Subtopic: Power transmission systems – gear trains, differential, final drives and power take-off
Question Statement:
In a tractor rear axle differential, the bevel pinion has 12 teeth and bevel/crown gear has 42 teeth. The input speed and torque of the bevel pinion are 520 rpm and 1200 N m, respectively. There is a planetary gear between the differential unit and each half axle with 4:1 speed reduction. The left wheel encounters poor traction when the tractor is moving in straight path that causes 15% drop in the left axle torque. If the differential efficiency is 0.98, the right axle torque under locked differential condition in N m will be ______
Official Answer Key:
9320 to 9330
Step-by-Step Solution & Explanation
Bevel gear ratio = 42/12 = 3.5. Crown-wheel (differential case) torque = T_pinion×ratio×η = 1200×3.5×0.98 = 4116 N·m. In an (open) differential this splits equally to the two half-axle inputs: 4116/2 = 2058 N·m each. The planetary final drive (4:1) multiplies each side's torque ×4: normal axle torque = 2058×4 = 8232 N·m per side (total available = 16,464 N·m). Poor traction cuts the left axle torque to 85% of this: T_left = 8232×0.85 = 6997.2 N·m. Under the locked-differential condition the housing forces equal wheel speeds, so the total driving torque (16,464 N·m) is retained and the balance is carried by the right wheel: T_right = 16,464 − 6997.2 = 9466.8 N·m. Discrepancy note: the official accepted range is 9320–9330 N·m, about 1.5% below our value. This gap is consistent with a slightly different point of application for the 0.98 differential efficiency or the 15% traction loss (e.g. applying η only once at the final total, or applying the 15% reduction to the post-planetary torque before redistribution) — both variants shift the result by a similar small margin, but we could not pin down the exact intended sequence from the transcribed question. We show our clearly-stated method and flag the key value.
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