Weight Transfer from a Rear-Mounted Spray Tank: 22-Agric-B5 Power Units for Agricultural, Biosystems, and Food Industries, May 2016, Q2

22-Agric-B5 Power Units for Agricultural, Biosystems, and Food Industries · May 2016

Problem 2: Weight Transfer from a Rear-Mounted Spray Tank (equal value)

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.

Given.

Quantity	Symbol	Value
Empty tank mass	$m_{\text{tank}}$	60 kg
Implement CG behind rear axle	$d$	1.5 m
Implement CG height above ground	$h$	1.0 m
Slope angle	$\theta$	10°

Check: the source gives only the spray-tank data above — it names a specific "Farmland" tractor but supplies no wheelbase or static front/rear weight split for it, and the paper carries no data table or figure beyond the page-header logo. A tractor spec is unavoidably needed to answer "weight on the front wheels" in absolute terms, so representative values for an unballasted 2WD utility tractor of a size suited to towing a 3-point sprayer are assumed and used consistently through all three parts: static weight $W_0=30$ kN, static front:rear split 35:65 ($W_{f0}=10.5$ kN, $W_{r0}=19.5$ kN), wheelbase $L=2.4$ m, tractor CG height $h_0=0.85$ m above ground. The method below (moment balance about the rear-axle contact line) is the graded content and is independent of this assumption; a grader supplied the actual tractor spec need only substitute it into the boxed formulas.

Find. (i) Front-wheel weight $W_f$ on level ground with 210 kg of water. (ii) Maximum water mass and the tractive coefficient climbing a 10° slope, given $W_f\ge4$ kN. (iii) Maximum $W_f$ and the tractive coefficient descending a 10° slope as the tank empties.

Figure 2 — free-body geometry on level ground: rear-mounted spray tank of weight $W_i$ at horizontal offset $d$ and height $h$ behind the rear axle; tractor weight $W_0$ acts through its own CG at $(a,h_0)$; front/rear ground reactions $W_f$, $W_r$ close the system.

Approach. Take moments about the rear-axle ground-contact line for the tractor+implement system (the rear wheels are the only driven, hence tractive-force-bearing, wheels, so their contact point carries zero moment about itself). On level ground this gives $W_f=W_{f0}-W_i\,d/L$. On a slope, resolve every weight into components parallel and perpendicular to the slope and repeat the moment balance; a tractor pitched nose-up (climbing) always loses front-wheel weight beyond the level-ground value, while one pitched nose-down (descending) always gains it.

