FIGURE 8.1 below indicates a system of forces with three coplanar forces acting on the same point - NSC Mechanical Technology Fitting and Machining - Question 8 - 2020 - Paper 1

To find the distance X for the beam to be in equilibrium, take moments about point O:

1. Set Up Moment Equation:

   • Clockwise moments about O ($\Sigma RHM$): For the 500 N load acting at distance X: $500 \cdot X$
   • Anticlockwise moments ($\Sigma LHM$): For the 3000 N load, the moment due to this load is $3000 \cdot 1.5$

2. Balance the Moments:

   • Setting these moments equal: $500 \cdot X = 3000 \cdot 1.5$
   • Calculating gives: $500X = 4500$
   • Therefore: $X = \frac{4500}{500} = 9 \text{ m}$

The type of stress induced in the material is compressive stress.

Using the formula for stress ($\sigma = \frac{F}{A}$):

• The area $A = L \times B = 0.03 \text{ m} \times 0.016 \text{ m} = 0.00048 \text{ m}^2$.
• The force $F = 50 \times 10^3 \text{ N}$: $\sigma = \frac{50 \times 10^3}{0.00048} = 104,166.67 \text{ Pa} = 104.17 \text{ MPa}$.

Using the formula for change in length ($\Delta L = \frac{\sigma \cdot L}{E}$):

• Where: $\Delta L = \frac{104.17 \times 10^6 \cdot 0.08}{90 \times 10^9} \approx 0.0000933 \text{ m} = 0.0933 \text{ mm}$.

Using the formula for safe working stress: $\text{Safe Working Stress} = \frac{\text{Break Stress}}{\text{Safety Factor}} = \frac{600 \text{ MPa}}{4} = 150 \text{ MPa}.$