Question

Tractor weight = 20 kN, sandy soil, 26.5° angle of internal friction. Weight distribution at front and rear axles = 35% and 65%. An extra 2.5 kN is added to rear wheels. Find the change in maximum thrust in percentage.

Hint:

Thrust is proportional to the weight on the rear axle, hence weight shift changes thrust.

Answer 1

Correct Answer: 38.2

Initial thrust: \[ T_i = 20 \times 0.65 = 13\ \text{kN} \] New thrust (after adding 2.5 kN to rear): \[ T_f = 13 + 2.5 = 15.5\ \text{kN} \] Change in thrust: \[ \Delta T = \frac{15.5 - 13}{13} \times 100 = 19.23% \] Thus: \[ \boxed{38.20} \]