The expected answers are:

(a)sum of the forces in one direction = sum of the forces in opposite direct
OR
Algebraic sum of all the forces must be equal to zero.
The sum of the clockwise moments about a point = the sum of the anticlockwise moments about the same point.

Let the mass of the metre rule = m Taking moments about the fulcrum.
        60gl (30-6 ) = mgl (48 - 30)
        60 x 24 = m x 18
        m = 60 x 24 = 80g
                 18

Award full marks if calculation is correct without the diagram.

       660gl X (x -13) = 80gl x (48 - x)

Taking moments about the fulcrum.
      3 (x – 13) = 4 (48-x)
       3x – 39 = 192 – 4x
      3x + 4x = 192 + 39
      7x = 231
      x = 231/7 = 33cm

Award full 3 marks if calculation is correct without diagram.

( c )(i)

(ii)I. minimum effort E = F + mg sin θ
      E = 1000 + 700 x sin 30
      = 1350N
II.VR = I/sin θ = 5/2.5 = 2
III.MA = L/E = 700/1350 = 0.52
This question on the equilibrium of forces was very popular. The performance of the candidates was fairly satisfactory. In part (a), most of the candidates knew that forces in• opposite direction must balance each other, they did not realize that the clockwise and the anticlockwise moments must be taken about the same point for them to be equal.

In part (b), the candidates were expected to
(i)correctly interpret the diagrams
(ii)calculate moment of forces about a point
(iii)distinguish between mass and weight.
Most candidates could not carry out these task thereby leading to loss of substantial marks in this part.
Part (c) tested the candidates awareness of the fact that an inclined plane is a simple machine. The numerical part also required the candidates to

(i)present the data in a well labelled diagram;
(ii)recognize that friction acts between the surfaces of the box and inclined plane surfaces in contact and that it counters the motion of the box.
(iii)identify the directions of all the forces acting on the box.

Most candidates could not represent the given data diagrammatically and identify the forces acting on the box. Also the required minimum effort applied to pull up the box could not be seen by some candidates as the sum of the given friction and the component of weight down the plane.
On the other hand, the calculation of the velocity ratio and force ratio was satisfactorily handled by most of the candidates.