Mechanical Comprehension: Will the Winch Lift the Crate?
A maintenance team is rigging a temporary winch. A motor turns at 1,200 rpm and delivers 2.0 N-m of torque. Its shaft carries a 12-tooth gear driving a 36-tooth gear; the 36-tooth gear shares a shaft with a 10-tooth gear, which drives a 40-tooth gear. The final gear carries a cable drum of radius 0.10 m. The crew wants to hoist a crate weighing 250 N. A tag on the motor states that the gear train loses about 15 percent of the transmitted torque to friction. A junior technician argues the winch will lift the crate easily because 'the gears multiply force by twelve'.
1. Compute the drum's rotational speed and the ideal force available at the cable, ignoring friction.
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Speed through the train: rpm_out = 1200 x (12/36) x (10/40) = 1200 x (1/3) x (1/4) = 100 rpm.
Torque multiplication: each stage multiplies torque by its tooth ratio: stage 1: 36/12 = 3; stage 2: 40/10 = 4; total = 12. Ideal torque at the final gear = 2.0 x 12 = 24 N-m.
Force at the cable: F = torque / drum radius = 24 / 0.10 = 240 N.
Note that the 240 N is already less than the 250 N crate, even before friction. The junior technician's factor of twelve is correct as far as it goes, but it says nothing about the drum radius, which is what converts torque into cable tension.
2. Including the stated 15 percent loss, what is the maximum load the winch can lift, and can it lift the crate?
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Apply the loss to the transmitted torque: T_available = 24 x 0.85 = 20.4 N-m.
Convert to cable tension: F_max = 20.4 / 0.10 = 204 N.
Verdict: 204 N < 250 N, so the winch cannot lift the crate. The shortfall is about 46 N.
Cross-check from the load side: lifting 250 N on a 0.10 m drum requires 250 x 0.10 = 25 N-m at the drum. Supplying that through a 15 percent loss requires 25 / 0.85 = 29.4 N-m at the final gear, but the train delivers only 24 N-m - about a 23 percent torque deficit. Both routes give the same answer, which is the check that the loss was applied to the right quantity.
3. Recommend a change that makes the lift work with minimal new hardware, and state the trade-off it introduces.
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Option A - shrink the drum. Required radius r = T_available / load = 20.4 / 250 = 0.0816 m, so a drum of about 0.08 m radius works (20.4 / 0.08 = 255 N, a small margin).
Trade-off: line speed drops. Angular speed is 100 rpm = 10.47 rad/s, so v = 10.47 x 0.10 = 1.05 m/s now versus 10.47 x 0.08 = 0.84 m/s - the crate rises about 20 percent slower. For a maintenance hoist that is normally acceptable, and it costs a drum swap rather than new gearing.
Option B - change a gear ratio, e.g. replace the 10-tooth gear with an 8-tooth, giving 40/8 = 5 and a total multiplier of 15: torque becomes 2.0 x 15 x 0.85 = 25.5 N-m, or 255 N at a 0.10 m drum. Same 255 N result, but it also slows the drum to 1200 x (1/3) x (1/5) = 80 rpm.
Both options trade speed for force - the gear train cannot create power, only exchange one for the other - so the crew must confirm the slower lift still fits the job.