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∘
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.