Four students are pulling on ropes in a four-person tug of war. The relative sizes of the forces acting on the various ropes are \(F_{ W }=200 \text{ N} , F_{ X }=240 \text{ N} , F_{ Y }=180 \text{ N}\) and \(F_{ Z }=210 \text{ N}\). The situation is shown in the diagram below.
What is the resultant force vector \((F_{\vec{R}})\) acting at the centre of the tug-of-war ropes? (3 marks)
→ By resolving the vertical vectors of \(F_{ W }=200 \text{ N}\) up and \(F_{ Y }=180 \text{ N}\) down, the net force in the vertical direction is \(F_v=20\ \text{N}\) up.
→ Similarly, the net force in the horizontal direction is \(F_h=30\ \text{N}\) to the right.
→ Thus the resultant vector \((F_{\vec{R}})\) can be calculated using the vector diagram below.
→ The magnitude of \(F_{\vec{R}}=\sqrt{20^2+30^2}\) and \(\theta=\tan^{-1}\Big{(}\dfrac{20}{30}\Big{)}\)
→ Hence \(F_{\vec{R}}=36.1\ \text{N}, 33.7^{\circ}\) above the horizontal.