Stuntman Physics Question

[Hugh R]

In the motion picture and television industry we used to let stuntmen do a stunt of attaching a line to one stuntman the line would go over a pulley attached to the underside of an aerial boom lift and then the other end of the line would go to a larger piece of rope with two stuntmen about the same weight would be on a ten foot A-frame ladder. The two stuntmen would grab the rope and jump off the ladder, and the other guys flies up in the air. In the Industry this is called a "Hong Kong Pull".

We no longer all the use of aerial boom lifts for two reasons; 1) OSHA doesn't like them because they aren't rated for lifting people and 2) we think they're underrated for this purpose. A stuntman wanted to know if I had any "math" to prove their underrated, let alone illegal. So here is what I came up with from a few websites that seem to be in agreement with each other, but intuitively I think differently, and at the end I'd like the PPOT Brain Trust to weigh in.

Stunt man 2 (mass 2) is 175 lbs and is the one going up. The two stuntmen on the ladder each weigh 175 lbs. [350 lbs for mass 1]

Force=mass x acceleration, lets assume a frictionless pulley, rope and air drag.

So, (M1 +M2)G = M1*G -M2*G

Rearranging: M1*G-M2*G/M1+M2

350*g-175*G/350+175

11,200-5,200/525=11.42 which is actually acceleration.

mass of M1 and M2 is 525 and multiplying that by actual acceleration of 11.42 we get 525 x 11.42= 6,000 lbs. The boom lift is rated at a distributed load of 500 lbs. In this scenario the load that the Pulley, and hence the boom lift would see is 6,000 pounds; 12 times the rated load. This is why we require the use of a rated crane.

This whole thing can therefore be summarized by the following:

Force = M2*G-M1*G; or 350*32 ft/sec/sec-175 * 32 ft/sec/sec or

11,200-5,200= 6,000 lbs.

Here is what I don't get, you subtract the mass that is rising from the mass that is descending. By my logic, they are both loads that the pulley "sees". One is rising against gravity and one is being pulled down by gravity. Wouldn't you want to add the two loads times G; rather than subtract one from the other?

Hope this makes sense.

[RPKESQ]

Vector analysis.

[widebody911]

Monkey roll.

[rick-l]

mass in this is a stone

edit or maybe a slug

[rick-l]

what I mean is the force is 175 and 350 pounds force. that 525 pulls the pulley shaft down and the pulley shaft exerts 525 pounds on its mount since the position of the pulley does not move.

Now the force on the rope is 175 in one direction and 350 (for a sum of 175) in the other so it moves in the direction of the heavy weight.

[masraum]

Originally Posted by rick-l

mass in this is a stone

edit or maybe a slug

slugs are the American unit that would go with 32fps^2

Hugh, you're over working this, but you've hit on the answer. Once the guys are all in the air, assuming that the pulley isn't moving up or down, the crane is supporting the weight of all three guys as gravity pulls down on them. So the downward force on the spot where the pulley is attached is the gravitational force of all three guys. Whether the guys are moving up or down, gravity is always pulling on all three, so the force acting on the top of the pulley would be the same as

[jyl]

two masses hang from a connecting rope over a pulley
rope has no stretch, pulley has no friction, no air resistance
the larger mass is M, the smaller mass is m, M > m
when the masses are released, they both accelerate
mass M has acceleration - a and mass m has acceleration a
the difference in the signs is because mass M is accelerating down and mass m is accelerating up
but the absolute value of the accelerations is the same, a, because they are connected by the rope

the forces acting on mass M to produce acceleration - a are gravity (down) and the force being transmitted by the rope (up)
gravity force is M * - g the negative sign is because gravity is pointing down
rope force call it T for tension and it is positive because the rope tension is pointing up
their sum produces the net force on mass M which is M * - a
so M * - a = M * - g + T
so M * a = M * g - T
so T = M * g - M * a

the forces acting on mass m to produce acceleration a are also gravity and the force being transmitted by the rope
gravity force is m * - g the negative sign is because gravity is pointing down
rope force call it T for tension and it is positive because the rope tension is pointing up
their sum produces the net force on mass m which is m * a
so m * a = m * - g + T
so m * a = T - m * g
so T = m * a + m * g

Since T = T
M * g - M * a = m * a + m * g
( M + m ) * a = ( M - m ) * g
a = [ (M - m) / ( M + m ) ] * g

