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∘
136 exam-style questions on CCEA A Level Maths 2.3 Forces and Newton's laws, covering 2.3.1 Forces and Newton's laws, 2.3.2 Forces and Newton's laws, 2.3.3 Forces and Newton's laws, 2.3.4 Forces and Newton's laws, 2.3.5 Forces and Newton's laws, 2.3.6 Forces and Newton's laws, 2.3.7 Forces and Newton's laws, 2.3.8 Forces and Newton's laws, 2.3.9 Forces and Newton's laws, 2.3.10 Forces and Newton's laws, 2.3.11 Forces and Newton's laws, 2.3.12 Forces and Newton's laws, and 2.3 Forces and Newton's laws. Each one has a worked solution and a mark scheme showing where the marks go.