# Need some help, force to move an object

1. Dec 13, 2005

### 9000_red

this is my first post here, so hopefully this is the correct forum. any help is appreciated...

i'm a business major, and i'm working on a project that involves saving a particular company money. i work for coca-cola, and i figured that i could use the fact that people i work with have trouble moving full pallets of coca-cola, due to the weight of the pallet. the pepsi vendor where i work receives pallets that are stacked 3 layers lower than the ones we receive, and moves them with much less effort. i figured i could come up with a formula for how much more work would get done, by shaving a few lbs off the pallets we pull, making the work easier.

i figured that there are 12 cans in a single 12pk, 16 12pks in a layer and 13 layers in a pallet. the weight of the cokes themselves minus the pallet and the cardboard containers is roughly 1872 lbs. shave off 3 layers of that and save 432lbs. that saves alot of effort and people will throw more product if it takes less effort, right?

THE QUESTION :yuck:

what are some good equations or formulas to justify this. i was thinking how much work would it take a 180 lbs male (me) to move this 1872 lbs pallet roughly 50 yards?

2. Dec 13, 2005

### 9000_red

any comments at all would be appreciated.

3. Dec 13, 2005

### 9000_red

anyone? anyone?

4. Dec 13, 2005

### 9000_red

just pointing me in a general direction would be great...

5. Dec 13, 2005

### kp

geesh...I don't know about that one. metabolic work and mechanical work can not be measured the same way.

search "Power Factoring" by Pete Sisco, he attempts to incorporate mechanical work with metabolic work when lifting weights, and has some pretty interesting formulas,
maybe some from here, much smarter than me, could derive something from his methods.

6. Dec 14, 2005

### -Christastic-

like what kp said, mechanical and metabolic work is completely different...technically speaking if you're talking about putting a stack of pallets on a cart and moving the cart some distance x, the total mechanical work done by you to move the cart 50 ft, 50 m, or 50 km is the same; 0 (considering moving across a level surface). The only mechanical work performed in this system would be actually lifting the pallets up onto the cart. So as far as that goes I'm clueless on how to help you.

The only thing I can think of off the top of my head is to comparethe force it takes to start the pallets moving. All you need to know for this is the coeffecient of friction for your cart or whatnot and the mass of your pallets which you've already got.

Hope this helps

7. Dec 15, 2005

### mybsaccownt

"what are some good equations or formulas to justify this. i was thinking how much work would it take a 180 lbs male (me) to move this 1872 lbs pallet roughly 50 yards?"

*DISCLAIMER*

maybe I am simplifying this more than I should be and I could be very wrong, but...this makes sense to me...

upon reviewing my post i realized that i am almost completely wrong (all these finals are killing my brain)...but i'll leave it up in case it helps you get started

1 Calorie = 4186 J

the force you need to do to move a pallet of weight w, a distance d would be

rolling resistance * weight of object = Force

there is no way for me to know the rolling resistance of the pallet lifters you are using so i'll assume it would be close to that of a train wheel on a steel track (since pallet lifters have hard wheels that resist deformation, don't they? and warehouse floors are pretty hard too)

so we'll say it is 0.001

(http://72.14.203.104/search?q=cache...htm+rolling+resistance*weight+of+object&hl=en)

so, if you want to move 1872lbs (roughly 8327.1 N) , 50 yards (45.72 m)

so, .001*8327.1= 8.3271 N

force *distance = 8.3271*45.72 = 381 J

and that's awfully low, maybe I made a mistake

ok, it can still be salvaged, i'll assume you're pushing these pallets instead of using a pallet lifter, (this can be useful because you want to see how much more work can be done in smaller chunks as opposed to bigger ones, so the ratio of work done on heavy vs lighter will be the same regardless of the actual force that is being applied)

ok, so the work needed to move a weight w, a distance d

the force will be counteracted by the friction force...ff = N*muK = 8327.1*(i can't find a coefficient for friction of plastic on cement or wood on cement...)

W = w*d

W = 8327.1*45.72

it will require 380715 J or 91 Calories

how much power you use will depend on the time it takes you to do this task

W/t = Power

i'm going to stop here, because i'm either horribly off track or this will give you a good starting point to compare how much effort is required for any case

ok, i went back over all of this and i have to say...this is iffy at best, there's a lot of things i didn't take into account, and i probably did a lot of this wrong...but

Last edited by a moderator: Apr 21, 2017
8. Dec 15, 2005

### FredGarvin

I would recommend, as a first pass analysis, to look at the time required to move one of the pallets with and without the added weight. The notion of work in this case may be a bit deceiving (unruly might be a better word). However, if you can base your argument on the facts that the heavier stack has to be moved at a slower speed and requires a longer time for the person doing the move to set up, you could have a savings based simply on the time required to do the move of a single pallet.

If you ever had a chance to talk to one of your factory's manufacturing engineers, they would be a good source for how to look at this because it is exactly what they deal with...a process.

I would do some trials measuring the time it takes to do the following:
1. Time to hook up the pallet prime mover (can you possibly use a smaller one with less weight?)

2. Time to maneuver the pallet from point A to point B.

3. Time required to unhook the prime mover.