This set of Farm Machinery Multiple Choice Questions & Answers (MCQs) focuses on “Traction and Traction Theory – 1”.

1. A traction wheel having 600 mm diameter was tested in a soil bin and following data was obtained, angular speed=10rpm, input torque to wheel angle=60 Nm, drawbar pull=150 N, normal load on wheel axle=500 N, wheel forward speed=0.25 ms^{-1}. What will be the slippage?

a) 19.47%

b) 20.18%

c) 20.43%

d) 19.98%

View Answer

Explanation: μ = P/R

_{2}= 150/500 = 0.3

V

_{t}= \(\frac{0.6}{2}*\frac{2π*10}{60*10}\) = 0.3142 ms

^{-1}

Slippage = \(\frac{0.3142-0.25}{0.3142}\)*100 = 20.43%

2. A two-wheel drive tractor is pulling a load of 12 KN horizontally on levelled surface at forward speed of 5km/hr. The rolling radius of traction wheel and wheel slip are 0.65m and 20% respectively. If rear axle torque is 9KNm the tractive efficiency is ______

a) 69.31%

b) 23.90%

c) 51.49%

d) 10.24%

View Answer

Explanation: Slip, Vt – Va = 0.20 Vt

0.80 Vt = Va

Vt = \(\frac{V_a}{0.80}\) = 1.736 ms

^{-1}

η = \(\left[\frac{12*\frac{25}{18}}{\frac{9}{0.65}*1.736}\right] = \frac{16.66}{24.036}\) = 69.31%

3. The line of pull on an implement is 20° above the horizontal and in a vertical plane which is at an angle 10° with the direction of travel. What drawbar power would be required at 5km/h

a) 13.8 KW

b) 11.5 KW

c) 12.5 KW

d) 9.87 KW

View Answer

Explanation: Draft = Pcos20° = 8.328 KN

Side draft = (Psin10°) cos20° = 1.468 KN

Drawbar power = 9 * 5 * 5/18 = 12.5 KW

4. 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 tractor in km/h.

a) 12.09 kmh^{-1}

b) 11.31 kmh^{-1}

c) 14.90 kmh^{-1}

d) 9.99 kmh^{-1}

View Answer

Explanation: P = \(\frac{2πNτ}{60} \)

τ = \(\frac{35*746*60}{2π*1200}\) = 207.776 Nm

On each axle = 207.776/2 = 103.88 Nm

V = \(\frac{πDN}{60} = \frac{π*1.5*1200}{30*60}\) = 3.14 ms

^{-1}*5/18 = 11.31 kmh

^{-1}

5. What is steady state turn?

a) μ = \(\sqrt{\frac{g*A*r}{Zcg}} \)

b) μ = \(\sqrt{\frac{Zcg}{g*A*r}} \)

c) μ = \(\sqrt{\frac{g-A-r}{Zcg}} \)

d) μ = \(\sqrt{\frac{g+A+r}{Zcg}} \)

View Answer

Explanation: If angle of steering is continuously changing it is called dynamic state turn. If steering angle is fixed angle is fixed to some amount the turning is called steady state turn.

6. A traction wheel having 600 mm diameter was tested in a soil bin and following data was obtained, angular speed=10rpm, input torque to wheel angle=60 Nm, drawbar pull=150 N, normal load on wheel axle=500 N, wheel forward speed=0.25 ms^{-1}. What will be the tractive efficiency?

a) 69.96%

b) 59.67%

c) 24.48%

d) 51.49%

View Answer

Explanation: μ = P/R

_{2}= 150/500 = 0.3

V

_{t}= \(\frac{0.6}{2}*\frac{2π*10}{60*10}\) c = 0.3142 ms

^{-1}

Slippage = \(\frac{0.3142-0.25}{0.3142}\)*100 = 20.43%

Tractive efficiency = \(\frac{drawbar \, power}{axle \, power} = \frac{pull*Va}{F*Vt} = \frac{150*0.25}{\frac{60}{0.3}*0.3142}\) = 59.67%

7. What is wheel’s force?

a) Force = wheel torque * wheel radius

b) Force = wheel torque + wheel radius

c) Force = wheel torque/wheel radius

d) Force = wheel torque – wheel radius

View Answer

Explanation: Traction is a force which generated between tyre and soil surface and due to this force only, tractor pull loads and move forward. More technically, when powered wheel try to move on soil, this wheel’s force (wheel torque/wheel radius) try to break the soil.

8. How much traction would be generated by the force of 1000N applied by tyre on soil and soil capacity is 2000 N?

a) 3000 N

b) 20,000 N

c) 2 N

d) 1000 N

View Answer

Explanation: Amount of traction generated depends both on power in wheels as well as on soil strength. In this case the traction will be 1000 N because here wheel power is less so wheel can’t use soil strength fully, thus wheel power limits the amount of traction generated.

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