7.1 Define the following terms: 7.1.1 A force 7.1.2 Forces in equilibrium 7.1.3 Resultant of a system of forces 7.2 A load of 40 kN causes a tensile stress of 20 MPa in a round brass bar - NSC Mechanical Technology Automotive - Question 7 - 2017 - Paper 1

Forces in equilibrium refer to a situation where the net force acting on an object is zero, meaning all forces are balanced and there is no change in motion.

The resultant of a system of forces is the single force that produces the same effect as all the individual forces acting together.

To find the diameter of the bar, we start with the relation of stress and area:

$ext{Stress} = \frac{F}{A}$

Rearranging gives:

$A = \frac{F}{\text{Stress}}$

The force, F = 40 kN = 40,000 N and stress = 20 MPa = 20 x 10^6 N/m². Substituting values:

$A = \frac{40,000}{20 \times 10^6} = 0.002$ m².

Thus, the diameter can be calculated using:

$D = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4 \times 0.002}{\pi}} \approx 0.05045$ m or 50.45 mm.

Strain is defined as the change in length per unit length. It can be calculated using:

$E = \frac{\text{Stress}}{E_{Young}}$

With a stress of 20 x 10^6 N/m² and Young's Modulus for brass = 90 GPa = 90 x 10^9 N/m², we have:

$E = \frac{20 \times 10^6}{90 \times 10^9} = 2.222 \times 10^{-4}$.

The change in length can be calculated using:

$\Delta L = \text{Strain} \times \text{Original Length}$

Substituting in: $$\Delta L = (2.222 \times 10^{-4}) \times 800 \text{mm} = 0.17776 \text{mm} \approx 0.178 ext{mm}.$

To solve this, we first find the components of the forces:

• X-component = 280 N + 300 \cos(30°) - 400 \cos(120°)
• Y-component = 300 \sin(30°) + 400 \sin(120°) - 170 N.

Calculating:

• X-component = 280 + 300 \cdot 0.866 - 400 \cdot (-0.5) = 133.93 N
• Y-component = 300 \cdot 0.5 + 400 \cdot 0.866 - 170 = 264.95 N.

The resultant R is given by:

$R = \sqrt{(X^2 + Y^2)} = \sqrt{(133.93^2 + 264.95^2)} \approx 298.10 N$.

The direction can be found using:

$$\tan(\theta) = \frac{Y}{X} \implies \theta = \tan^{-1}\left(\frac{264.95}{133.93}\right) \approx 63.60°.$

For the reactions at the supports, we can use the equilibrium equations. Let R_A and R_B be the reactions at A and B:

• Sum of vertical forces = 0:

$R_A + R_B = 800 + 350 + 80 \cdot 6.2$

Calculating the total load:

• The load due to the uniform distribution = 80 N/m * 6.2 m = 496 N.
• Thus:

$R_A + R_B = 800 + 350 + 496 = 1646 N.$

Taking moments about point A:

$R_B \cdot 6.2 - (350 \cdot 1.7) - (496 \cdot 3.1) = 0$

Solving gives: $R_B = 693.96 N$.

Now substituting back: $$R_A = 1646 - 693.96 = 952.03 N.$