1. 1
   A technician uses a spanner to turn a rusted nut on a workbench. The spanner is 0.24 m long, and the technician applies a force perpendicular to the spanner at its very end.

   (a)

   Which quantity, along with the applied force, is needed to calculate the moment of the force about the nut?

   [1 mark]

   • AThe mass of the spanner
   • BThe perpendicular distance from the line of action of the force to the pivot
   • CThe time for which the force is applied
   • DThe speed at which the spanner is turned

   (b)

   The technician applies a force of 60 N perpendicular to the spanner at its very end, 0.24 m from the nut. What is the moment of this force about the nut?

   [1 mark]

   • A250 N m
   • B60.24 N m
   • C14.4 N m
   • D0.004 N m

   (c)

   The technician finds the nut will not turn with a force of 60 N at 0.24 m from the nut. Explain, using the idea of moments, one change the technician could make to the spanner or the way it is used so that the same 60 N force produces a larger moment, and calculate the new moment if the distance is increased to 0.40 m.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   Two children sit on a playground see-saw that is balanced on a central pivot. One child sits 1.5 m from the pivot on the left side, and the other sits on the right side. The see-saw is perfectly horizontal and balanced.

   (a)

   Which principle explains why the see-saw does not rotate?

   [1 mark]

   • AThe law of reflection applied to the see-saw surface
   • BNewton's first law applied only to the pivot
   • CThe conservation of momentum of each child
   • DThe principle of moments: the sum of the clockwise moments equals the sum of the anticlockwise moments

   (b)

   The child on the left weighs 300 N and sits 1.5 m from the pivot. The child on the right weighs 450 N. For the see-saw to remain balanced, at what distance from the pivot must the child on the right sit?

   [1 mark]

   • A1.0 m
   • B1.5 m
   • C2.25 m
   • D3.0 m

   (c)

   A third, lighter child wants to join the child on the left side of the see-saw shown above, sitting at the same 1.5 m distance from the pivot. Explain what would happen to the see-saw immediately after the lighter child sits down, and describe one way the see-saw could be rebalanced without anyone changing seats.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   A uniform steel beam of weight 800 N and length 6 m rests horizontally on two supports, one at each end. A workman of weight 700 N stands on the beam 2 m from the left-hand support.

   (a)

   Taking moments about the left-hand support, calculate the upward force provided by the right-hand support. (Assume the beam's weight acts at its centre.)

   [3 marks]

   (b)

   Using your answer to part (a) and the fact that the beam is in equilibrium, calculate the upward force provided by the left-hand support, and state the two conditions that must both be true for the beam to be in equilibrium.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   A wheelbarrow is used to move a pile of bricks. The wheelbarrow has a wheel at the front (acting as the pivot), a load of bricks placed in the tub above the wheel, and the gardener lifts and pushes on two handles at the back, 1.6 m from the wheel. The bricks weigh 400 N and are positioned 0.4 m from the wheel.

   (a)

   Explain, using the principle of moments, how the wheelbarrow acts as a lever to allow the gardener to lift the bricks using a much smaller force than the weight of the bricks, and calculate the minimum force the gardener must apply at the handles to just lift the load, assuming the wheelbarrow's own weight is negligible.

   [6 marks]

   (b)

   The gardener notices that a pair of long-handled bolt cutters works on a similar principle to the wheelbarrow but is used to cut through a metal bolt rather than lift a load. Explain how both the wheelbarrow and the bolt cutters use the principle of moments to gain a mechanical advantage, and evaluate why gears, rather than a simple lever, are used in a bicycle to transmit the rotational effect of the rider's pedalling force to the rear wheel.

   [6 marks]

   Total for question 4: 12 marks

End of questions