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FIGURE 9.1 below shows a gear drive system - NSC Mechanical Technology Automotive - Question 9 - 2017 - Paper 1

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FIGURE 9.1 below shows a gear drive system. Driver gear A on the shaft of an electric motor has 30 teeth and mesh with gear B with 40 teeth on a counter shaft. On th... show full transcript

Worked Solution & Example Answer:FIGURE 9.1 below shows a gear drive system - NSC Mechanical Technology Automotive - Question 9 - 2017 - Paper 1

Step 1

9.1.1 Rotation frequency of the output shaft if the electric motor rotates at 2 300 r/min

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Answer

To find the rotation frequency of the output shaft, we'll use the gear ratio formula. The formula is given by:

NF=NATATBTCTDTETFN_{F} = N_{A} \cdot \frac{T_{A}}{T_{B}} \cdot \frac{T_{C}}{T_{D}} \cdot \frac{T_{E}}{T_{F}}

Where:

  • NAN_{A} = input speed = 2300 r/min
  • TAT_{A} = teeth of gear A = 30
  • TBT_{B} = teeth of gear B = 40
  • TCT_{C} = teeth of gear C = 20
  • TDT_{D} = teeth of gear D = 60
  • TET_{E} = teeth of gear E = 50
  • TFT_{F} = teeth of gear F = 70

Plugging in the values:

NF=2300304020605070N_{F} = 2300 \cdot \frac{30}{40} \cdot \frac{20}{60} \cdot \frac{50}{70}

Calculating step-by-step:

  1. First calculate the gear ratios:
  • 3040=0.75\frac{30}{40} = 0.75
  • 2060=13=0.3333\frac{20}{60} = \frac{1}{3} = 0.3333
  • 5070=0.7143\frac{50}{70} = 0.7143
  1. Now, substituting these back into our equation:

NF=23000.750.33330.7143410.71 r/minN_{F} = 2300 \cdot 0.75 \cdot 0.3333 \cdot 0.7143 \approx 410.71 \text{ r/min}

Therefore, the rotation frequency of the output shaft is approximately 410.71 r/min.

Step 2

9.1.2 Velocity ratio between the input and output shaft

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Answer

The velocity ratio (VR) can be calculated using the formula:

VR=NOUTPUTNINPUTVR = \frac{N_{OUTPUT}}{N_{INPUT}}

Where:

  • NOUTPUTN_{OUTPUT} = output frequency = 410.71 r/min
  • NINPUTN_{INPUT} = input frequency = 2300 r/min

Substituting the values:

VR=410.7123000.178VR = \frac{410.71}{2300} \approx 0.178

Thus, the velocity ratio between the input and output shaft is approximately 0.178.

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