Calculate the: 9.1.1 Rotation frequency of the driver pulley in r/min (4) 9.1.2 Power transmitted in this system (4) 9.2 Figure 9.2 shows a gear-drive system - NSC Mechanical Technology Automotive - Question 9 - 2016 - Paper 1

First, we can find the tension difference in the belt using the ratio: $\frac{T_1}{T_2} = 2.5$ Given:

• Force in the slack side (T2) = 140 N

So: $T_1 = 2.5 \times T_2 = 2.5 \times 140 = 350 N$

To calculate power transmitted:

Using the gear ratio formula: $N_{INPUT} = N_{OUTPUT} \times \left( \frac{T_{OUT} \times T_{IN}}{T_{IN} \times T_{OUT}} \right)$ Given:

• Output frequency $N_{OUTPUT} = 160$ r/min Calculating the input frequency: $\frac{36 \times 18 \times 60}{36 \times 40 \times 60} \implies N_{INPUT} = 1380 \text{ r/min}$

The velocity ratio (VR) can be calculated as: $VR = \frac{N_{INPUT}}{N_{OUTPUT}}$ Given:

• $N_{INPUT} = 1380$ r/min
• $N_{OUTPUT} = 160$ r/min

$VR = \frac{1380}{160} \approx 8.625 \\ \implies VR \approx 8.63$

The fluid pressure can be calculated using the area of piston A: $A_a = \frac{\pi D^2}{4} = \frac{\pi (0.04)^2}{4} \approx 1.2566 \times 10^{-3} m^2$

Using the force on piston A: $P_a = \frac{F_a}{A_a} = \frac{275 N}{1.2566 \times 10^{-3}} \approx 21884.00 \, Pa$

Calculating the area of piston B: $A_b = \frac{\pi D_b^2}{4} = \frac{\pi (0.075)^2}{4} \approx 4.42 \times 10^{-3} m^2$

Using the pressure calculated: $P_b = P_a = 21884.00 \, Pa$ $F_b = P_b \times A_b = 21884.00 imes 4.42 \times 10^{-3} \approx 967.32 \, N$

To get the mass: $Mass = \frac{F_b}{g} = \frac{967.32}{10} \approx 96.73 \, kg$

The purpose of traction control is to enhance vehicle stability by preventing wheel spin during acceleration. It helps maintain grip on the road, improving overall safety and control.

Passive safety features, such as airbags, are designed to protect vehicle occupants during a collision without requiring any action from them. These systems automatically activate to reduce injuries, making them crucial for vehicle safety.