9.1 Calculate the: 9.1.1 Rotation frequency of the output shaft if the electric motor rotates at 2 300 r/min 9.1.2 Velocity ratio between the input and output shaft 9.2 FIGURE 9.2 below shows a belt drive system with a 260 mm driver pulley - NSC Mechanical Technology Welding and Metalwork - Question 9 - 2017 - Paper 1

The velocity ratio (VR) can be calculated using the formula: $VR = \frac{N_{input}}{N_{output}}$ Substituting the known values: $VR = \frac{2300}{410.71} \approx 5.61$

For the driven pulley, the rotation frequency $n_r$ is given by: $n_r = \frac{v}{D_R}$ Substituting the speed and diameter:

• $v = 32$ m/s
• $D_R = 0.26$ m

Calculating: $n_r = \frac{32}{0.26} \times 60 \approx 738.46\text{ r/min}$

Given that the tension in the slack side $T_2$ is 140 N, and the ratio of the tension in the tight side to that in the slack side is 2.5: $T_1 = 2.5 \times T_2 = 2.5 \times 140 = 350 \text{ N}$

The power transmitted can be calculated using: $P = (T_1 - T_2) \cdot v$ Substituting known values: $P = (350 - 140) \cdot 32$ $P = 210 \cdot 32 = 6720\text{ Watts}$

To find the fluid pressure $P_A$ in the hydraulic system, we use: $A_A = \frac{\pi D^2}{4}$ Where $D = 0.02$ m. Thus, $A_A = \frac{\pi (0.02)^2}{4} = \approx 3.14 \times 10^{-4} m^2$ Substituting into the pressure formula: $P_A = \frac{F}{A_A}$ Assuming a force $F = 300 N$, we find: $P_A = \frac{300}{3.14 \times 10^{-4}} \approx 967741.94 \text{ Pa} = 0.97 \text{ MPa}$

To calculate the stroke at Piston B: $V_B = A_B \cdot L_B$ Where:

• $V_A = V_B$ Assuming cross-sectional area $A_B = \frac{\pi D_B^2}{4} = \frac{\pi (0.075)^2}{4} \approx 4.42 \times 10^{-3} m^2$Thus:$L_B = \frac{A_A \cdot L_A}{A_B}$Substituting:$L_B = \frac{3.14 \times 10^{-4} \cdot 185}{4.42 \times 10^{-3}} \approx 12.98 \text{ mm}$$

The purpose of traction control in a motor vehicle is to prevent wheel spin during acceleration. It does this by adjusting the power delivered to the wheels, ensuring better grip and control on various surfaces.

A safety belt is regarded as an active safety feature because it is designed to actively restrain the occupants during a collision, reducing the risk of injury. It must be fastened by the driver or passengers to be effective, thereby promoting safety through user engagement.