7.1 Four forces of 100 N, 200 N, 300 N and 400 N respectively are acting on the same point, as shown in FIGURE 7.1 below. Determine, by means of calculations, the magnitude and direction of the resultant force for the system of forces in FIGURE 7.1. 7.2 A load of 40 kN is exerted onto a solid brass bar with a diameter of 56 mm. The original length of the bar is 0,85 m. Young's modulus for brass is 90 GPa. Calculate the: 7.2.1 Stress in the bar 7.2.2 Strain 7.2.3 Change in length 7.3 FIGURE 7.3 below shows a uniform beam that is supported by two vertical supports, A and B. A uniformly distributed force of 80 N/m is exerted over the whole length of the beam. Determine, by means of calculations, the magnitudes of the reactions in supports A and B.

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7.1 Four forces of 100 N, 200 N, 300 N and 400 N respectively are acting on the same point, as shown in FIGURE 7.1 below - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2017 - Paper 1

Question 7

7.1-Four-forces-of-100-N,-200-N,-300-N-and-400-N-respectively-are-acting-on-the-same-point,-as-shown-in-FIGURE-7.1-below-NSC Mechanical Technology Welding and Metalwork-Question 7-2017-Paper 1.png

7.1 Four forces of 100 N, 200 N, 300 N and 400 N respectively are acting on the same point, as shown in FIGURE 7.1 below. Determine, by means of calculations, the ma... show full transcript

Worked Solution & Example Answer:7.1 Four forces of 100 N, 200 N, 300 N and 400 N respectively are acting on the same point, as shown in FIGURE 7.1 below - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2017 - Paper 1

Step 1

7.1 Determine the Resultant Force

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Answer

To find the resultant force, we need to calculate the horizontal and vertical components of the forces:

Horizontal Components:
- For 200 N:
  - $200 \cos(30^\circ) = 173.21 \text{ N}$
- For 300 N:
  - $300 \cos(50^\circ) = 192.84 \text{ N}$
- For 400 N:
  - $400 \cos(0^\circ) = 400 \text{ N}$
- For 100 N:
  - 100 N is vertical (no horizontal component).
Summing up:
- $\Sigma H = -200 + 173.21 + 192.84 + 400 = 165.05 \text{ N}$
Vertical Components:
- For 200 N:
  - $200 \sin(30^\circ) = 100 \text{ N}$
- For 300 N:
  - $300 \sin(50^\circ) = 229.81 \text{ N}$
- For 400 N:
  - 0 N (horizontal force).
- For 100 N:
  - 100 N.
Summing up:
- $\Sigma V = 100 + 229.81 - 100 = 229.81 \text{ N}$
Resultant Magnitude and Angle:
- Magnitude: $R = \sqrt{(\Sigma H)^2 + (\Sigma V)^2} = \sqrt{(165.05)^2 + (229.81)^2} = 278.71\text{ N}$
- Angle: $\theta = \tan^{-1}\left( \frac{\Sigma V}{\Sigma H} \right) = \tan^{-1}\left( \frac{229.81}{165.05} \right) = 287.71^\circ \text{ from east}$

Step 2

7.2.1 Stress in the bar

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Answer

To calculate stress, we use the formula: $\sigma = \frac{F}{A}$ where:

$F = 40 \text{ kN} = 40 \times 10^3 \text{ N}$
$A = \frac{\pi \times d^2}{4} = \frac{\pi \times (0.056)^2}{4} = 2.46 \times 10^{-3} \text{ m}^2$ .

Thus, the stress is: $\sigma = \frac{40 \times 10^3}{2.46 \times 10^{-3}} = 16.26 \times 10^6 \text{ Pa}$

Step 3

7.2.2 Strain

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Answer

Strain ( $\epsilon$ ) is calculated as: $\epsilon = \frac{\sigma}{E}$ where:

$\sigma = 16.26 \times 10^6 \text{ Pa}$
Young's modulus, $E = 90 \text{ GPa} = 90 \times 10^9 \text{ Pa}$ .

Therefore, strain is: $\epsilon = \frac{16.26 \times 10^6}{90 \times 10^9} = 0.18 \times 10^{-3}$

Step 4

7.2.3 Change in length

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Answer

The change in length ( $\Delta L$ ) can be calculated using: $\Delta L = \epsilon \times L_0$ where:

$L_0 = 0.85 \text{ m}$ .

Thus, the change in length is: $\Delta L = (0.18 \times 10^{-3}) \times 0.85 = 0.15 \text{ mm}$

Step 5

7.3 Calculate Reactions in Supports A and B

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Answer

To find the reactions at supports A and B:

Moments about B:
- Setting the sum of moments about B to zero: $\Sigma M_B = (A \times 12) - (960 \times 6) - (750 \times 6) = 0$
- Solve for A: $A = \frac{5760 + 6000}{12} = 980 \text{ N}$
Moments about A:
- Setting the sum of moments about A to zero: $\Sigma M_A = (B \times 12) - (750 \times 4) - (960 \times 6) - (300 \times 12) = 0$
- Solve for B: $B = \frac{750 \times 4 + 960 \times 6 + 300 \times 12}{12} = 1030 \text{ N}$

7.1 Four forces of 100 N, 200 N, 300 N and 400 N respectively are acting on the same point, as shown in FIGURE 7.1 below - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2017 - Paper 1

Worked Solution & Example Answer:7.1 Four forces of 100 N, 200 N, 300 N and 400 N respectively are acting on the same point, as shown in FIGURE 7.1 below - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2017 - Paper 1

7.1 Determine the Resultant Force

7.2.1 Stress in the bar

7.2.2 Strain

7.2.3 Change in length

7.3 Calculate Reactions in Supports A and B

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