which is mass not weight. But taking in gravity at 9.8 m/s^2 times 95.2 kg = 932.96 kg m/s^2 or simply 932.96 Newtons. So now we know we need to lift an object 933 Newtons a distance of 15 feet, or 4.57 meters. So 933 newtons times 4.57 meters = 4.264 kilojoules. If 6 feet (1.82m), then 1.698 kilojoules.

No, the 210 pounds (95 kg) is his weight, not his mass. You should be dividing by g instead of multiplying to get his mass.

So let's assume we want to lift this weight almost instantaneously, I'll use 0.1 seconds. So for the 15 feet figure, 4264 joules is a little over 42 kilowatts. The 6 is about 17 kilowatts.

You should use simple calculus to determine his initial velocity if you know the maximum height. If he's launched upward at 150 feet per second, gravity alone won't stop him at 15'.

I just reread my post and it appears that in an effort to not disclose too much, my description was probably too vague. Some of the facts leading to the alleged accident differ from those below, but this is the best comparative fact pattern of what allegedly happened:

A 30K lb. piece of heavy construction equipment was being used to lift and hold a 2K lb. object 1 foot off the ground. The object was being held by the bucket of the piece of equipment. A 210 lb. person was 9 feet off the ground laying on top of the bucket. The 2K lb. object fell off of the bucket and released all of its weight instantaneously. As a result, the bucket suddenly moved up/rared back/moved in some way and caused the person to fall off. Based on all of that, how much force would it take to throw the 210 pound person off of the bucket and cause him to land 15 feet away from his original location on the bucket? I assume the weight of the bucket and the angle the boom/arm was at when the weight was released are relevant factors in the calculations, but unfortunately I don't have that info.

Does that fact pattern make my question clearer?

It would've taken him 3/4 of a second to fall straight down 9'. To cover 15' horizontally in that span, he'd need to be moving at 20 feet per second. I don't remember physics, though.

[jayjohnson600]

210 object at 9ft high ending up at the ground level and 15ft away.

Need to know the angle of the motivating force providing the horizontal component force.

If the horizontal force was exactly horizontal(no vertical component vector), and the 210 lb object was at the edge of the 9ft high platform(no loss of force to start the free fall), then that will simplify the math some. Also, assume that the platform doesn't move(no whipping action or lever action to change the component forces etc).

Calculate the airtime
Time to freefall from a stop to the ground(9ft distance) in earth's gravity: d=1/2 (g) t*t
9ft = 1/2 (32ft/s*s) t*t
18ft = (32ft/s*s) t*t
18ft/32ft/s*s = t*t
.5625 s*s = t*t
.75s = t
The object fell from a stop at 9ft high, to the ground in .75 seconds

The 210 lb object also moved a distance of 15ft horizontally in .75 seconds

If the 210 lb oject moved 15ft horizontally in the air(negligible friction and drag, plus no roll) then the horizontal velocity(assume instantaneous and no deformation) imparted is:
v=d/t
v=15ft / .75 seconds
v= 20ft / second

Calculate the motivating force
What force is required to move 210 lbs 15ft(horizontal and let gravity do its thing)in .75 seconds ?

F=m*V/(t*t)
F=210 lbs * (20ft/sec) / (.75sec*sec)
F=210 lbs * (20ft) / (.75sec)
F=4200 lbs * ft / (.75sec)
F=5600 lb ft /sec

So, a 5600 lb ft /second force was imparted on a 210 lb object located 9ft in the air, to motivate it to land 15ft away on the ground.

My best guess, but please confirm the math for yourself.

[CWO]

[GhostTX]

210 object at 9ft high ending up at the ground level and 15ft away.

Need to know the angle of the motivating force providing the horizontal component force.

If the horizontal force was exactly horizontal(no vertical component vector), and the 210 lb object was at the edge of the 9ft high platform(no loss of force to start the free fall), then that will simplify the math some. Also, assume that the platform doesn't move(no whipping action or lever action to change the component forces etc).

...

Here's what I did.

For a given height, we can find the velocity at impact using mgh = 1/2mv^2
The masses cancel and you get v^2=2gh, which I calculate to 24 ft/s.

Finding the velocity, we find the fall time to be 0.375s (d=vt, 9'= 24ft/s * t)

Circling back to the "instantaneous" velocity once the force was imparted, using d=vt, 15'=v*0.375 gets 40ft/s.

Back to the KE=1/2mv^2 -> KE=1/2 * 210lbs * (40ft/s)^2 -> 168000 lb-ft^2/s^2

My $0.02 (and check my math)

[Ruffdaddy]

Can you draw a diagram of the initial state of the platform, and the final state.

Good work jay!

[jayjohnson600]

And looking at your problem in a different way...

Calculate with energy equations, potential and kinetic energy.

Say the load was static, calculate the potential energy of the 2000 lbs and the 210 lb object 1 ft up(start with load on the ground, then delta as both would be released after the event). This is making some assumptions and removing some neglible values that could be included but are not significant.
E=m g z
2980 j

Then kinetic energy to motivate 210 lb object to a velocity of 20ft/sec:
K=1/2 m v*v
1770 j

It's possible to have occurred in the energy equations with cardinal vectors, with room to spare.
Since the 210 lb object was laying on the object, I would assume that the horizontal force vector was imparted by the snap-back angular rotation/accelleration of the boom arm when the 2000 lb load was released. So, there are more equations that would take up additional amount of the spare energy that is availble from the static potential.

A court case cannot really go into the equation deeper because the angles, weights and heights cannot be shown/accepted as evidence on hearsay.

The doctor's medical opinion on the accident will carry much more weight then attempting to prove points through the unfounded(hearsay) physics of the situation. IMHO.

[Nash B.]

Here's what I did.

For a given height, we can find the velocity at impact using mgh = 1/2mv^2
The masses cancel and you get v^2=2gh, which I calculate to 24 ft/s.

Finding the velocity, we find the fall time to be 0.375s (d=vt, 9'= 24ft/s * t)

Circling back to the "instantaneous" velocity once the force was imparted, using d=vt, 15'=v*0.375 gets 40ft/s.

Back to the KE=1/2mv^2 -> KE=1/2 * 210lbs * (40ft/s)^2 -> 168000 lb-ft^2/s^2

My $0.02 (and check my math)

24 ft/s is his velocity when he hits the ground. It's not a constant velocity: he's accelerating at 32 feet per second squared. You have to use calculus (or remember how acceleration, velocity, and position are related):

Acceleration = -32 feet per second squared
Velocity = a * t = -32 * t feet per second
Position = 1/2a * t^2 + s0= -16 * t^2 + 9 feet from the ground
Find the time when position = 0: 0 = -16 * t^2 + 9
16 * t^2 = 9
t = sqrt(9/16) = 3/4 seconds