FIGURE 8.1 below shows a system of four forces acting onto the same point - NSC Mechanical Technology Fitting and Machining - Question 8 - 2021 - Paper 1

The horizontal forces are calculated as follows:

$\Sigma HC = 120\cos(0) - 185\cos(45)$

Calculating each component gives:

• For the first force: $120\cos(0) = 120$
• For the second force: $185\cos(45) \approx 130.82$

Adding these gives:

$\Sigma HC = 120 - 130.82 \approx -10.82 \text{ N}$

To find the magnitude of the resultant force, use:

$R = \sqrt{(\Sigma VC)^2 + (\Sigma HC)^2} = \sqrt{(50.82)^2 + (-10.82)^2} \approx 52.58 \text{ N}$

For the angle of the resultant, use:

$\theta = \tan^{-1}\left(\frac{\Sigma HC}{\Sigma VC}\right) = \tan^{-1}\left(\frac{-10.82}{50.82}\right) \approx -12.06^\circ$

The point load representing the uniformly distributed load (UDL) can be calculated using the formula:

$\text{Point Load} = \text{UDL} \times \text{Length} = 16 \text{ kN/m} \times 5 \text{ m} = 80 \text{ kN}$

To calculate the magnitude of the reaction at support A, we take moments about support B:

$A \times 10.25 = 9.5 \times 0 + 80 \times 9.5 + 7.75 \times 5\n\Rightarrow A = \frac{23.625 + 0 + 97.375}{10.25} \approx 7.41 \text{ kN}$

To find the reaction at support B, we can again take moments about support A:

$B \times 10.25 = 9.5 \times 0 + 80 \times 2.5 + 4.5 \times 5\n\Rightarrow B = \frac{222.5}{10.25} \approx 21.71 \text{ kN}$