Particles – Resultant Force in 2D)
Solution A: Trigonometry Method
i.
ii.
Solution B: Rectangular Method
i.
ii.
Answer
𝐹
𝐹1
𝐹2
𝐹3
Σ𝐹
𝑥 component
400 cos 60
4
−100 ( )
5
−𝐹3 sin 30
0
𝑦 component
400 sin 60
3
100 ( )
5
−𝐹3 cos 30
𝑅
Answer
[2 marks]
For 𝐹2 to be minimum, 𝜃 = 90° [1 mark]
𝐹𝑥
400 sin 30
𝐹2 cos 90
0.5𝐹2 cos 40
𝑅𝑥
[4 marks]
Forces
𝐹1
𝐹2
𝐹3
𝑅
𝐹𝑦
400 cos 30
−𝐹2 sin 90
0.5𝐹2 sin 40
0
Σ𝐹𝑦 = 𝑅𝑦 = 0
400 cos 30 − 𝐹2 sin 90 + 0.5𝐹2 sin 40 = 0 [1 mark]
0.6786𝐹2 = 400 cos 30
𝐹2 = 510.4777 𝑁 [1 mark]
𝐹3 = 255.2389 𝑁 [1 mark]
Σ𝐹𝑥 = 𝑅𝑥
400 sin 30 + 𝐹2 cos 90 + 0.5𝐹2 cos 40 = 𝑅𝑥 [1 mark]
𝑅𝑥 = 400 sin 30 + 510.4777 cos 90 + 255.2389 cos 40
𝑅𝑥 = 𝑅 = 395.5243 𝑁 [1 mark]
Particles – Equilibrium in 2D
Lecture 4- Rigid Bodies- Moment 2D
Q1
Q2