Calling All Mechanical Engineers - HELP

[B-assman]

I need to figure out how to calculate the force in lbs exerted by dropping a 500lb weight approx 4ft. For simplicity sake - figure you're at sea level and don't worry about centrifical force exerted by the spinning earth.

FYI - Im replacing a part of a drop hammer and I need to know how much force is being exerted by the drop.

Thanks
Dave

[Slipknot]

force equals mass times acceleration

[B-assman]

Yes - but exactly how do you apply that.

If M = 500lbs
A = 9.8 Meters per Sec ^2
does that mean that the force is 4900lbs??

Doesn't the height of the drop need to figure in there somewhere?

[spence]

Are the parts sized by actual force?

-spence

[B-assman]

The part has a breaking strength of about 28,000 lbs. The worst case scenario is if the weight drops and it needs to be stopped just before it bottoms out - the part needs to be able to stop the weight. So if It's dropping about 4 ft and the part will break at 28000lbs of force - will it be able to stop the weight??

[wheresmy50]

It's been a while since physics 100, but I'll take a stab at this.

Your units are screwed up. You can't use 500 pounds (US) and 9.80m/s^2 (metric) for gravity. Well, you can I guess but it doesn't make sense.

You need to first calculate the speed at impact.

Velocity = the square root of 2*9.8m/s^2 * height (I had to look that one up)

So, velocity = 4.84 m/s

Momentum = m* v = 1292 kg m/s

Calculating the impact force is more difficult since you need to know what happens when the weight stops. F=ma is true, but the "a" isn't acceleration due to gravity, it's deceleration at impact. In other words, you have to know how long (in time) the impact lasts.

If we assume (and this is one hell of an assumption) the impact is one millisecond, which might be true if nothing is deformed in the impact.

Then the force of impact is 1292280 Newtons, which is 290,000 pounds of force.

The rub here is that it's highly dependent on the time of impact. Double our time to 2 miliseconds, and you cut the force in half.

To be honest, I really didn't think it would be this complicated when I started. Hope I didn't screw it up.

Interesingly enough though, this is why recoil pads on guns and shock absorbers work - the force is directly related to the amount of time over which that force is applied. If you can double the time, you cut the force in half.

[Swimmer]

Wow! Thats neat, really.

[scoobe]

Flash backs to highschool!

[B-assman]

Thanks for the replies everyone - in the end practice won out over theory. Nothing beats a real world test. The part withstood the worst case force with no obvious signs of stress - so I guess in theory the time of "impact" must be more at least 10 miliseconds

Thanks
Dave

[striprman]

What is the average force F with which a drop hammer of weight W that falls through a distance h strikes the work, if it compresses it by amount d.

F= Wh/d : where h and d are the units of measurement

W=500 lbs
h= 4 feet=48 inches
d= compression of struck material (lets assume .25 inches)

F=500*48/.25=96000 pounds

[thortum]

Wow I love the people on this site.