(a) Explain with the aid of a diagram what is meant by the moment of a force about a point.
(b) State the conditions of equilibrium for a number of coplanar parallel forces.
A metre rule is found to balance at the 48cm mark. When a body of mass 60g is suspended at the 6cm mark, the balance point is found to be at the 30cm mark. Calculate:
(i) the mass of the metre rule; (ii) the distance of the balance point from the zero end, if the body were moved to the 13cm mark.
(c) Show that the efficiency E, the force ratio M.A and the velocity ratio V.R of a machine are related by the equation \(E = \frac{M.A}{V.R} \times 100%\)
The efficiency of a machine is 80%. Determine the work done by a person using this machine to raise a load of 200kg through a vertical distance of 3.0m.
[Take g = 10ms\(^{-2}\)]
(a) Moment of a force is the product of force and perpendicular distance from the turning point.
Moment = F x D
The S.I unit is Nm.
(b) Conditions: (i) The sum of forces in one direction must be equal to sum of forces in the opposite direction.
(ii) Sum of clockwise forces must be equal to sum of anti-clockwise forces about the same point.
Sum of clockwise moment = Sum of anti- clockwise moment.
\(M \times 18 = 60 \times 24\)
Mass of metre rule \(M = \frac{60 \times 24}{18} = 80g\)
Clockwise moment = anti- clockwise moment
\(80 \times 48 - (13 + x) = 60 \times x\)
\(80 \times 35 - x = 60x\)
\(2800 - 80x = 60x \implies 2800 = 60x + 80x\)
\(2800 = 140x \implies x = 20cm\)
Distance of balance point from the zero end = 20cm + 13cm = 33cm.
(c) \(Efficiency = \frac{\text{Work output}}{\text{Work input}} \times 100%\)
= \(\frac{\text{Load x Distance moved by load}}{\text{Effort x Distance moved by effort}}\)
= \(M.A \times \frac{1}{V.R} \times 100%\)
(ii) \(80% = \frac{200 \times 3.0}{\text{Work done by effort}} \times100% \)
\(0.8 = \frac{600}{\text{Work done by effort}} \implies \text{Work done by effort} = \frac{600}{0.8} = 750J\)
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