Three forces F₁, F₂ and F₃ acting on a particle P are given by F₁ = (7i - 9j) N F₂ = (5i + 6j) N F₃ = (pi + qj) N where p and q are constants - Edexcel - A-Level Maths Mechanics - Question 3 - 2012 - Paper 1

Now we need to determine the resultant R from forces F₂ and F₃. We first express R as:

$R = F₂ + F₃$

Substituting the values:

$R = (5i + 6j) + (pi + qj)$

$R = (5 + p)i + (6 + q)j$

Substituting the values of p and q:

$R = (5 - 12)i + (6 + 3)j$

$R = -7i + 9j$

To find the magnitude of R, we use:

$|R| = \\sqrt{(-7)^2 + 9^2} = \\sqrt{49 + 81} = \\sqrt{130} \\\approx 11.4 \\text{ N}$

To find the angle θ that R makes with the j-axis, we utilize the tangent function:

$\tan(θ) = \frac{\text{opposite}}{\text{adjacent}}$

Here, the opposite side is the component in the i direction (-7) and the adjacent is the component in the j direction (9):

$\tan(θ) = \frac{-7}{9}$

We calculate θ:

$θ = \tan^{-1}\left(-\frac{7}{9}\right)$

Using a calculator, we find:

$θ ≈ -39.2°$

To express the angle with respect to the positive j-axis, we find the angle with respect to j, which is:

$θ_{j} = 90° + (-39.2°) ≈ 50.8°$

Thus, the angle to the nearest degree is:

$θ ≈ 51°$