A locomotive has four pairs of driving wheels (four axles), each axle carrying 24 × 10^4 N. If the coefficient of friction is 1/6, what is the maximum tractive load the locomotive can pull on level track?

Introduction / Context:
Locomotive tractive effort on level track is limited by adhesion (friction) between driving wheels and rails. The maximum horizontal pull without slipping equals the product of the normal load on the driving axles and the coefficient of friction.

Given Data / Assumptions:

• Number of driving axles = 4 (four pairs of driving wheels).
• Axle load = 24 × 10^4 N per axle.
• Coefficient of friction μ = 1/6; level track (no grade resistance considered).

Concept / Approach:
Total normal load on driving wheels: N_total = axle_load * number_of_axles. Maximum tractive force without slip: T_max = μ * N_total.

Step-by-Step Solution:
Compute N_total = 4 * (24 × 10^4) = 96 × 10^4 N.Compute T_max = (1/6) * 96 × 10^4 = 16 × 10^4 N.Therefore, the maximum load that can be pulled (neglecting rolling resistance) is 16 × 10^4 N.

Verification / Alternative check:
Check units and proportionality: doubling μ would double T_max; doubling axle load doubles T_max, consistent with adhesion theory.

Why Other Options Are Wrong:

• 32 × 10^4 N overestimates the available adhesion with μ = 1/6.
• 8 × 10^4 N and 4 × 10^4 N underestimate; they would imply fewer axles or lower μ.
• 24 × 10^4 N mixes single-axle load with total tractive capacity.

Common Pitfalls:
Confusing wheel pairs with axles; forgetting to multiply by the number of axles carrying the load.

Final Answer:
16 × 10^4 N