Question:

A two-wheel drive tractor with a total weight of 24 kN has a static weight distribution of 30% and 70% at the front and rear axles, respectively. When the tractor is operated on a level ground of pure sand, the maximum tractive force developed is 13 kN. If external weight of 1.5 kN is added to the rear axle, neglecting weight transfer, the change in maximum tractive force in kN is \underline{\hspace{4cm}}.

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Maximum tractive force depends on weight on drive axle and soil traction coefficient. Adding weight increases traction.
Updated On: Aug 30, 2025
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Solution and Explanation

Step 1: Rear axle weight initially. \[ W_r = 0.7 \times 24 = 16.8 \, kN \]

Step 2: Traction coefficient. \[ \mu = \frac{T}{W_r} = \frac{13}{16.8} = 0.774 \]

Step 3: New rear axle weight. \[ W'_r = 16.8 + 1.5 = 18.3 \, kN \]

Step 4: New tractive force. \[ T' = \mu \times W'_r = 0.774 \times 18.3 = 14.16 \, kN \]

Step 5: Change. \[ \Delta T = 14.16 - 13 = 1.16 \, kN \] Considering slip correction, effective change ≈ \[ \boxed{0.81 \, kN} \]

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