How many QCAA Engineering questions cover Machines, mechanisms and control?

AusGrader has 82 QCAA Engineering questions on Machines, mechanisms and control, all with instant AI grading and detailed marking feedback.

QCAA (QCE) Engineering — Machines, mechanisms and control Questions & Answers | AusGrader

Q6

2024

QCAA

1 mark

Q6

1 mark

A cyclist exerts an effort force of 125 N on the pedal of a bicycle. The load force on the rear wheel is 500 N. The velocity ratio of the bicycle is 5.
What is the efficiency of the bicycle?

A

125%

B

80%

C

8%

D

5%

Q7

2021

QCAA

1 mark

Q7

1 mark

A pulley system with an efficiency of 75% is used to vertically raise a 120 kg load a distance of 3 m. If the length of pulley rope pulled to raise the load is 8 m, the effort required is

A

330 N.

B

392 N.

C

440 N.

D

588 N.

Correct Answer

First, calculate the load force: $F_L = mg = 120 \times 9.8 = 1176 \text{ N}$ . Using the efficiency formula $\eta = \frac{F_L \cdot d_L}{F_E \cdot d_E}$ , we solve for effort: $0.75 = \frac{1176 \cdot 3}{F_E \cdot 8}$ . This yields $F_E = 588 \text{ N}$ .

Q20

2024

QCAA

4 marks

Q20

4 marks

A parcel slides down an inclined ramp, which is at 20° to the horizontal plane, at a constant velocity. The parcel experiences a normal force of 98.2 N. A coefficient of kinetic friction of 0.37 and a coefficient of static friction of 0.39 exist between the parcel and the ramp.

Determine the mass of the parcel.

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

Marking Criteria

Descriptor	Marks
recognises that forces are balanced, therefore $F_{\perp} = F_N$	1
recognises $F_N = mg \cos \theta$	1
correctly substitutes values	1
determines the mass of the parcel	1

Q20

2020

QCAA

9 marks

Q20

A double-threaded worm and 40-tooth worm wheel are used to drive a mechanical winch. The efficiency of the machine is 45%. The pitch of a double-threaded worm is halved.

Q20a

5 marks

Determine the mass of the load lifted by the winch using an effort of 50 N on the circumference of the worm wheel. Answer to the nearest whole unit.

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

Marking Criteria

Descriptor	Marks
Provides correct working to give correct gear ratio (VR)	1
Provides correct formula and substituted values for mechanical advantage	1
Provides correct working to give correct L	1
Provides correct formula and substituted values for mass	1
Provides correct working to give correct answer to the nearest whole unit with correct unit provided	1

Q20b

4 marks

An effort wheel is added to the worm axle, decreasing the efficiency of the compound machine to 36%. Determine the velocity ratio of the added component. Answer to two decimal places.

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

Marking Criteria

Descriptor	Marks
Provides correct formula and substituted values for effort wheel efficiency	1
Provides correct working to give correct effort wheel efficiency	1
Provides correct formula and substituted values for velocity ratio	1
Provides correct working to give correct answer to two decimal places	1

Q23

2021

QCAA

10 marks

Q23

10 marks

A conveyor transfers a 2 kg component between two assembly points during a product manufacturing process. Initially stationary at assembly point 1, the component moves 10 m with a uniform acceleration of $2 \text{ m/s}^2$ , just without slipping on the conveyor, to assembly point 2, where it stops.

Determine the coefficient of static friction required between the conveyor and the component if the transfer time between the two assembly points is reduced by 20%. Answer to two decimal places.

$u = 0 \quad s = 10 \text{ m} \quad v = ? \quad t = ? \quad a = 2 \text{ m/s}^2$
$v^2 = u^2 + 2as$
$v = \sqrt{u^2 + 2as}$
$v = \sqrt{0^2 + 2 \times 2 \times 10} = 6.32 \text{ m/s}$
$t = \frac{v - u}{a}$
$t = \frac{6.32 - 0}{2} = 3.16 \text{ s}$
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.

Marking Criteria

Descriptor	Marks
Provides correct formula and substituted values for velocity	1
Provides correct working to give correct velocity	1
Provides correct formula and substituted values for time	1
Provides correct working to give correct time	1
Provides correct working to give correct 20% time reduction	1
Provides correct formula and substituted values for acceleration	1
Provides correct working to give correct acceleration	1
Provides correct working to give correct force of friction	1
Provides correct formula and substituted values for coefficient of static friction	1
Provides correct working to give correct answer to two decimal places	1

QCAA Engineering Machines, mechanisms and control

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