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

Next, we calculate the vertical components:

• For the 120 N force: 120 N * sin(0°) = 0 N
• For the 280 N force at 45°: 280 N * sin(45°) = 197.99 N
• For the 250 N force at 210°: 250 N * sin(210°) = 250 * (-1/2) = -125 N
• For the 150 N force at 165°: 150 N * sin(165°) = 150 * sin(15°) ≈ 38.57 N

Summing these results gives: 0 + 197.99 - 125 + 38.57 ≈ 111.56 N.

The magnitude of the equilibrium force (E) can be found using the formula:

$E = ext{sqrt}(VC^2 + HC^2)$

Substituting the calculated values:

E ≈ 127.39 ext{ N}.$$

To find the direction (θ) of the equilibrium force, we use:

$tan(θ) = \frac{VC}{HC}$

Thus:

$θ = tan^{-1} \left( \frac{111.56}{101.48} \right) ≈ 37.19°.$ The direction of the equilibrium force is approximately 37.19°.

To calculate the reactions at supports A and B, we will take moments about each support. For support A:

$\sum R_{OM} = \sum R_{LM}$

Substituting in the values:

$A \cdot 6 = (285 \times 2.7) + (165 \times 4.2) + (345 \times 5.4)$

Calculating:

$A \cdot 6 = 769.5 + 693 + 1863 \Rightarrow A = 3325.5/6 ≈ 554.25 N.$

Repeat for support B:

$\sum R_{OM} = \sum R_{LM}$

Finally, from equilibrium: B can be calculated as:

$B = 1444.75/6 ≈ 240.75 N.$

The resistance area (A) of the brass bush can be calculated using:

$A = \frac{\pi(D^{2}-d^{2})}{4}$ Where D = 58 mm and d = 42 mm.

Calculating:

$A \approx 1.26 \times 10^{-3} m^{2}.$

To calculate the stress (σ) in the material:

$σ = \frac{F}{A}$ Where F = 50 kN and A is from the previous calculation.

$σ = \frac{50 \times 10^{3}}{1.26 \times 10^{-3}} = 39.68 MPa.$

Strain (ε) can be calculated by:

$ε = \frac{L - L_0}{L_0}$ where L = 68,975 mm and L_0 = 68 mm.

Calculating:

$ε = \frac{68.975 - 68}{68} ≈ 0.01434.$

Young's modulus (E) is calculated using:

$E = \frac{σ}{ε}$ Substituting in the stress and strain values:

$E = \frac{39.68 \times 10^{6}}{0.01434} ≈ 2.77 \times 10^{10}\text{ Pa} (or 27.7 GPa).$