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Moments, Equilibrium and Centre of Gravity: Question 7

Syllabus 1.5.2

Structured Core 7 marks

A builder uses a straight, rigid steel bar as a lever to prise up the edge of a heavy paving slab. The bar rests across a small wooden block, which acts as the pivot (fulcrum). The builder pushes down on one end of the bar with a force of 80 N80\text{ N}, at a perpendicular distance of 1.2 m1.2\text{ m} from the pivot. This is just enough to lift the edge of the slab, which pushes back up on the other end of the bar with a force of 960 N960\text{ N}.

(a) Calculate the moment of the builder's 80 N80\text{ N} force about the pivot. [2]

(b) The bar is in equilibrium at the instant the slab just begins to lift. Use the principle of moments to calculate the perpendicular distance, dd, between the pivot and the point where the slab's 960 N960\text{ N} force acts on the bar. [3]

(c) Explain, in terms of perpendicular distances from the pivot, why the builder is able to lift the heavy slab using a downward force very much smaller than the slab's own resistance force. [2]

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Worked solution

Part (a): Moment of the builder’s force

moment=force×perpendicular distance\text{moment} = \text{force} \times \text{perpendicular distance}

moment=80 N×1.2 m\text{moment} = 80\text{ N} \times 1.2\text{ m}

moment=96 N m\text{moment} = \boxed{96\text{ N m}}

Part (b): Finding the unknown distance dd

The bar is in equilibrium, so by the principle of moments, the moment of the builder’s force about the pivot equals the moment of the slab’s resistance force about the pivot:

960 N×d=96 N m960\text{ N} \times d = 96\text{ N m}

Rearrange to make dd the subject:

d=96 N m960 Nd = \frac{96\text{ N m}}{960\text{ N}}

d=0.10 md = \boxed{0.10\text{ m}}

So the slab’s force acts only 10 cm10\text{ cm} from the pivot.

Part (c): Explaining how the lever works

The builder’s 80 N80\text{ N} force acts at a perpendicular distance of 1.2 m1.2\text{ m} from the pivot, while the slab’s 960 N960\text{ N} resistance force acts at a perpendicular distance of only 0.10 m0.10\text{ m} from the pivot. Because moment == force ×\times distance, the much larger distance of the builder’s force compensates for its much smaller size, so the two moments can still be equal (96 N m96\text{ N m} each). This is why a lever allows a relatively small applied force, acting far from the pivot, to balance (and just overcome) a much larger resistance force acting close to the pivot.

Final answers

  • (a) Moment of the builder’s force == 96 N m96\text{ N m}
  • (b) d=d = 0.10 m0.10\text{ m} (10 cm10\text{ cm})
  • (c) The applied force acts at a much greater distance from the pivot than the resistance force does, so its moment can match the resistance force’s moment despite being much smaller