Question 6 of 6: Gradeability

Question 6: Gradeability (32 marks total, 6.1–6.4)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

6.1 — Free body diagram at the point of wheel spin

Take $x$ along the road surface (positive up-slope) and $y$ perpendicular to the road surface (positive away from the surface). Three forces act on the wheel at the point where it is just about to spin: the weight $W$ acting vertically downward through the wheel centre, resolved into a down-slope component $W\sin\theta$ (along $-x$) and an into-surface component $W\cos\theta$ (along $-y$); the normal reaction $N$ from the road surface (along $+y$); and the tractive (friction) force $F_t$ the road surface exerts on the tire (along $+x$, up-slope), which reaches its maximum available value $f N$ exactly at the point of incipient spin.

Sum of forces, $y$ (perpendicular to the surface): $$\sum F_y = 0: \quad N - W\cos\theta = 0 \quad\Rightarrow\quad N = W\cos\theta$$ Sum of forces, $x$ (along the surface, at the point of incipient spin): $$\sum F_x = 0: \quad F_t - W\sin\theta = 0 \quad\Rightarrow\quad F_t = W\sin\theta$$ with $F_t$ at its maximum value $fN$ exactly at this threshold.

6.2 — Deriving $\tan\theta = f$

At the instant the wheel is on the verge of spinning, the tractive force the tire can generate is at its maximum, $F_t = fN$, and the two equilibrium equations from 6.1 hold simultaneously: $N = W\cos\theta$ and $F_t = W\sin\theta$. Substituting the first into $F_t = fN$ gives $F_t = fW\cos\theta$, and setting this equal to the $x$-equation's $F_t = W\sin\theta$: $$fW\cos\theta = W\sin\theta \quad\Rightarrow\quad f = \frac{\sin\theta}{\cos\theta} = \tan\theta$$ The weight $W$ cancels, so the threshold grade at which a wheel begins to spin is independent of vehicle weight and depends only on the available tire-road friction coefficient $f$: any gradient with $\tan\theta$ still below $f$ can be climbed (traction exceeds the required down-slope force), while $\tan\theta > f$ means the required tractive force exceeds what friction can supply and the wheel spins.

6.3 — The friction coefficient $f$, and a reasonable value for a forest road

$f$ is the coefficient of traction (friction) mobilised between the tire and the road surface -- the same physical quantity as an ordinary friction coefficient, governing how much tangential (driving) force a given normal load can transmit before the tire loses grip and spins. It depends on the road surface material and condition (compacted gravel vs. loose or wet gravel, native soil, ice), tire tread and inflation, and vehicle speed. Unlike a paved highway (dry asphalt typically $f\approx0.7$–$0.9$), a forest road running on compacted gravel or native subgrade offers meaningfully lower and more variable traction, and drops further when wet or loosely surfaced. A reasonable design value for a well-built, dry-to-moist gravel forest road is $f\approx0.5$ (the low end of typical values, $f\approx0.3$, better reflects a wet or poorly maintained surface, and would be the conservative choice for a road that must remain climbable in all conditions).

6.4 — Maximum allowable road gradient

Given. $\tan\theta \le f$, assumed $f = 0.5$ (Q6.3, dry-to-moist compacted gravel forest road).

Find. The maximum gradient, expressed as a percent grade, that a wheel with this traction coefficient can climb before spinning.

1. Convert the threshold condition to a percent grade. Road grade as a percent is $100\tan\theta$, so the traction-limited threshold $\tan\theta = f$ becomes: $$\text{Grade}_{max} = 100f = 100\times0.5 = \boxed{50\%}$$