# Mechanical Tipping Ratios

*Most recent answer: 04/20/2011*

Q:

Imagine a MagLIGHT brand flashlight...
http://bit.ly/hcxlyI
...which has its batteries in its long barrel/handle section...
http://bit.ly/fwlunr
...and so is much heavier in its barrel/handle than it is in the wider section where is the light bulb and reflector.
Now imagine that said flashlight, with its batteries in it, is sitting vertically on a table with its wider (and lighter) bulb and reflector section down against/touching the table, likd this:
http://bit.ly/huAIF1
Now imagine pushing horizontally against its topmost end, tipping it past its tipping point until it falls over and is lying horizontally on the table like this:
http://bit.ly/gJ9jN0
Now imagine putting your fingers around the very same tip of the flashlight against which you pushed to tip it over, and then picking it up and tipping it back upright again, so that it's once again vertical on the table like this:
http://bit.ly/huAIF1
MY QUESTION: What is the amount of energy needed to tip it over from vertical to horizontal compared with (versus) the amount of energy needed to pick/tip it back up and make it vertical again?
I'm not looking for a precise number of foot pounds or any other measurement. Rather, I'm looking for a ratio, in effect. I can intuit that it requires more energy to lift the tip of the flashlight from its tipped-over horizontal position up to its vertical position, than it does to tip it over from its vertical position to its tipped-over horizontal position.
But, assuming I'm correct about that, I just want to know approximately how much MORE energy is required to pick it up from horizontal back to vertical, than than is required to tip it over from vertical to horizontal.

- HarpGuy (age I'm no kid)

Vallejo, California

- HarpGuy (age I'm no kid)

Vallejo, California

A:

I can't give precise ratios because I don't know precisely how the weight is distributed in the flashlight. So I'll work out a simpler problem which should get all the key ideas across.

Say you have a uniformly dense rectangular block, with height r times the width. The tipping point occurs when the center of mas is straight up from the contact point. Draw the picture and you'll see that if you call the gravitational potential energy (mgh) of the flat block 1 unit, the tipping point has energy sqrt(1+r

The ratio of the maximum force needed is given by the ratio of the torques needed to counteract the gravitational torque around the pivot point. That's even simpler, since the gravitational torque just depends on how far the center of mass is displaced from the vertical line through the pivot. That ratio is just r.

Mike W.

Say you have a uniformly dense rectangular block, with height r times the width. The tipping point occurs when the center of mas is straight up from the contact point. Draw the picture and you'll see that if you call the gravitational potential energy (mgh) of the flat block 1 unit, the tipping point has energy sqrt(1+r

^{2}) and the upright energy is r. So the ratio of the work done to tip the block from the flat to the upright position to the reverse process is (sqrt(1+r^{2})-1)/(sqrt(1+r^{2})-r). If, for example, r=5, that comes out around 40.The ratio of the maximum force needed is given by the ratio of the torques needed to counteract the gravitational torque around the pivot point. That's even simpler, since the gravitational torque just depends on how far the center of mass is displaced from the vertical line through the pivot. That ratio is just r.

Mike W.

*(published on 04/20/2011)*