(a) (i) What is a machine?
(ii) State two uses of gears.
(iii) Define the velocity ratio for a pair of gear wheels.
(iv) How can the mechanical advantage of a gear system be increased?

The diagram above illustrates the gears system of a bicycle.
(i) Determine its velocity ratio.
(ii) If the bicycle has an efficiency of 90%, calculate the effort required to overcome a load of 70N.
(iii) Why is the calculated effort less than the actual effort required?
(a)(i) What is a machine? A machine is a device in which an effort applied at one point is used to overcome a load at another point, thereby making work easier to do.
(a)(ii) Two uses of gears:
- To change the speed of rotation (and to change the direction of motion).
- To magnify force, i.e. to increase the turning effect (torque) transmitted from one shaft to another.
(a)(iii) Velocity ratio of a pair of gear wheels. It is the ratio of the number of teeth on the driven wheel to the number of teeth on the driving wheel:
\[ V.R. = \frac{\text{number of teeth on driven wheel}}{\text{number of teeth on driving wheel}} \]
(a)(iv) How the mechanical advantage is increased. Make the driving gear smaller, with fewer teeth than the driven gear; this raises the velocity ratio and hence the mechanical advantage.
(b)(i) Velocity ratio of the bicycle gear system. The driven gear has 12 teeth and the driving gear has 18 teeth, so:
\[ V.R. = \frac{12}{18} = \frac{2}{3} \]
(b)(ii) Effort to overcome a load of 70 N at 90% efficiency.
\[ \text{Efficiency} = \frac{M.A.}{V.R.} \Rightarrow M.A. = \text{Efficiency} \times V.R. = 0.90 \times \frac{2}{3} = 0.60 \]
Since \(M.A. = \dfrac{\text{Load}}{\text{Effort}}\):
\[ \text{Effort} = \frac{\text{Load}}{M.A.} = \frac{70}{0.60} = 116.7\ \text{N} \]
(b)(iii) Why the calculated effort is less than the actual effort required. The calculation ignores the extra force lost to friction between the chain and the gears and between the tyres and the load; in practice this friction means a larger effort is actually needed.
(a)(i) What is a machine? A machine is a device in which an effort applied at one point is used to overcome a load at another point, thereby making work easier to do.
(a)(ii) Two uses of gears:
- To change the speed of rotation (and to change the direction of motion).
- To magnify force, i.e. to increase the turning effect (torque) transmitted from one shaft to another.
(a)(iii) Velocity ratio of a pair of gear wheels. It is the ratio of the number of teeth on the driven wheel to the number of teeth on the driving wheel:
\[ V.R. = \frac{\text{number of teeth on driven wheel}}{\text{number of teeth on driving wheel}} \]
(a)(iv) How the mechanical advantage is increased. Make the driving gear smaller, with fewer teeth than the driven gear; this raises the velocity ratio and hence the mechanical advantage.
(b)(i) Velocity ratio of the bicycle gear system. The driven gear has 12 teeth and the driving gear has 18 teeth, so:
\[ V.R. = \frac{12}{18} = \frac{2}{3} \]
(b)(ii) Effort to overcome a load of 70 N at 90% efficiency.
\[ \text{Efficiency} = \frac{M.A.}{V.R.} \Rightarrow M.A. = \text{Efficiency} \times V.R. = 0.90 \times \frac{2}{3} = 0.60 \]
Since \(M.A. = \dfrac{\text{Load}}{\text{Effort}}\):
\[ \text{Effort} = \frac{\text{Load}}{M.A.} = \frac{70}{0.60} = 116.7\ \text{N} \]
(b)(iii) Why the calculated effort is less than the actual effort required. The calculation ignores the extra force lost to friction between the chain and the gears and between the tyres and the load; in practice this friction means a larger effort is actually needed.