Three coplanar forces of magnitudes 10N, 25 N and 20 N act at a point O in the directions shown in the diagram.
- (a)Given that the component of the resultant force in the x-direction is zero, find α, and hence find the magnitude of the resultant force.[4]
- (b)Given instead that α = 45, find the magnitude and direction of the resultant of the three forces.[5]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.
- [M1] 20 cos 30=25 cos 60+10cos α For resolving forces horizontally, all relevant terms included
- [A1] [17.32=12.5+10cosα, → cosα = 0.4821] From α ≈ 61.18
- [M1] Resultant = 20 sin 30+10sin 61.2-25 sin 60 For resolving forces vertically, all relevant terms included
- [A1] Magnitude of resultant force = 2.89 N A0 for -2.89 N or for ±2.89 Ν. Allow 2.89 N downwards
- [M1] X = 25 cos 60+10cos 45-20cos 30 For either horizontal or vertical component, correct number of relevant terms. Allow ±X and/or ±Y
- [A1] Y = 20sin 30+10sin 45-25 sin 60 For both correct, allow unsimplified
- [M1] R = √X²+Y² OE. Using a method to find the resultant force, using expressions for X and Y with at least 5 relevant terms.
- [M1] α = tan⁻¹ Y/X OE. A method to find the direction, using expressions for X and Y with at least 5 relevant terms.
- [A1] Resultant = 5.10 N, Direction = 63.8° below positive x-axis For both correct, angle clearly explained. May use a diagram with a correct arrow and arc for angle. Allow angle 296° (measured anticlockwise from +ve x-axis)
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