The 33 1/3 second mystery #26

[kazoo]

The world famous Yomama diamond is displayed in a room measuring 15m x 15m x 15m. The room is constructed of concrete blocks with a floor covered by carpeting. The diamond is displayed on a pedestal 1m high covered by a glass shroud. They are the only objects in the room, and are located centrally. Pressure sensitive wires run under the carpeting in parallel fashion on 8cm centers, and will trip if 4kg pressure is applied to an individual wire.

Burglar Pat McGroin, who weighs 74kg was able to steal the diamond using a 5m 2x4, 4 leather straps (2m x 5cm ea.), a hammer, and a saw.

How did Pat pull off the heist?

[SaltPork]

He beat all the security guards to death with the 2x4 and stole the frickin' diamond.

[nash]

Originally posted by wrkime
He beat all the security guards to death with the 2x4 and stole the frickin' diamond.

We have a winner! Good job wrkime!!

[kazoo]

Sorry, but correct answers don't involve pounding people with pieces of wood, much as I'd love to engage in that practice with some folk I know.

[yotta]

The answer has to do with distributing weight, or causing preassure not to be applied to the wires.

If he cut the 2x4 in half, he'd have 2 250 cm halves, each of which would span 31 of the wires. Strapping these to his feet, the weight would be distributed to about 2.4kg per wire, assuming he's careful, and the 2x4s do not bend at all.

Probably wouldn't quite work in the real world, though, as the wood would bend, and not distribute weight evenly....

[kazoo]

You have the concept for solving the problem, yotta, but your math is flawed.

Following your suggestion of cutting the 2 x 4 in half, there are two 250 cm pieces, theoretically spanning 31 wires when approaching them from the perpendicular. However, Pat has two legs, and so long as he slowly shuffles his feet, his 74 kg mass is quasi-evenly distributed between both slats, thereby reducing the point load over span to 1.18 kg/wire.

Before anybody says it, yes, if Pat were to approach in fashion parallel to the wires, he'd trip the alarm, but Pat had checked that out.

yotta wins a fuzzy bunny holding a broken calculator.

[yotta]

Yes, he would have to have them perpendicular to the wires.

My math was not wrong, but I did not fully explain myself. I was assuming he'd have the 2x4s parallel to one another, long sides next to one another. Pat can't put the boards end to end (as per your calculation), because then his feet would need to be 250 centimetes apart (his feet need to be in the center or each 2x4 for weight to be distributed evenly, especialy since he's on carpet), which would hurt, unless he's very tall.

[kazoo]

True enough. I was envisioning more of a cross-country skiing mode for Pat, yet the more I ponder this, the static loading to a given wire would constantly be increasing or decreasing, so long at Pat is in motion. While standing still with ankles side by side, your calculation is valid. Once Pat's feet move apart, the pressure would drop as a direct function of Pat's inseam measurement but wouldn't reach 1.18 unless Pat was an NBA wannabee.

Keep the fuzzy bunny, and give back the broken calculator.