M = 350 lb, m = 175 lb, g = 32 ft / sec^2
so a = 10.6667 ft / sec^2

Now solve for T
T = m * a + m * g
T = 175 lb * 10.6667 ft / sec^2 + 175 lb * 32 ft / sec^2
= 7466.6667 ft * lb / sec^2

The force on the pulley, which we'll call P for pull, is 2 * - T
P = 2 * - T
= -14,933.3333 ft * lb / sec^2

The pulley is attached to a platform, which must support "X pounds". That means X pounds in Earth's gravity g, so

-14,933.333 ft * lb / sec^2 = X lb * 32 ft / sec^2
X = [ -14,933.333 ft * lb / sec^2 ] / [32 ft / sec^2 ]
= -466.6667 lb

So if the platform must support 467 lb, and for man-lifting stuff you want a 3X safety factor (I'm just guessing), the platform needs to be rated for
467 lb * 3 = 1,400 lb.

That ignores shock loading, like if the stuntmen allow any slack in the rope, and elasticity of the rope. Guys who use elastic ropes to yank stuck vehicles from mud know that these are a huge deal.

Sorry for all the tedious units, that's how I was always taught to do physics. Write out the units explicitly and carry them through the whole calculation, if you end up with a mass in lb / sec you know you screwed up.

By the way, you see from the above that the platform is not necessarily supporting the combined mass of M and m. If M = m, it is the case, X = M + m and neither mass is moving, a = 0. But if M > m, the X < M + m, and the platform supports less weight as M / m gets larger. When m = 0, a = g and T = 0, P = 0, X = 0.

Imagine an extreme case, and you can see it. Suppose M = 350 lb and m = 0.0625 lb, the two stuntmen are lifting just 1 ounce. The stuntmen free-fall as if they were not holding a rope ( a = g appx ) , the rope flies through the pulley with no resisting weight on the other side, and the pulley is supporting almost no weight (X = 0 appx )

So, if your stunt involves two 175 lb stuntmen pulling a midget, this might all be okay.

[slodave]

I think my brain just froze. Can some reset it for me?

[Pazuzu]

Originally Posted by Hugh R

Stunt man 2 (mass 2) is 175 lbs and is the one going up. The two stuntmen on the ladder each weigh 175 lbs. [350 lbs for mass 1]

NO.

175 pounds and 350 pounds is the FORCE already, not the mass. It has taken the "g" value into account.
So, right off the bat your numbers are off by 32. I didn't delve into any deeper than that.

EDIT: gyl got it wrong as well. Sorry!

[Pazuzu]

I was wrong! I'm drinking beer, so sue me...

The factor of "g" does need to be removed, but it's also removed in the carry capacity of the crane, so it washes out.

The crane must be able to hold the weight of the three men to be safe, which it does (well, not quite, it's 25 pounds overweight).

However! There is another factor, and that is the "jerk". The change in acceleration.

When the men jump off the ladder, the jerk felt by the crane increases substantially (a factor of three, if they all weigh about the same). The crane might not be rated for that jerk, or dynamic load.

[GWN7]

Another consideration should be the rope. I'm sure you have a rope wrangler who keeps track if the rope used is dynamic or static and how many times it has been used.

[mattdavis11]

Take out an extra insurance policy and roll film.

[jyl]

Now, how would you calculate the max transient load on the platform?

[Mrmerlin]

you could verify this with load cube, it would attach between the pulley and the arm
( they have a similar thing in the crane industry where the cube is placed , so to verify that the load isnt still connected to the support structure or the load isnt more than the lifting device is able to carry)

then have the jumpers and the jumpee do their thing, the load cube will provide the force reading it takes.

[Zeke]

Like the kid who said let the air out of the tires of the trailer stuck under the bridge, get a bigger crane.

[VINMAN]

Is that a laden or unladen stuntman?.... About three fitty.....

[Hugh R]

Wow, thanks guys, esp. JYL and Mike Bradshaw. I new in my gut you had to include the weight on both sides of the pulley since one is pulling against gravity and one is being pulled down by gravity. I guess what I don't understand is if one side of the pulley has 175 lbs, and the other side has 350 lbs, for a total of 525 lbs, how can the pulley only see 467 pounds? I'll pencil out your math and see if I can better figure it out.

I think next time we do this, I'm going to get the weights of all stuntmen involved and hook and accelerometer to the top of the pulley and see what we get.

