(a) State two; (i) laws of solid friction; (ii) advantages of friction; (iii) methods of reducing friction. (b) Draw and label a diagram of a pulley system ...

Answer Details

(a)(i) Two laws of solid friction:

• Friction opposes the relative motion (or the tendency to relative motion) between two surfaces in contact.
• The limiting (frictional) force \(F\) is directly proportional to the normal reaction \(R\) between the surfaces, so that \(F = \mu R\); it is also independent of the area of contact and depends only on the nature of the surfaces.

(a)(ii) Two advantages of friction:

• It provides the grip that prevents a person from slipping when walking and enables vehicle tyres to hold the road.
• It allows belts and pulleys to transmit power in machines and enables brakes to bring moving vehicles to rest.

(a)(iii) Two methods of reducing friction:

• Lubrication of the moving surfaces with oil or grease.
• Use of ball bearings or roller bearings between the moving parts (smoothing or polishing the surfaces also helps).

(b) Pulley system with velocity ratio of 5:

A block-and-tackle whose velocity ratio is 5 has five rope segments (strings) supporting the movable (lower) block. This is arranged with 3 pulleys in the fixed upper block and 2 pulleys in the movable lower block. The load \(W\) hangs from the lower block and the effort \(E\) is applied to the free end of the rope.

Because 5 strings support the lower block, when the effort end is pulled through a distance of 5 units the load rises through only 1 unit, giving

\[ \text{V.R.} = \frac{\text{distance moved by effort}}{\text{distance moved by load}} = \frac{5}{1} = 5. \]

(c)(i) Relationship between efficiency, mechanical advantage and velocity ratio:

Efficiency is the ratio of useful work output to work input, expressed as a percentage:

\[ E = \frac{\text{work output}}{\text{work input}} \times 100\% = \frac{\text{load} \times \text{distance moved by load}}{\text{effort} \times \text{distance moved by effort}} \times 100\%. \]

Rearranging the factors,

\[ E = \left(\frac{\text{load}}{\text{effort}}\right) \times \left(\frac{\text{distance moved by load}}{\text{distance moved by effort}}\right) \times 100\%. \]

Now \(\dfrac{\text{load}}{\text{effort}} = \text{M.A.}\) and \(\dfrac{\text{distance moved by effort}}{\text{distance moved by load}} = \text{V.R.}\), so \(\dfrac{\text{distance moved by load}}{\text{distance moved by effort}} = \dfrac{1}{\text{V.R.}}\). Therefore

\[ E = \frac{\text{M.A.}}{\text{V.R.}} \times 100\%. \]

(c)(ii) Work done by the person (work input):

Useful work output in raising the load:

\[ W_{\text{out}} = mgh = 300 \times 10 \times 4 = 12000\ \text{J}. \]

Since efficiency \(E = \dfrac{W_{\text{out}}}{W_{\text{in}}} \times 100\%\), the work input (work done by the person) is

\[ W_{\text{in}} = \frac{W_{\text{out}}}{E} = \frac{12000}{0.80} = 15000\ \text{J}. \]

The work done by the person is \(W_{\text{in}} = 15000\ \text{J} = 15\ \text{kJ}\).