Traction on an upgrade – tractive effort required on a slope 1:S A vehicle of weight W ascends an upgrade of 1 : S on a track with coefficient of rolling resistance μ. What tractive force T must be developed at the wheels to just maintain motion uphill (neglecting air resistance)?

Difficulty: Medium

Correct Answer: T = W * (μ + 1/S)

Explanation:


Introduction / Context:
Vehicle tractive effort on grades must overcome both rolling resistance and the component of weight acting along the slope. Estimating this demand is fundamental for traction calculations, locomotive design, and roadway operational analysis on steep grades.



Given Data / Assumptions:

  • Weight of vehicle = W.
  • Grade = 1 : S (rise 1 in S horizontally).
  • Coefficient of rolling resistance = μ (dimensionless).
  • Air resistance and drivetrain losses neglected.


Concept / Approach:
The tractive effort required is the sum of resistances along the line of motion. On an upgrade, two main components oppose motion: rolling resistance μW and grade resistance W * sinθ. For small road gradients, tanθ ≈ sinθ ≈ 1/S, so grade resistance ≈ W/S. Therefore, the total required tractive effort equals W(μ + 1/S).



Step-by-Step Solution:

Resolve weight along slope → grade resistance R_g ≈ W * (1/S).Compute rolling resistance → R_r = μ * W.Total tractive effort → T = R_g + R_r = W * (μ + 1/S).


Verification / Alternative check:
Dimensional sense check: μ and 1/S are both nondimensional; multiplying by W gives force units. For a level road (S → ∞), expression reduces to T = μW, as expected.



Why Other Options Are Wrong:
W(μ − 1/S) would predict decreasing effort on steeper upgrades, which is non-physical. W(μ + S) is dimensionally inconsistent. W/(μ + 1/S) is incorrect summation and inversion. neglects grade resistance.



Common Pitfalls:
Using tanθ instead of 1/S without considering small-angle equivalence; forgetting to add rolling and grade resistances; sign mistakes when descending versus ascending.



Final Answer:
T = W * (μ + 1/S)

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