Coplanar forces of magnitudes 30N, 15N, 33N and PN act at a point in the directions shown in the diagram, where tan α = 4/3. The system is in equilibrium. [Figure 3.1]

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. 3(a)
2. Resolving either direction.
3. [M1] Correct number of terms, allow sign errors, allow sin/cos mix. Do not allow with just sinα and cosα.
4. (33+15)×(3/5) = P cosθ + 30 cosθ
5. OR (33+15) cos(tan⁻¹(4/3)) = P cosθ + 30 cosθ
6. OR 19.8+9 = P cosθ + 30 cosθ
7. [A1] OE, but see note for final A1. Allow: 28.8 = (P+30)cosθ, (33+15)cos53(.1) = Pcosθ +30 cos θ, 19.81+9.01 = Pcosθ + 30 cosθ, 19.86+9.03 = P cosθ +30 cos θ.
8. 15×(4/5) + 30 sinθ = 33×(4/5) + P sinθ
9. OR 15sin(tan⁻¹(4/3)) + 30sinθ = 33sin(tan⁻¹(4/3)) + Psinθ
10. OR 12+30 sinθ = 26.4 + P sinθ
11. [A1] OE, but see note for final A1. Allow: 14.4 = (30-P) sinθ, 15sin 53(.1) +30sin θ = 33 sin 53(.1) + Psin θ, 12.00 +30 sin θ = 26.39 + P sin θ, 11.98+30 sin θ = 26.35 + Psin θ.
12. [Use cos²θ + sin²θ = 1 with] cosθ = 28.8/(P+30) and sinθ = 14.4/(30-P) to get (14.4)²/(30-P)² + (28.8)²/(P+30)² = 1
13. [A1] AG. Must have evidence of where 28.8 and 14.4 come from. A0 for any error seen. A0 if use of inexact angles seen. Any inexact decimals seen for force components, i.e. if 14.4 and/or 28.8 have come from rounding to 3sf, scores M1A1A1A0 max 3/4. If exact values of sinα and cosα not shown (e.g. 28.8 = (P+30)cosθ or 14.4 = (30-P)sin θ from no working), this scores M1A1A1A0 max 3/4 marks.
14. 3(b)
15. Sub P=6 into (14.4)²/(30-P)² + (28.8)²/(P+30)² = 1 to get (14.4/24)² + (28.8/36)² = (3/5)²+(4/5)² = 0.36+0.64=1
16. [B1] Must see either (3/5)² + (4/5)² = 1 or 0.36+0.64 = 1 as minimum working.
17. θ = 36.9
18. [B1] AWRT 36.9.