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∘
106 exam-style questions on WJEC A Level Maths 4.9 Forces and Newton's laws (A-level only), covering 4.9.1 Forces and Newton's laws (A-level only), 4.9.2 Forces and Newton's laws (A-level only), 4.9.3 Forces and Newton's laws (A-level only), 4.9.4 Forces and Newton's laws (A-level only), and 4.9.5 Forces and Newton's laws (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.