Question Bank • QB_FP_ADV_074 NAT • 2 Marks

Section 3: Farm Power

Tractors and Power Tillers

Subtopic: Power transmission systems – gear trains, differential, final drives and power take-off

Question Statement:

A constant-mesh transmission gearbox of a tractor connects an input shaft running at $N_1 = 1800\text{ rpm}$ to a countershaft through a gear pair with $Z_1 = 18$ and $Z_2 = 36$ teeth. The second reduction gear pair from the countershaft to the output shaft has $Z_3 = 15$ and $Z_4 = 45$ teeth. The rotational speed of the transmission output shaft in $\text{rpm}$ is ________ (answer in integer).

Answer Key:

300

Step-by-Step Solution & Explanation

``` Compound Gear Train: Input Shaft (N_1 = 1800 rpm) ---> Z_1 = 18 teeth Countershaft: Z_2 = 36 teeth, Z_3 = 15 teeth Output Shaft: Z_4 = 45 teeth ``` **Step-by-step Calculation:** 1. First reduction gear ratio ($i_1$): $i_1 = \frac{Z_2}{Z_1} = \frac{36}{18} = 2.0$ $N_{\text{counter}} = \frac{N_1}{i_1} = \frac{1800}{2} = 900\text{ rpm}$ 2. Second reduction gear ratio ($i_2$): $i_2 = \frac{Z_4}{Z_3} = \frac{45}{15} = 3.0$ $N_{\text{output}} = \frac{N_{\text{counter}}}{i_2} = \frac{900}{3} = 300\text{ rpm}$ **Overall Speed Ratio Shortcut:** $i_{\text{total}} = i_1 \times i_2 = 2.0 \times 3.0 = 6.0$ $N_{\text{output}} = \frac{N_1}{i_{\text{total}}} = \frac{1800}{6.0} = 300\text{ rpm}$
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