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 clockwise 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 06:56:57 · 4 answers · asked by ** 3 in Science & Mathematics ➔ Engineering
Please tell me how you got your answer
2007-11-15 07:12:50 · update #1
Like I said last time:-
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, so they are synchronized again. 7x12 = 84 and the others divide in to 84 with no remainder so we're done!
2007-11-15 08:27:23 · answer #1 · answered by Chris H 6 · 0⤊ 0⤋
Look for the lowest common denominator -
If the 6 tooth gear makes one turn, the 12 goes 1/2 & the 14 tooth only goes 3/7ths of a turn.
So the common denominator is 14*6 = 84.
6 tooth gear 14 turns
12 tooth gear = 7 turns
14 tooth gear = 6 turns
And all are back to their starting position
2007-11-15 07:28:23 · answer #2 · answered by Doug B 3 · 0⤊ 0⤋
The largest gear will have to make 14 turns, middle gear 12 turns and the smallest gear will have to make 6 turns. No logic reason , just a simple guess.
2007-11-15 07:04:50 · answer #3 · answered by a new world 2 · 0⤊ 1⤋
6 tooth-14
12 tooth-7
14tooth-1
2007-11-15 07:08:23 · answer #4 · answered by Radar 3 · 0⤊ 1⤋