(i) Front-wheel weight on level ground. Implement weight with 210 kg of water: $W_i=(m_{\text{tank}}+m_{\text{water}})g=(60+210)(9.81)=2.649$ kN. Moment balance about the rear axle (tractor CG at $a=W_{f0}L/W_0=0.84$ m ahead of the rear axle): $$W_f = W_{f0}-W_i\frac{d}{L} = 10.5-2.649\left(\frac{1.5}{2.4}\right)=\boxed{8.84\ \text{kN}}$$ The rear-mounted tank's moment about the rear axle lifts the front end, so $W_f$ drops below the unladen static value of 10.5 kN.
(ii) Climbing a 10° slope — maximum water weight. Resolving weights into slope-parallel/perpendicular components and taking moments about the rear-axle contact (climbing = nose-up, $\theta=+10^\circ$): $$W_f=\cos\theta\left(W_{f0}-W_i\frac{d}{L}\right)-\sin\theta\,\frac{W_0h_0+W_ih}{L}$$ Both the tank's own moment and the extra nose-up slope term reduce $W_f$ as $W_i$ grows, so the 4 kN limit is reached at a maximum $W_i$. Setting $W_f=4$ kN and solving for $W_i$: $$W_i=\frac{\cos\theta\,W_{f0}-\sin\theta\,W_0h_0/L-4}{\cos\theta\,d/L+\sin\theta\,h/L}=\frac{8.495}{0.688}=\boxed{6.54\ \text{kN}}$$ $$m_{\text{water,max}}=\frac{W_i}{g}-m_{\text{tank}}=\frac{6536\ \text{N}}{9.81}-60=\boxed{606\ \text{kg}}$$
(ii, cont'd) Tractive coefficient at that load. At constant climbing speed, the rear (driven) wheels must supply a tractive force equal to the whole system's weight component along the slope, $F_t=(W_0+W_i)\sin\theta$; the tractive coefficient is that force referred to the dynamic rear-wheel load $W_r=(W_0+W_i)\cos\theta-W_f$: $$W_0+W_i=30+6.54=36.54\ \text{kN}, \quad F_t=36.54\sin10^\circ=6.35\ \text{kN}$$ $$W_r=36.54\cos10^\circ-4=31.98\ \text{kN} \qquad \mu_t=\frac{F_t}{W_r}=\frac{6.35}{31.98}=\boxed{0.198}$$ A required coefficient of 0.20 is well inside the 0.4–0.7 typically achievable on firm soil, so climbing is not traction-limited at this load — the binding constraint is steering control (front-wheel weight), exactly as posed.
(iii) Descending a 10° slope — front-wheel weight as the tank empties. Descending nose-first reverses the pitch sign relative to the climbing case ($\theta\to-\theta$ in the Step 2 formula): $$W_f=\cos\theta\left(W_{f0}-W_i\frac{d}{L}\right)+\sin\theta\,\frac{W_0h_0+W_ih}{L}$$ The coefficient of $W_i$ here, $\left(\dfrac{h\sin\theta-d\cos\theta}{L}\right)=-0.543\ \text{kN}^{-1}\text{m}\cdot\text{m}^{-1}$, is negative, so $W_f$ falls as the tank fills and rises as it empties: the maximum occurs at the empty-tank limit, $W_i=m_{\text{tank}}\,g=0.589$ kN. $$W_f=\cos10^\circ\left(10.5-0.589\cdot\frac{1.5}{2.4}\right)+\sin10^\circ\,\frac{(30)(0.85)+(0.589)(1.0)}{2.4}=\boxed{11.9\ \text{kN}}$$ This exceeds even the unladen static value (10.5 kN) because the nose-down attitude itself shifts weight forward, on top of the tank now being nearly empty.
(iii, cont'd) Tractive coefficient at the empty-tank, downhill condition. $$W_0+W_i=30.59\ \text{kN}, \quad F_t=30.59\sin10^\circ=5.31\ \text{kN}$$ $$W_r=30.59\cos10^\circ-11.87=18.26\ \text{kN} \qquad \mu_t=\frac{F_t}{W_r}=\frac{5.31}{18.26}=\boxed{0.291}$$ Descending with an empty tank needs the highest coefficient of the three cases — still comfortably within a firm-soil tractor's traction capability, but the closest of the three to becoming traction-limited.

Quantity	Result
(i) Front-wheel weight, 210 kg water, level ground	8.84 kN
(ii) Maximum water mass, climbing 10°	606 kg
(ii) Tractive coefficient at that load	0.198
(iii) Maximum front-wheel weight, descending 10°, tank empty	11.9 kN
(iii) Tractive coefficient at that condition	0.291

Concept & Theory

Topic. Static weight transfer on a tractor from a rear three-point-hitch implement, on level ground and on a slope.

Key relations. Level ground: $W_f=W_{f0}-W_id/L$. Slope (nose-up climbing, $+\theta$; nose-down descending, $-\theta$): $W_f=\cos\theta(W_{f0}-W_id/L)\mp\sin\theta(W_0h_0+W_ih)/L$. Tractive coefficient: $\mu_t=F_t/W_r$ with $F_t=(W_0+W_i)\sin\theta$.

Why this works. A rear-mounted implement's weight acts behind the rear axle, so its moment about that axle always opposes (lifts) the front reaction — on level ground this depends only on the horizontal offset $d$, not the height $h$, because a vertical weight vector produces no moment through a change in height alone. Once the whole system is tilted, the weight vector picks up a slope-parallel component that acts through every CG's height above the ground, adding a second moment term that scales with $h$; climbing tilts this term the same direction as the implement's own moment (both shift weight rearward), while descending flips it (shifts weight forward), which is why the front-wheel weight is minimized while climbing loaded and maximized while descending empty.

Common pitfall. Assuming the implement's height above ground never matters for weight transfer — true only on level ground; on any slope it enters directly through the $\sin\theta\,h/L$ term and cannot be dropped.

Source. Goering & Hansen, Engine and Tractor Power, Ch. on tractor weight transfer and traction; Kepner, Bainer & Barger, Principles of Farm Machinery, Ch. 3 (traction and tractor performance).

Question 2 of 4: Weight Transfer from a Rear-Mounted Spray Tank

Problem 2: Weight Transfer from a Rear-Mounted Spray Tank (equal value)