The three forces F₁, F₂ and F₃ are acting on a particle - AQA - A-Level Maths Mechanics - Question 13 - 2017 - Paper 2

To find the angle θ that F makes with the horizontal, we can use the tangent function:

$\tan(θ) = \frac{F_y}{F_x} = \frac{-21}{33}$

Calculating the angle: $θ = \tan^{-1} \left(\frac{-21}{33}\right) = -30.9°$

Since we need the acute angle, we take the absolute value: $θ = 30.9°$

So, the acute angle that F makes with the horizontal is approximately 30.9° (to the nearest 0.1°).

For the forces to be in equilibrium, the sum of all forces must equal zero:

$F₁ + F₂ + F₃ + F₄ = 0$

Thus, $F₄ = - (F₁ + F₂ + F₃)$

Substituting the values we calculated earlier: $F₄ = - (33i - 21j) = -33i + 21j$

Therefore, the fourth force F₄ is (-33i + 21j) N.