Tractor Systems and Components Questions and Answers

Answer: C

The statement that is not appropriate is It is dimensionless.

Cone index is the force needed to push a standard cone penetrometer into the soil divided by the area of the cone's base. It is therefore a pressure, reported in kPa or MPa, and has dimensions of force per unit area.

The other statements are accepted. Cone index is used as an index of soil strength for trafficability and tillage work. It is a composite parameter, because the penetration resistance mixes shear, compression and soil-metal friction rather than measuring one soil property. The ASAE standard penetrometer is pushed at a constant rate of about 30 mm/s (1829 mm/min).

Answer: B

The maximum pull is about 14.46 kN.

The front wheels lift when the moment of the pull about the rear wheel contact point equals the restoring moment of the weight. The centre of gravity lies 0.30 × 2160 = 648 mm ahead of the rear axle. The pull acts 12° below the horizontal: its horizontal part acts at the hitch height of 0.70 m, and its downward part acts 0.12 m behind the rear axle, so both tend to tip the tractor backwards:

\[ \begin{aligned} P(0.70\cos 12^{\circ} + 0.12\sin 12^{\circ}) &= 15.84 \times 0.648 \\ P(0.6847 + 0.0249) &= 10.264 \\ P &= 10.264 / 0.7097 \\ &= 14.46\ \text{kN} \end{aligned} \]

1.48 kN is far too small to lift the front of a 15.84 kN tractor through a 0.7 m hitch, and 39.04 or 85.54 kN are several times the tractor's own weight. Note that the downward component helps tip the tractor here only because the hitch is behind the rear axle.

Answer: A

The categories are based on maximum drawbar power.

The ASAE-SAE three-point hitch standard sizes the hitch (pin diameters, link spacing, mast height) in Categories I to IV according to the range of tractor drawbar power each is meant for, from small tractors in Category I to the largest in Category IV.

Drawbar pull depends on speed and soil conditions as well as on the tractor, so it is not used for the classification. Engine brake power and PTO power are related figures, but the standard's ranges are stated in terms of drawbar power, the power actually delivered to a hitched implement.

Answer: D

The diameter is 45 mm.

The ASAE standard for rear power take-off shafts specifies three types. Type 1 runs at 540 rpm and has a 35 mm (1 3/8 in) shaft with 6 splines. Type 2 runs at 1000 rpm on a 35 mm shaft with 21 splines. Type 3 runs at 1000 rpm on a 45 mm (1 3/4 in) shaft with 20 splines.

So a 1000 rpm shaft with 20 splines is the 45 mm Type 3. 35 mm is the shaft of Types 1 and 2, which have 6 and 21 splines, and 30 mm and 40 mm are not standard PTO shaft sizes.

Answer: B

The line of pull passes above the virtual hitch point and bending force exists on lower links.

In free-link operation the links can float, so each acts as a two-force member carrying only axial load, and equilibrium forces the line of pull through the virtual hitch point, where the link lines meet. That describes the last option (through the point, tensile force in the lower links), not restrained operation.

When the links are restrained, the restraint supplies a moment, so the line of pull no longer has to pass through the virtual hitch point, and the moment shows up as bending in the lower links. That rules out the option through the point with bending (no moment would be needed) and the option below the point with no bending (an off-point line of pull cannot be balanced without it).

Answer: D

The differential lock ensures distribution of higher amount of torque to the wheel under better traction.

An open differential always sends equal torque to both wheels, and that torque is limited by the wheel with the poorer grip: when one wheel slips, it spins and the other gets little. Locking the differential makes both wheels turn together, so the wheel on firm ground can now take as much torque as its traction allows, far more than the slipping wheel.

Equal torque is what the open differential gives, the condition the lock is meant to escape. Equal power is not a design feature of either. Sending more torque to the wheel with poor traction would only make it spin faster.

Answer: A

The change in rear-wheel reaction is 1.30 kN.

The draft acting at the hitch height creates a moment that lifts the front of the tractor and transfers weight to the rear wheels. Taking moments about the front wheel contact, the extra rear reaction is the draft times hitch height divided by the wheel base:

\[ \Delta R = \frac{6 \times 0.39}{1.8} = 1.30\ \text{kN} \]

The tractor's weight and the position of its centre of gravity set the static reactions but do not enter the change caused by the pull. 3.90 kN divides by the 0.6 m from the rear axle to the centre of gravity and 1.95 kN by the 1.2 m from the centre of gravity to the front axle; neither is the wheel base that the moment acts over.

Answer: C

The peak torque is 180 N m.

Torque reserve (torque backup) is the rise from the torque at rated (maximum) power to the peak torque, expressed as a percentage of the rated-power torque:

\[ \begin{aligned} \text{reserve} &= \frac{T_{peak} - T_{rated}}{T_{rated}} \\ 0.20 &= \frac{T_{peak} - 150}{150} \\ T_{peak} &= 150 \times 1.2 = 180\ \text{N m} \end{aligned} \]

120 is 150 reduced by 20% instead of increased; peak torque always exceeds the torque at maximum power, since torque rises as the engine is lugged down. 100 and 210 do not follow from a 20% reserve.

Answer: B

The instability is avoided by adding weight on the front axle.

When a tractor pulls a heavy drawbar load, weight transfers from the front axle to the rear. If the dynamic load on the front axle falls below about 20% of the tractor's weight, the front end becomes light: steering control is lost, and in the extreme the tractor rears up and overturns backwards. Ballast (front weights) added to the front axle restores enough load there to keep the tractor stable and steerable.

Adding weight on the rear axle, or removing it from the front, moves more load to the rear and makes the instability worse. Reducing rear weight would lose traction without directly restoring the front-axle load.

Answer: A

The key's answer is increased by \(W\sin\theta\).

On a slope the weight W splits into a component \(W\cos\theta\) normal to the surface and a component \(W\sin\theta\) along it. Rolling the wheel up the slope, the towing force must overcome the rolling resistance on the surface plus the downslope component of the weight:

\[ \begin{aligned} R_{\text{level}} &= \rho W \\ R_{\text{slope}} &= \rho W\cos\theta + W\sin\theta \end{aligned} \]

where \(\rho\) is the rolling resistance coefficient. The key treats the total resistance to rolling the wheel up the slope as the rolling resistance, so it rises by the grade term \(W\sin\theta\). Strictly, the part due to rolling alone changes only by the small factor \(\cos\theta\); the large change comes from the slope.

The options 'increased by \(W\cos\theta\)' and 'decreased by \(W\cos\theta\)' use the normal component, which supports the wheel rather than opposing its motion. A decrease by \(W\sin\theta\) would describe rolling down the slope.