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If there are 2 gears meshed together and they are of different sizes, then rotating the smaller one will make the larger one spin with a smaller angular velocity but with more torque. And the opposite happens when you spin the larger one. Using a lower gear ratio in a bicycle for example, makes it easier to go uphill. How does the increased torque from the lower gear ratio help in this? Like how does the higher torque equate to a greater force to move the bike forward?

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Changing to an easier gear reduces the ratio of torque required at the crankset over torque at the wheel. Given constant torque at the wheel (riding up a hill of a given grade at a given speed) this means the torque, and pedal forces, required at the crankset are lower. The rider thus feels it is easier to turn the pedals.

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I do not want to put too much focus on the gear systems.

In general, $P_{in} = P_{out}$ (assuming no power loss). Using, $P = F v$, one gets $F_{in} v_{in} = F_{out} v_{out}$. That is, one can "scale up" the output force by moving through a greater distance per unit time (i.e. since $F_{out} = F_{in} (v_{in}/v_{out})$, increase $(v_{in}/v_{out})$)

What with gear systems? Replace forces with torques, velocities with angular velocities etc.

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  • $\begingroup$ So for a gear system, how does the increased torque make the bicycle easier to move uphill? $\endgroup$ – BlurryPic Nov 18 '15 at 7:01

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