[Pazuzu]

Originally Posted by Hugh R

We no longer all the use of aerial boom lifts for two reasons; 1) OSHA doesn't like them because they aren't rated for lifting people and 2) we think they're underrated for this purpose. A stuntman wanted to know if I had any "math" to prove their underrated, let alone illegal. So here is what I came up with from a few websites that seem to be in agreement with each other, but intuitively I think differently, and at the end I'd like the PPOT Brain Trust to weigh in.

As for the 467 pounds, I think it must be a derivation error coming from the fact that jyl forgot that the measurement "pounds" is a force, not a mass, so he has an extra factor of "g" floating around.

As for the original question, it's true that OSHA does not like people under booms and cranes (in fact, it's against the rules to be under a load, no matter how light the load and how big the crane). Also, the crane rated at 500 pounds static lift IS undersized for 525 pounds of humans in a dynamic lift situation. That doesn't even take into account the lateral forces on the crane (since the people are not all standing directly under the boom, there's some angle on the rope, and that creates lateral "swinging" forces on the boom which is a BIG no-no).

Now, if you had a big static platform that was engineered, you'd be fine for doing this. Problem is, big static engineered platforms are not particularly mobile, so all of your stunts would have the exact same background...

[jyl]

Originally Posted by Pazuzu

As for the 467 pounds, I think it must be a derivation error coming from the fact that jyl forgot that the measurement "pounds" is a force, not a mass, so he has an extra factor of "g" floating around.

"lb" is not a unit of force and I'm not calculating that the force on the pulley is "467 pounds" or that the pulley "sees" a force of 467 lb. "lb" is a unit of mass. Force is mass x acceleration ( F = m *a ) and the unit of force is "lb * ft / sec^2". When we use "pounds" as a shorthand expression for force, we are implicitly leaving out the relevant acceleration.

I calculated the force vector on the pulley is -14,933 ft lb / sec^2. But no manlift is load rated for a force expressed in ft lb / sec^2. The manlift is rated for a given mass expressed in lb and that rating applies if stationary in Earth gravity.

So I'm saying the manlift (that the pulley is mounted to) has to be rated for "467 lb" of mass, to withstand the force of -14,933 ft lb / sec^2 on the pulley. (And then I'm assuming you'd want a big safety factor which I guessed at 3X, because you're lifting humans and because we're ignoring shock/dynamic loading.)

If I were simply off by a factor of "g", then my scalar 467 would be 467 * 32 or 467 / 32, both of which would be obviously wrong, plus the units would be obviously wrong.

From above post

The force on the pulley, which we'll call P for pull, is 2 * - T
P = 2 * - T
= -14,933.3333 ft * lb / sec^2

The pulley is attached to a platform, which must support "X pounds". That means X pounds in Earth's gravity g, so

-14,933.333 ft * lb / sec^2 = X lb * 32 ft / sec^2
X = [ -14,933.333 ft * lb / sec^2 ] / [32 ft / sec^2 ]
= -466.6667 lb

Hugh, if your intuition is having a hard time w/ the idea that the force on the pulley depends on the ratio of the weights (M / m ), not merely on the total weight ( M + m ), maybe do an experiment. With a nylon string (rope), a pencil (pulley), and some fruit in plastic grocery bags (stuntmen). Hang 5 oranges on one side and 4 on the other, have your helper let go of the bags, you feel the force on the pencil that you're holding as the heavier side descends slowly. Now hang 8 oranges on one side and 1 on the other, and feel the difference in the force as the heavier side drops rapidly. In the second case, the force should feel smaller. I haven't done this (we don't eat enough fruit) but that's what the math says. You can even make the ratio more and more extreme (get a knife and make it 8.75 oranges vs 0.25 oranges). If you can discern a difference at an extreme ratio (must be non-zero and non-infinity), then you know there is a difference at a less extreme ratio, because the system is linear (the formulas have no polynomials).

Now, it's been >30 years since the last physics class, so I could certainly have gotten this wrong.

If so, it'd be helpful for me to point it out specifically, since I'll have to be helping my daughter through high school physics pretty soon so I actually need to review/relearn all this stuff over the next year - basic mechanics, electromag, thermodynamics, etc. My copy of Halliday & Resnick is falling apart so badly that I need to buy a new one.

[jyl]

When you next do the stunt, with a scale hooked between pulley and platform, it would be interesting to see the difference between max force when there is no slack in the rope as the stuntmen step off the ladder, and when the stuntmen leave, say, 1 foot of slack.