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∘
87 exam-style questions on OCR (MEI) A Level Maths 3.4 Forces, covering 3.4.1 Language relating to forces, 3.4.2 Acceleration due to gravity, 3.4.3 Identify forces and draw force diagrams, 3.4.4 Resultant of concurrent forces, 3.4.5 Equilibrium of a particle, 3.4.6 Resolve a force into components (A-level only), 3.4.7 Equilibrium condition for resultant zero (A-level only), 3.4.8 Forces in equilibrium sum to zero (A-level only), 3.4.9 Equations for a particle in equilibrium (A-level only), 3.4.10 Frictional force and normal contact force (A-level only), 3.4.11 Model friction with F ≤ μR (A-level only), and 3.4.12 Apply Newton's Laws to friction problems (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.