Do we have any physics gurus in here?

[Ruffdaddy]

The answer to your problem lies almost solely on the geometry and equipment details. The moment applied to the equipment changes with geometry,as does the persons trajectory. This is a much more complex problem than the simple physics 1 chapter 3 answers here.

This involves understanding the suspension affects From the loading as well.

Simple mechanics tells you an instantaneous force of 2,000 lbf was applied to the person if it was purely suspension flex from the loading. Common sense tells you laying on the bucket is dumb as shit.

[jayjohnson600]

I've been having to assume that the 210 lb object is getting motivated with a pure horizontal instantaneous force, and it greatly simplifies the math. But, this is not a reasonable assumption because the 210 lb object was said to be laying on the platform/bucket. If that's the case, then the platform/bucket would be nearly horizontal and a snap-back event would cause the platform/bucket to have basically only a verticle vector force to impart on the 210 lb object. That would completely refute the assumption of motivation by a purely horizontal instantaneous force.

Good exercise, as I haven't thought about physics with ANY kind of depth for quite a long time.

[BERNIE MOSFET]

I'm not a physics guru and my answer may in fact be wrong, but I'll give it a shot. I read your description as trying to lift an object 15 feet into the air, so I'm disregarding the 9 foot height you mentioned. I can come back to it later if in fact you meant 6 feet.

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.

Edit: I just realized a lot of folks probably don't remember or know what joules means. Joules is basically a unit of force over a given amount of time. The last part is the key - over a certain period of time. This could be instantaneous, several seconds or whatever. Most people tend to understand power, such as in electricity we use watts. 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.

To be pedantic: joules is a unit of energy, or work.

[Ruffdaddy]

If you have the equipment model, we can probably come up with geometric limits base on the bucket linkage. There is a very solid geometric range it can even operate in.

[BERNIE MOSFET]

If you have the equipment model, we can probably come up with geometric limits base on the bucket linkage. There is a very solid geometric range it can even operate in.

That's what i was thinking.

I think there are a limited number of moment arms that can be derived given the limited range of motion of the equipment. The equipment could probably be reduced to a static free body diagram, and then dynamic loads applied once weights and cg are determined. Then, force magnitude and direction could be derived.

[Ruffdaddy]

That's what i was thinking.

I think there are a limited number of moment arms that can be derived given the limited range of motion of the equipment. The equipment could probably be reduced to a static free body diagram, and then dynamic loads applied once weights and cg are determined. Then, force magnitude and direction could be derived.

Absolutely, then just go with a worst case analysis to see if it's even possible.

Was the machine in motion?

[Mysticcobrakilla]

Buncha dumb motherfuckers in here!







That is a horrible representation of a CAT 365c

[71chevellejohn]

Sounds like somebody was operating without all of their PPE. Once he was over 6ft, he should have been fully harnessed with fall protection gear.

[turboford]

All this proves it gravity is still a mother fucker.