Question

A 2WD tractor with a total weight of 20 kN has a static weight distribution of 25% and 75% at the front and rear axles, respectively. When the tractor is operated on a level ground, the maximum tractive force developed is 10 kN. If external weight of 5 kN is added to the rear axle, neglecting weight transfer, the change in maximum tractive force in kN is

• A. 3.33
• B. 5.33
• C. 7.33
• D. 10.33

Hint:

Since tractive force varies linearly with axle weight:
\[ \Delta H = \mu \times \Delta W_r \] Using \(\mu = 2/3\) and \(\Delta W_r = 5\text{ kN}\):
\[ \Delta H = \frac{2}{3} \times 5 = 3.33\text{ kN} \] This avoids calculating intermediate totals.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a two-wheel drive (2WD) tractor, tractive force is developed entirely by the driving (rear) wheels, proportional to the normal load acting on that axle and the coefficient of traction.

Step 2: Key Formula or Approach:
The maximum tractive force \(H\) is:
\[ H = \mu \times W_r \] where \(\mu\) is the coefficient of traction and \(W_r\) is the vertical weight acting on the drive (rear) axle.

Step 3: Detailed Explanation:
1. Find the initial static weight on the rear axle (\(W_{r1}\)):
\[ W_{r1} = 75\% \text{ of } 20\text{ kN} = 0.75 \times 20 = 15\text{ kN} \] 2. Use the initial maximum tractive force (\(H_1 = 10\text{ kN}\)) to determine \(\mu\):
\[ \mu = \frac{H_1}{W_{r1}} = \frac{10}{15} = \frac{2}{3} \] 3. Find the new rear axle weight (\(W_{r2}\)) after adding the 5 kN ballast:
\[ W_{r2} = 15\text{ kN} + 5\text{ kN} = 20\text{ kN} \] 4. Calculate the new maximum tractive force (\(H_2\)):
\[ H_2 = \mu \times W_{r2} = \frac{2}{3} \times 20 = 13.33\text{ kN} \] 5. Calculate the change in maximum tractive force (\(\Delta H\)):
\[ \Delta H = H_2 - H_1 = 13.33 - 10 = 3.33\text{ kN} \]

Step 4: Final Answer:
The correct option is 1, which corresponds to 3.33.