An anchoring force P\mathbf{P}P of magnitude 120 N is applied to a structural joint. The force acts in the third quadrant relative to the standard unit vectors i\mathbf{i}i (pointing right) and j\mathbf{j}j (pointing upwards). The direction of the force makes an angle of 25∘ 25^\circ\,25∘ with the negative yyy-axis.
The force can be expressed as a vector [PxPy] N\begin{bmatrix} P_x \\ P_y \end{bmatrix} \text{ N}[PxPy] N.
Find the correct expression for PyP_yPy.
Py=120cos25∘P_y = 120 \cos 25^\circPy=120cos25∘
Py=−120sin25∘P_y = -120 \sin 25^\circPy=−120sin25∘
Py=−120cos25∘P_y = -120 \cos 25^\circPy=−120cos25∘
Py=120sin25∘P_y = 120 \sin 25^\circPy=120sin25∘
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.