(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...
(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 about a point
The moment of a force about a point is the turning effect of the force about that point. It is equal to the product of the force and the perpendicular distance of its line of action from the point.
\[\text{Moment}=F\times d\]
where \(F\) is the force and \(d\) is the perpendicular distance from the pivot to the line of action of the force. The SI unit is newton metre, \(\text{N m}\).
Moment of a downward force \(F\) about pivot \(O\): \(\text{moment}=F\times d\), where \(d\) is the perpendicular distance from \(O\) to the line of action of \(F\).
(b) Conditions for equilibrium of coplanar parallel forces
The algebraic sum of the forces must be zero, that is, total upward force equals total downward force.
The algebraic sum of moments about any point must be zero, that is, total clockwise moment equals total anticlockwise moment.
The metre rule balances at the 48 cm mark, so its weight acts at the 48 cm mark.
(i) Mass of the metre rule
Taking moments about the balance point at 30 cm:
\[60(30-6)=M(48-30)\]
\[60\times24=M\times18\]
\[M=\frac{60\times24}{18}=80\text{ g}\]
Therefore, the mass of the metre rule is 80 g.
(ii) New balance point
Let the new balance point be \(x\) cm from the zero end. The 60 g body is at 13 cm and the rule's weight acts at 48 cm.
Taking moments about the new balance point:
\[60(x-13)=80(48-x)\]
\[60x-780=3840-80x\]
\[140x=4620\]
\[x=33\text{ cm}\]
Therefore, the balance point is 33 cm from the zero end.
(c) Relation between efficiency, mechanical advantage and velocity ratio
The moment of a force about a point is the turning effect of the force about that point. It is equal to the product of the force and the perpendicular distance of its line of action from the point.
\[\text{Moment}=F\times d\]
where \(F\) is the force and \(d\) is the perpendicular distance from the pivot to the line of action of the force. The SI unit is newton metre, \(\text{N m}\).
Moment of a downward force \(F\) about pivot \(O\): \(\text{moment}=F\times d\), where \(d\) is the perpendicular distance from \(O\) to the line of action of \(F\).
(b) Conditions for equilibrium of coplanar parallel forces
The algebraic sum of the forces must be zero, that is, total upward force equals total downward force.
The algebraic sum of moments about any point must be zero, that is, total clockwise moment equals total anticlockwise moment.
The metre rule balances at the 48 cm mark, so its weight acts at the 48 cm mark.
(i) Mass of the metre rule
Taking moments about the balance point at 30 cm:
\[60(30-6)=M(48-30)\]
\[60\times24=M\times18\]
\[M=\frac{60\times24}{18}=80\text{ g}\]
Therefore, the mass of the metre rule is 80 g.
(ii) New balance point
Let the new balance point be \(x\) cm from the zero end. The 60 g body is at 13 cm and the rule's weight acts at 48 cm.
Taking moments about the new balance point:
\[60(x-13)=80(48-x)\]
\[60x-780=3840-80x\]
\[140x=4620\]
\[x=33\text{ cm}\]
Therefore, the balance point is 33 cm from the zero end.
(c) Relation between efficiency, mechanical advantage and velocity ratio
