1. A 2-wheel drive 35 HP tractor has 1.5 m rear wheel diameter. The engine runs at 1200 rev/min. The total reduction of speed is 30:1. Find the travelling speed of the tractor in km/h and tractive force at each driving wheel.
a) 3.896 KN
b) 4.157 KN
c) 7.437 KN
d) 8.315 KN
Discussion
Explanation: Naxle = 1200/30 = 40 rpm
Velocity = πNa * Da
= π * 40/60 * 1.5 = 3.14 m/s
Tractive force = P/V = \(\frac{(35*0.746)}{3.14}\) = 8.315 KN
Tractive at each driving wheel = 8.315/ 2
= 4.157 KN.
2. A tractor having weight as 1500 KGF is brought to a speed of 10 KMPH in 10 seconds by applying a uniform acceleration. Calculate initial fore on tractor.
a) 416.67 N
b) 372.98 N
c) 747.8 N
d) 219.84 N
Discussion
Explanation: Acceleration: \(\frac{v-u}{t}=\frac{10*5}{18}-\frac{0}{10}=\frac{50}{18*10}=\frac{5}{18}\)
Initial Force = ma
= 1500*\(\frac{5}{18}\)
= 416.67 N.
3. What HP of tractor is suitable for a 40 hectares farm?
a) 20-25 HP
b) 10-15 HP
c) 30-35 HP
d) 40-45 HP
Discussion
Explanation: Generally, 1.5 hectare/HP has been recommended where adequate irrigation facilities are available and more than one crop is taken. So, a 30-35 tractor is suitable for 40 hectares farm.
4. A tractor with a mass of 1600 kg develops a maximum tractive effort of 3000 N in top gear on a concrete surface. The total rolling resistance is 160 N. If the gradient angle is defined as the angle that the inclined plane makes with the horizontal, the maximum gradient can be negotiated by the tractor is _________
a) 10.44°
b) 11.63°
c) 29.4°
d) 34.29°
Discussion
Explanation:
Referring to figure
In the limiting case:
3000 = mtg sin θ + 160
mtg sin θ = 3000-160
mtg sin θ = 2840
sin θ = \(\frac{2840}{mtg}=\frac{2840}{1600*9.8} \)
sin θ = 0.18112
θ = sin-1(0.18112)
θ = 10.44°.
5. The complete path of power from the engine to the wheels is known as _______
a) Muffler
b) Fender
c) Power train
d) Hitch system
Discussion
Explanation: Transmission is the speed reducing mechanism, equipped with several gears. It may be called a sequence of gears and shafts, through which the engine power is transmitted to the tractor wheels. It consists of tractor to suit different field conditions. The complete path of power from engine to the wheels is called power train.
6. In between which years did power take off was introduced?
a) 1975-1979
b) 1992-1996
c) 1999-2003
d) 1915-1919
Discussion
Explanation: Experimental power take offs were tried as early as 1878, and various homemade versions arose over the subsequent decades. International harvester company was first to market with a PTO on production tractor, with its model 8-16 introduced in 1918.
7. In between which years did hydraulic controls and three-point linkage were developed?
a) 1989-1993
b) 1947-1951
c) 1937-1941
d) 1920-1924
Discussion
Explanation: Harry Ferguson patented the three-point linkage for agricultural tractors in Britain in 1926. He and his colleagues continued to developed several innovations to this device which made the system workable, effective, desirable. In 1938, after almost two decades of trying to sell Henry Ford on using Ferguson’s system on tractors mass-produced by Ford, Ferguson finally convinced Ford. The American mass market debut was via the Ford Ferguson 9N in 1939.
8. If the inner and outer steering angle in a tractor are 9° and 9.8° respectively then for correct steering conditioning the value of the ratio of pivot axis spacing to track length is equal to ______
a) 0.256
b) 0.525
c) 0.990
d) 0.756
Discussion
Explanation:θ = 9.8°
Φ = 9.0°
We know that,
Cot θ – Cot Φ = \(\frac{c}{b} \)
-Cot (9.8) + Cot 9 = \(\frac{c}{b} \)
-5.789 + 6.314 = \(\frac{c}{b} \)
\(\frac{c}{b}\) = 0.525.
9. While turning a 4-wheel tractor equipped with the Ackermann steering mechanism the axis of the inner front wheel makes an angle of 45°, whereas the angle made by the axis of the outer front wheel is 30° with horizontal line parallel to the front axle. The kingpins of the axle are 1300 mm apart. The spindle genths are essentially zero. What should be the wheel base of the tractor if there is no slippage of any of the wheels?
a) 1886.78 mm
b) 1775.83 mm
c) 1336.59 mm
d) 2000.01 mm
Discussion
Explanation: Cot θ – Cot Φ = \(\frac{c}{b} \)
Cot30° – Cot45° = \(\frac{c}{b} \)
b = \(\frac{1300}{Cot30° – Cot45°} \)
b = 1775.83mm.
10. A 2-wheel drive 35hp tractor has 1.5 rear wheel diameter. The engine runs at 1200 rev/min. the total reduction of speed is 30:1. Find the travelling speed of the tractor in km/h and the tractive force at each driving wheel.
a) 3.9925 KN
b) 2.1992 KN
c) 1.6784 KN
d) 4.1977 KN
Discussion
Explanation: Va = \(\frac{2πN*Dr*60}{2*30*1000}=\frac{2\pi*1.5*60*1200}{2*30*1000}\)=11.304
Va = 3.11 m/s
P = F * v
F = \(\frac{P}{v}=\frac{35*746}{3.11}\)=8.3955 KN
Therefore, force on each wheel = \(\frac{8.3955}{2}\)=4.1977.