A-Level Mathematics May/June 2024 Q2(b): (b) Given instead that F = 10, find the magnitude and direction of the resultant force.

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  accepted ✓

  Let the positive x-direction be horizontal and the positive y-direction be vertical.

  First, resolve the forces into horizontal (X) and vertical (Y) components.

  Horizontal component: $X = 20 \cos(60^\circ)$ $X = 20 \times 0.5$ $X = 10$ N

  Vertical component: The 20 N force has an upward component, and the force F = 10 N acts downwards. $Y = 20 \sin(60^\circ) - 10$ $Y = 20 \times \frac{\sqrt{3}}{2} - 10$ $Y = 10\sqrt{3} - 10$ $Y \approx 17.3205... - 10 = 7.3205...$ N

  Now, find the magnitude of the resultant force, R, using Pythagoras' theorem. $R = \sqrt{X^2 + Y^2}$ $R = \sqrt{10^2 + (7.3205...)^2}$ $R = \sqrt{100 + 53.590...}$ $R = \sqrt{153.590...}$ $R = 12.393...$ N

  Magnitude $R = 12.4$ N (to 3 s.f.)

  Next, find the direction of the resultant force, $\theta$, using trigonometry. The angle is measured from the positive horizontal axis. $\tan(\theta) = \frac{Y}{X}$ $\tan(\theta) = \frac{7.3205...}{10}$ $\theta = \tan^{-1}(0.73205...)$ $\theta = 36.206...^\circ$

  Direction $\theta = 36.2^\circ$ above the horizontal (to 1 d.p.).

  Final Answer: The magnitude of the resultant force is $12.4$ N and its direction is $36.2^\circ$ above the positive horizontal axis.

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  How the marks are awarded

  • M1 — For attempting to resolve forces in perpendicular directions. The expressions $20 \cos(60^\circ)$ and $20 \sin(60^\circ) - 10$ show a correct method for finding horizontal and vertical components.
  • A1 — For both correct component values. The horizontal component $X=10$ and the vertical component $Y=10\sqrt{3}-10$ (or $7.32...$) are both calculated correctly.
  • DM1 — For using Pythagoras' theorem on the calculated components to find the magnitude of the resultant: $R = \sqrt{10^2 + (7.3205...)^2}$. This mark is dependent on the first M1 mark.
  • DM1 — For using trigonometry to find the direction of the resultant from the components: $\theta = \tan^{-1}(\frac{7.3205...}{10})$. This mark is also dependent on the first M1 mark.
  • A1 — For the final correct magnitude of $12.4$ N and a clear, correct direction of $36.2^\circ$ above the positive horizontal axis. Both values must be correct and appropriately rounded.

  Common mistakes

  • Mixing up sine and cosine when resolving the 20 N force, leading to incorrect horizontal and vertical components.
  • Incorrectly adding the downward force F=10 N instead of subtracting it, resulting in a vertical component of $20\sin(60^\circ) + 10$.
  • Failing to provide a clear description of the direction, for example just writing '36.2°' without specifying that it is 'above the horizontal' or 'anti-clockwise from the positive x-axis'.
  • Prematurely rounding component values, which can lead to an inaccurate final answer for the magnitude or direction.

  Examiner tip: Always resolve forces into two perpendicular components (usually horizontal and vertical) before combining them to find a resultant.

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