Here are 3 interlocking gears. The largest gear has 14 teeth, the middle gear has 12 teeth, and the smallest gear has 6 teeth. The smallest gear can ONLY move in a counterclockwise direction, while the middle gear can move counter clockwise.
Q: How many turns will each gear have to make in order to turn all gears in their original starting direction??
2007-11-15 03:56:53 · 4 answers · asked by ** 3 in Science & Mathematics ➔ Engineering
6x14, 7x12, 14x6
The clockwise/counter clockwise thing is a red herring. The smallest gear would have to be in the middle because the outer pair would have to turn clockwise, but it doesn't affect the result.
6x14 is 84 teeth 84/6=14 84/12=7
It doesn't make any difference how they are connected.
How did I do that? The 6 and 12 are 2:1 so they aren't the problem. The 12 and 14 are two teeth off, so the 14 is two teeth further behind after each revolution of the 12, so after 7 revolutions it's a full turn behind.
2007-11-15 04:03:32 · answer #1 · answered by Chris H 6 · 1⤊ 0⤋
1st, you may have miscopied the question. I do not think the smallest gear and the middle gear can both turn counter clockwise. If the middle gear is in the middle (duh) the other two gears will turn in the opposite direction from the middle gear.
2nd, the 14 tooth gear will turn 6 turns, the 12 tooth gear will turn 7 turns, and the 6 tooth gear will turn 14 times.
3rd try using the formula: {(teeth per gear) X Revolutions} / (teeth in the other gear) = (other gear's Revolutions)
Fun problem. Thank you
2007-11-15 12:13:42 · answer #2 · answered by lil'oleJewler 2 · 0⤊ 0⤋
Large - 6 turns
Middle - 7 turns
Small - 14 turns
2007-11-15 12:38:22 · answer #3 · answered by Anonymous · 0⤊ 0⤋
tricky problem. research with search engines like google. just that may help!
2014-11-05 23:08:53 · answer #4 · answered by Anonymous · 0⤊ 0⤋