Moments

Moments KUS objectives BAT understand moment and solve problems using moments on a Non- Uniform rod and/or when a rod is at a tipping point Starter: A wooden beam is supported by two ropes. A man stands at one end of the beam. Critique these models: Tension in the ropes is equal. The beam is modelled as a light horizontal rod. The man is a particle Tension in the ropes is equal. The beam is modelled as a light uniform rod. The man has weight acting vertically Tension in the ropes is not equal. The beam and man are modelled as particles.

Find two mistakes in Bill’s working and explain the errors made WB1 Bill and Helen are sitting on either end of a horizontal balance beam AB that is held up by one support at the midpoint as shown. The Beam has length 6 m, a mass of 40 kg and is non-uniform. Bill mass a mass of 50 kg and stands at A. Helen has a mass of 60 kg and stands at a point C such that the beam is in equilibrium. Bill attempts to work out the distance AC as shown below Find two mistakes in Bill’s working and explain the errors made Is there a solution to this problem? R A 40 50 60 First mistake there are two unknowns – the distance from A to the weight 40g is unknown. This is the significant mistake as no marks would be awarded – he is solving the wrong problem. Second mistake – has missed out g from the 450 force. Although he has the correct answer for his working – this is not a significant mistake R = 50g+ 40g+ 60g Forces in equilibrium 3R = 2 x 40 g + 60 g x d moments at B 450 = 80 g + 60 gd one unknown   d = 450 – 80g = 37/6 = 6.2 m solution 60g

Find the distance of the centre of mass of the rod from A WB2 A non-uniform rod AB of length 12 m and mass 30 kg rests in a horizontal position on a support at its midpoint. When a particle of mass 24 kg is attached to the rod at A and a particle of mass 36 kg is attached at B the rod is in equilibrium Find the distance of the centre of mass of the rod from A A R 24g 36g 30g B 𝟔 𝒎 𝟔 𝒎 𝒅 Resolve Forces: 𝑹=𝟑𝟎𝒈+𝟐𝟒𝒈+𝟑𝟔𝒈=𝟗𝟎𝒈 Take moments at A: (𝟗𝟎𝒈)×𝟔=𝟑𝟎𝒈 ×𝒅+𝟑𝟔𝒈 ×𝟏𝟐 First mistake there are two unknowns – the distance from A to the weight 40g is unknown. This is the significant mistake as no marks would be awarded – he is solving the wrong problem. Second mistake – has missed out g from the 450 force. Although he has the correct answer for his working – this is not a significant mistake Solving gives 𝒅= 𝟓𝟒𝟎𝒈 −𝟒𝟑𝟐𝒈 𝟑𝟎𝒈 = 𝟏𝟖 𝟓 =𝟑.𝟔 𝒎 How do we know the 30g is on this side?

Resolve Forces: 𝑹=𝟔𝟎𝒈+𝟐𝟓𝒈+𝑿𝒈=𝟏𝟑𝟕𝟐 gives X = 55 N WB3 A non-uniform plank PQ has length 10 m and mass 25 kg. The plank is in horizontal, in equilibrium and smoothly supported at its midpoint, M. A man of mass 60 kg stands on the plank at a distance of 2 m from P and a woman stands at a distance of 1 m from Q. Given that the reaction at M has magnitude 1372 N, find the distance of the centre of mass of the plank from P P R 60g Xg 25g Q 𝟐 𝒎 𝟑 𝒎 𝟒 𝒎 𝟏 𝒎 𝒅 Resolve Forces: 𝑹=𝟔𝟎𝒈+𝟐𝟓𝒈+𝑿𝒈=𝟏𝟑𝟕𝟐 gives X = 55 N Take moments at P: 1𝟑𝟕𝟐×𝟓=𝟔𝟎𝒈 ×𝟐+𝟐𝟓𝒈 ×𝒅+ 𝟓𝟓𝒈 ×𝟗 First mistake there are two unknowns – the distance from A to the weight 40g is unknown. This is the significant mistake as no marks would be awarded – he is solving the wrong problem. Second mistake – has missed out g from the 450 force. Although he has the correct answer for his working – this is not a significant mistake Solving gives 𝒅= 𝟔𝟖𝟔𝟎 −𝟏𝟐𝟎𝒈−𝟒𝟗𝟓𝒈 𝟐𝟓𝒈 = 𝟏𝟕 𝟓 =𝟑.𝟒 𝒎

Resolve Forces: 2𝑅=𝟏𝟖𝟎𝒈 so 𝑹=𝟗𝟎𝒈 WB4 A non-uniform plank of wood AB has length 9 m and mass 80 kg. The plank is smoothly supported at its two ends A and B, with A and B at the same horizontal level. An object of mass 100 kg is put on the plank at the point C , where AC = 3 m, as shown above. The plank is in equilibrium and the magnitudes of the reactions on the plank at A and B are equal. The plank is modelled as a non-uniform rod and the object as a particle. Find: (a) the magnitude of the reaction R on the plank at B. (b) the distance d of the centre of mass of the plank from A. A 𝑅 100g 80g B 𝒅 𝒎 3 𝒎 First mistake there are two unknowns – the distance from A to the weight 40g is unknown. This is the significant mistake as no marks would be awarded – he is solving the wrong problem. Second mistake – has missed out g from the 450 force. Although he has the correct answer for his working – this is not a significant mistake Resolve Forces: 2𝑅=𝟏𝟖𝟎𝒈 so 𝑹=𝟗𝟎𝒈 Take moments at A: 𝑅×𝟗 =𝟏𝟎𝟎𝒈 ×𝟑+𝟖𝟎𝒈×𝒅 𝒅= 𝟖𝟏𝟎𝒈 −𝟑𝟎𝟎𝒈 𝟖𝟎𝒈 = 𝟓𝟏 𝟖 =𝟔.𝟑𝟕𝟓 𝒎

One thing to improve is – KUS objectives BAT understand moment and solve problems using moments on a Non- Uniform rod and/or when a rod is at a tipping point self-assess One thing learned is – One thing to improve is –

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