(a) State the conditions of equilibrium for a number of coplanar parallel forces. (b) A metre rule is found to balance horizontally at the 48 cm mark. When ...

(a) Conditions of equilibrium for coplanar parallel forces

A number of coplanar parallel forces are in equilibrium when:

1. The algebraic sum of the forces is zero, i.e. the sum of the forces acting in one direction equals the sum of the forces acting in the opposite direction (the resultant force is zero).
2. The algebraic sum of the moments of the forces about any point is zero, i.e. the sum of the clockwise moments about the point equals the sum of the anticlockwise moments about that same point (the principle of moments).

(b) Balancing a metre rule

The rule balances by itself at the \(48\,\text{cm}\) mark, so the whole weight of the rule acts at the \(48\,\text{cm}\) mark.

(i) Mass of the metre rule

With the \(60\,\text{g}\) body hung at the \(6\,\text{cm}\) mark, the new balance point (fulcrum) is at the \(30\,\text{cm}\) mark. The body sits on one side of the fulcrum and the weight of the rule acts on the other side. Taking moments about the \(30\,\text{cm}\) fulcrum:

\[ 60\times(30-6)=m\times(48-30) \]\[ 60\times24=m\times18 \]\[ m=\frac{60\times24}{18}=\frac{1440}{18}=80\,\text{g} \]

The mass of the metre rule is 80 g.

(ii) New balance point with the body at the 13 cm mark

Let the new balance point be at the \(x\,\text{cm}\) mark. The body (\(60\,\text{g}\)) now acts at \(13\,\text{cm}\) and the rule's weight (\(80\,\text{g}\)) still acts at \(48\,\text{cm}\). Taking moments about the fulcrum at \(x\):

\[ 60\times(x-13)=80\times(48-x) \]\[ 60x-780=3840-80x \]\[ 140x=4620 \]\[ x=\frac{4620}{140}=33\,\text{cm} \]

The new balance point is 33 cm from the zero end.

(c) Pulling a box up an inclined plane

Box mass \(=70\,\text{kg}\), so its weight \(W=mg=70\times10=700\,\text{N}\); length of plane \(L=5\,\text{m}\); height of platform \(h=2.5\,\text{m}\); frictional force \(F=1000\,\text{N}\); angle of incline \(\theta=30^{\circ}\).

(i) Diagram of the forces acting on the box while in motion

The four forces acting on the box are: its weight \(W\) acting vertically downwards; the normal (reaction) force \(N\) acting perpendicular to the surface of the plane; the effort \(E\) applied up along the plane; and the frictional force \(F\) acting down along the plane, opposing the upward motion.

(ii) Calculations

I. Minimum effort applied in pulling up the box

Since the box moves up at uniform (constant) speed, the effort must balance the component of the weight along the plane together with the friction acting down the plane:

\[ E=W\sin\theta+F=700\sin30^{\circ}+1000 \]\[ E=700\times0.5+1000=350+1000=1350\,\text{N} \]

The minimum effort is 1350 N.

II. Velocity ratio of the plane

\[ V.R.=\frac{1}{\sin\theta}=\frac{1}{\sin30^{\circ}}=\frac{1}{0.5}=2 \]

(This agrees with \(\dfrac{\text{length}}{\text{height}}=\dfrac{5}{2.5}=2\).) The velocity ratio is 2.

III. Force ratio (mechanical advantage) of the plane

\[ \text{Force ratio}=M.A.=\frac{\text{load}}{\text{effort}}=\frac{700}{1350}=0.52 \]

The force ratio of the plane is 0.52.