A-LevelMathematicsMechanics9709/41 · Oct/Nov 2019 · Q58 Marks
A small ring P is threaded on a fixed smooth horizontal rod AB. Three horizontal forces of magnitudes 4.5 N, 7.5 N and F N act on P [Figure X.X].
- (i)Given that these three forces are in equilibrium, find the values of F and θ.[6]
- (ii)It is given instead that the values of F and θ are 9.5 and 30 respectively, and the acceleration of the ring is 1.5 m s⁻². Find the mass of the ring.[2]
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Mark scheme
Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.
- [M1] Resolve forces either horizontally or vertically
- [A1] 7.5cos60 + 4.5cos20 = Fcosθ [= 7.97861]
- [A1] 7.5sin60 - 4.5sin20 = Fsinθ [= 4.95609]
- [M1] Use Pythagoras or use the value found for θ to find F (F = √(7.98² + 4.96²))
- [M1] Use trigonometry or the value found for F to find θ (θ = tan⁻¹(4.96/7.98))
- [A1] F = 9.39 and θ = 31.8
- [M1] Apply Newton's second law to the ring along AB (4 terms) (9.5cos30 - 7.5cos60 - 4.5cos20 = m × 1.5)
- [A1] m = 0.166 kg
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