QCAA Engineering Machines, mechanisms and control

gear ratio of double-thread worm and wheel
$= \frac{\text{worm wheel teeth}}{\text{worm teeth}}$
$= \frac{40}{2} = 20 \therefore VR = 20$
$\eta = \frac{MA}{VR}$
$0.45 = \frac{MA}{20}$
$MA = 0.45 \times 20 = 9$
$MA = \frac{L}{E}$
$L = MA \times E$
$= 9 \times 50 = 450\ N$
$F = mg$
$m = \frac{F}{g} = \frac{450}{9.8} = 45.92 \approx 46\ kg$

$\eta_{total} = \eta_1 \times \eta_2$
$0.36 = 0.45 \times \eta_2$
$\eta_2 = \frac{0.36}{0.45} = 0.80$
$\eta = \frac{MA}{VR}$
$VR = \frac{MA}{\eta} = \frac{9}{0.80} = 11.25$

First pump
$W = \text{force} \times \text{distance} = Fd$
$W = mgh$
$W = 500 \times 9.8 \times 6$
$W = 29 400$ J

Power output: First pump
$P = \frac{\text{work done}}{\text{time taken}} = \frac{W}{t}$
$P = \frac{29 400}{60}$
$P = 490$ W

Power input: First pump
$\eta = \frac{\text{useful output}}{\text{input}}$
$\text{input} = \frac{\text{useful output}}{\eta}$
Power input = $\frac{490}{0.8}$
Power input = 612.5 W

Second pump
$W = \text{force} \times \text{distance} = Fd$
$W = mgh$
$W = 500 \times 9.8 \times 4$
$W = 19 600$ J

Power output: Second pump
$P = \frac{\text{work done}}{\text{time taken}} = \frac{W}{t}$
$P = \frac{19 600}{60}$
$P = 326.67$ W

Power input: Second pump
$\eta = \frac{\text{useful output}}{\text{input}}$
$\text{input} = \frac{\text{useful output}}{\eta}$
Power input = $\frac{326.67}{0.75}$
Power input = 435.56 W

Overall power input of two-pump system
Power = 612.5 + 435.56
Power = 1048.06 W

New single-pump system
Power input = 1048.06 W
Power output = $90\% \times 1048.06$
Power output = 943.25 W

New single-pump system: mass of water delivered per minute
$P = \frac{W}{t}$
$W = P \times t$
$W = 943.25 \times 60$
$W = 56 595$ J

$W = mgh$
$m = \frac{W}{gh}$
$m = \frac{56 595}{9.8 \times 10}$
$m = 577.5 \text{ kg} = 578$ litres per minute

$\therefore$ The new single-pump system will deliver 578 litres of water per minute directly to the main reservoir.

$PE = mgh$
$= 1000 \times 9.8 \times 10$
$= 98 \text{ kJ}$

$E = KE + PE$
$106 = KE + 98$
$KE = 106 - 98$
$= 8 \text{ kJ}$

$KE = \frac{1}{2}mv^2$
$8000 = \frac{1}{2}1000v^2$
$v = \sqrt{\frac{2 \times 8000}{1000}}$
$= \sqrt{16}$
$= 4 \text{ m/s}$

$v^2 = u^2 + 2as$
$4^2 = 0^2 + 2a(10)$
$16 = 20a$
$a = \frac{16}{20}$
$a = 0.8 \text{ m/s}^2$

$v = u + at$
$4 = 0 + 0.8t$
$t = \frac{4}{0.8}$
$t = 5 \text{ s}$

$PE = mgh$
$980 = 40 \times 9.8 \times h$
$h = \frac{980}{40 \times 9.8} = 2.5$ m

$VR = \frac{D_E}{D_L} = \frac{\pi D}{\text{thread pitch}} = \frac{\pi \times 30}{2.5} = 37.70$

$efficiency = \frac{MA}{VR}$
$MA = 0.62 \times 37.70 = 23.37$

$MA = \frac{F_L}{F_E}$
$F_E = \frac{40 \times 9.8}{23.37} = 16.77$ N

Number of rotations of input shaft = $\frac{2.5 \text{ m}}{0.0025} = 1000$

Distance travelled by effort = $1000 \times \pi \times 0.03 = 94.25$ m

$W = Fd = 16.77 \times 94.25 = 1580.57$ J

$P = \frac{W}{t} = \frac{1580.57}{15} = 105.4$ W

$u = 0 \quad s = 10 \text{ m} \quad v = ? \quad t = ? \quad a = 2 \text{ m/s}^2$
$v^2 = u^2 + 2as$
$v = \sqrt{u^2 + 2as}$
$v = \sqrt{0^2 + 2 \times 2 \times 10} = 6.32 \text{ m/s}$
$t = \frac{v - u}{a}$
$t = \frac{6.32 - 0}{2} = 3.16 \text{ s}$
20% time reduction required $= 3.16 - (0.2 \times 3.16)$
$= 2.53 \text{ s}$
$s = ut + \frac{1}{2}at^2$
$10 = 0 \times 2.53 + \frac{1}{2}a \times 2.53^2$
$10 = \frac{a \times 6.4}{2}$
$a = \frac{10 \times 2}{6.4} = 3.13 \text{ m/s}^2$
$F = ma$
$= 2 \times 3.13 = 6.26 \text{ N}$
$F_f = u_s \times F_N$
$u_s = \frac{F_f}{F_N}$
$= \frac{6.26}{2 \times 9.8} = \frac{6.26}{19.6} = 0.32$
$\therefore$ the coefficient of static friction required between the component and conveyor at a 20% time reduction is 0.32.

Forces perpendicular to the inclined plane are balanced, so
$F_{\perp} = F_N$
$F_N = mg \cos \theta$
$m = \frac{9.82}{9.8 \cos 20}$
$m = 10.66 \text{ kg}$

A NAND gate output is 0 when and only when all its inputs are at 1. Otherwise the output is 1.

$P = \frac{W}{t}$
$4355 = \frac{W}{90}$
$W = 4355 \times 90 = 391\ 950\ J$
$W = F \times d \approx PE = mgh$
$\therefore W = mgh$
$391\ 950 = m \times 9.8 \times 20$
$m = \frac{391\ 950}{9.8 \times 20}$
$= 1999.75 \approx 2000\ kg$
$\therefore$ The mass of the full scrap metal bin is 2000 kg.

KE on impact $= PE = mgh$
$= 5 \times 9.8 \times 20$
$\therefore KE = 980\ J$
$KE = \frac{1}{2} \times m \times v^2$
$v^2 = \frac{980}{0.5 \times 5}$
$v = \sqrt{392}$
$= 19.80 \approx 20\ m/s$
$\therefore$ The velocity of the scrap metal on impact with the ground is 20 m/s.