A force F\mathbf{F}F of magnitude 42 N42 \text{ N}42 N acts on a charged particle. The direction of the force relative to unit vectors i\mathbf{i}i (representing the positive horizontal direction) and j\mathbf{j}j (representing the positive vertical direction) is such that the force vector points into the second quadrant. The angle between the force vector and the positive j\mathbf{j}j direction is 62∘62^\circ62∘.
The force can be expressed as a vector [F1F2] N\begin{bmatrix} F_1 \\ F_2 \end{bmatrix} \text{ N}[F1F2] N.
Find the correct expression for F1F_1F1.
F1=42sin62∘F_1 = 42 \sin 62^\circF1=42sin62∘
F1=42cos62∘F_1 = 42 \cos 62^\circF1=42cos62∘
F1=−42sin62∘F_1 = -42 \sin 62^\circF1=−42sin62∘
F1=−42cos62∘F_1 = -42 \cos 62^\circF1=−42cos62∘
12 exam-style questions on Edexcel A Level Maths 5.1 Resolving Forces. Each one has a worked solution and a mark scheme showing where the marks go.