June 2024 Paper 3 Q9
9
Two horizontal forces of magnitudes 17 N and 12 N act at a point \(O\) along bearings of 050° and 340° respectively (see diagram).
A third horizontal force \(\mathbf{F}\) is now applied at \(O\). The three forces are in equilibrium.
| Scheme | Marks | AO |
|---|---|---|
| Resolve vertically or horizontally | M1* | 2.1 |
| \(R(\uparrow): \pm(12\cos 20 + 17\cos 50)\) | A1 | 1.1 |
| \(R(\rightarrow): \pm(17\sin 50 - 12\sin 20)\) | A1 | 1.1 |
| Magnitude is \(\sqrt{(12\cos 20 + 17\cos 50)^2 + (17\sin 50 - 12\sin 20)^2}\) | M1dep* | 3.4 |
| \(\tan\theta = \dfrac{17\sin 50 - 12\sin 20}{12\cos 20 + 17\cos 50}\) | M1dep* | 3.4 |
| Resultant force is 23.9 (N) on a bearing of 022(°) | A1 | 2.2a |
| [6] |
Notes
M1*: Resolve vertically or horizontally – correct number of relevant terms. Allow sign errors and sin/cos mix but must be using correct angles
A1: Need not be simplified
\(\pm 22.20370081\ldots\)
A1: Need not be simplified
\(\pm 8.918513813\ldots\)
M1dep*: Correct method to calculate magnitude from expressions with the correct number of relevant terms
M1dep*: Correct method to calculate a relevant angle (so allow reciprocal) from expressions with the correct number of relevant terms
A1: awrt 23.9 and awrt 022 (so 021.9 is okay but 21.9 is A0)
23.927896…
21.883744…
Alternative method
| Scheme | Marks |
|---|---|
| \(R^2 = 17^2 + 12^2 - 2(17)(12)\cos\alpha\) | M1* |
| \(R^2 = 17^2 + 12^2 - 2(17)(12)\cos 110\) | B1 A1 |
| \(\dfrac{\sin\theta}{12} = \dfrac{\sin\alpha}{R}\) or \(\cos\theta = \dfrac{R^2 + 17^2 - 12^2}{2 \times R \times 17}\) | M1dep* |
| Bearing \(= 50 - \theta\) | M1dep* |
| Resultant force is 23.9 (N) on a bearing of 022(°) | A1 |
M1*: Correct cosine rule with any angle \(\alpha\)
B1 A1: B1 for correct angle of 110 seen, A1 for correct expression for \(R\) or \(R^2\)
110 may be seen on a diagram
M1dep*: Correct sine rule with their \(\alpha\) or correct cosine rule
If correct \(\theta = 28.11625\ldots\)
M1dep*: Correct calculation for bearing
Dep. both M marks
A1: awrt 23.9 and awrt 022 (so 021.9 is okay but 21.9 is A0)
| Scheme | Marks | AO |
|---|---|---|
| Magnitude of \(\mathbf{F}\) is 23.9 (N) | B1FT | 1.2 |
| On a bearing of 202(°) | B1FT | 3.1b |
| [2] |
Notes
B1FT: Follow through their resultant in (a)
B1FT: Follow through the answer to (a) – if \(0 \lt \theta \lt 180\) then FT is \(180 + \theta\) and if \(180 \lt \theta \lt 360\) then FT is \(\theta - 180\)
As in part (a) their bearing must be at least three figures (with leading zeros if necessary) but allow greater accuracy