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∘
139 exam-style questions on AQA A Level Maths 3.3 R: Forces and Newton's laws, covering 3.3.1 Concept of a force and Newton's first law, 3.3.2 Newton's second law, 3.3.3 Weight and motion under gravity, 3.3.4 Newton's third law and equilibrium, 3.3.5 Addition of forces and resultants (A-level only), and 3.3.6 Friction (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.