Four pulling forces of 1.2 kN, 2 kN, 3.4 kN and 1.8 kN are acting on the same point, as shown in FIGURE 7.1 below - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2017 - Paper 1

The total horizontal force ($H_C$) is calculated as follows:

$H_C = 3.4 - (1.2 \cos(40^{\circ}) + 1.8 \cos(70^{\circ}))$

• Calculating values:
  • $1.2 \cos(40^{\circ}) \approx 0.92$ kN
  • $1.8 \cos(70^{\circ}) \approx 0.62$ kN
• Therefore: $H_C = 3.4 - (0.92 + 0.62) = 3.4 - 1.54 = 1.86 ext{ kN}$

The total vertical force ($V_C$) is calculated as:

$V_C = 1.2 \sin(40^{\circ}) + 2 + 3.4 \sin(50^{\circ}) - 1.8 \sin(70^{\circ})$

• Calculating values:
  • $1.2 \sin(40^{\circ}) \approx 0.77$ kN
  • $3.4 \sin(50^{\circ}) \approx 2.29$ kN
  • $1.8 \sin(70^{\circ}) \approx 1.69$ kN
• Therefore: $V_C = 0.77 + 2 + 2.29 - 1.69 = 3.37 ext{ kN}$

The magnitude of the resultant force (R) is calculated using the Pythagorean theorem:

$R = \sqrt{H_C^2 + V_C^2} = \sqrt{(1.86)^2 + (3.37)^2} \approx 3.86 ext{ kN}$

To find the direction (θ):

$\tan(θ) = \frac{V_C}{H_C} = \frac{3.37}{1.86}$

Thus, $θ \approx \tan^{-1}(1.81) \approx 61^{\circ}$