A number of forces act on a particle such that the resultant force is \( \begin{pmatrix} 6 \\ -3 \end{pmatrix} \) N - AQA - A-Level Maths Mechanics - Question 11 - 2020 - Paper 2

To find the total of the other forces acting on the particle, we can use the equation for the resultant force:

$\text{Resultant force} = \text{Force 1} + \text{Force 2}$

Where:

• Resultant force is ( \begin{pmatrix} 6 \ -3 \end{pmatrix} ) N
• Force 1 is ( \begin{pmatrix} 8 \ -5 \end{pmatrix} ) N

Let ( \text{Force 2} ) be the total of the other forces.

Therefore: $\begin{pmatrix} 6 \\ -3 \end{pmatrix} = \begin{pmatrix} 8 \\ -5 \end{pmatrix} + \text{Force 2}$

To isolate ( \text{Force 2} ), we rearrange the equation: $\text{Force 2} = \begin{pmatrix} 6 \\ -3 \end{pmatrix} - \begin{pmatrix} 8 \\ -5 \end{pmatrix}$

Calculating this gives: $\text{Force 2} = \begin{pmatrix} 6 - 8 \\ -3 + 5 \end{pmatrix} = \begin{pmatrix} -2 \\ 2 \end{pmatrix} \text{ N}$

Thus, the answer is ( \begin{pmatrix} -2 \ 2 \end{pmatrix} ) N.