View Single Post
Old 12-03-2009, 02:18 AM
JazzRoc JazzRoc is offline
Join Date: Nov 2009
Location: El Medano, Tenerife, Canary Isles
Posts: 104
Default Re: Are the Laws of Physics Wrong?

Originally Posted by galexander View Post
You're just trying to cloud the issue now.
Also you have just contradicted what you previously claimed. Earlier you said it was W.D. = Force x Mass x Distance and now its W.D. = Distance x Mass x Acceleration.
Make your mind up JazzRoc. Anyway your last equation still doesn't satisfactorily explain my example. The mass is fixed as is the force pushing the weights. And I don't dispute that the acceleration is different for each weight. Looks like I've got you on the run JazzRoc.
"W.D. = Force x Mass x Distance and W.D. = Distance x Mass x Acceleration"
Well. The latter is correct.

I had to dredge back some fifty years. You should try that sometime.

When a force acts on a body and causes it to move against a resistance the force is said to do work.
If the force does not vary in magnitude and is in the direction of motion, the amount of work done is equal to the product of the force and the distance moved, or -
Work = force x distance
If a force of one pound acts through a distance of one foot, the amount of work done is -
1 foot x 1 lb = 1 foot-pound
and one foot-pound is called the unit of work. If a weight of 1 lb is lifted through a vertical distance of 1 foot, 1 foot-pound of work is done. If a weight of, say, 7 lbs is lifted through a vertical distance of 9 feet, the work spent in lifting is -
9 x 7 = 63 foot-pounds

Now the "pound" is the force a pound mass exerts in the earth's gravitational field.
The "pound" force is the MASS of that pound times the ACCELERATION due to gravity.
hence W.D. = Distance x Mass x Acceleration

My stickiness with this subject is to do with Metric and Imperial units. Imperial (American) I find much easier to deal with: foot-pounds instead of G x newton-metres.

But the important point is that weight = mass x G, which is where confusion may spring.

Also work calculations done outside a gravity field have to take into account the acceleration involved from one speed to another to arrive at resolution: they no longer involve a return to a rest state.

Good luck.
Reply With Quote