A system of forces is shown in FIGURE 7.1 - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2016 - Paper 1

For the vertical components, we consolidate the vertical forces:

• The component from the 2.1 kN force:

$2.1 ext{ kN} \times ext{sin}(40°) \approx 0.96 ext{ kN}$

• From the 3.1 kN force:

$3.1 ext{ kN} \times ext{sin}(50°) \approx 2.37 ext{ kN}$

The overall vertical resultant then is:

$R_V = 0.96 + 2.37 \approx 3.33 ext{ kN}$

Next, the magnitude of the equilibrium force can be determined using:

$E = \sqrt{R_H^2 + R_V^2}$

Substituting the values:

$E = \sqrt{(4.32)^2 + (3.33)^2} \approx 5.43 ext{ kN}$

The equilibrium angle can be computed with:

$\tan(\theta) = \frac{R_V}{R_H}$

Calculating:

$\theta = \tan^{-1}\left(\frac{3.33}{4.32}\right) \approx 39.35°$

To calculate the stress in the bolt material, we first determine the force acting on the bolt:

$F = m \cdot g = 600 ext{ kg} \cdot 10 ext{ m/s}^2 = 6000 ext{ N}$

Next, the cross-sectional area for an M16 bolt is:

$A = \frac{\pi d^2}{4} = \frac{\pi (0.016)^2}{4} \approx 2.01 \times 10^{-4} ext{ m}^2$

Stress is then given by:

$\text{Stress} = \frac{F}{A} = \frac{6000}{2.01 \times 10^{-4}} \approx 29.84 ext{ MPa}$

One Pascal (Pa) is defined as one Newton force (1 N) acting on an area of one square meter (1 m²). Thus, it is expressed as:

$1 ext{ Pa} = 1 \frac{N}{m^2}$