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Four pulling forces of 1.2 kN, 2 kN, 3.4 kN and 1.8 kN are acting on the same point, as shown in FIGURE 7.1 below - NSC Mechanical Technology Automotive - Question 7 - 2017 - Paper 1

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Four-pulling-forces-of-1.2-kN,-2-kN,-3.4-kN-and-1.8-kN-are-acting-on-the-same-point,-as-shown-in-FIGURE-7.1-below-NSC Mechanical Technology Automotive-Question 7-2017-Paper 1.png

Four pulling forces of 1.2 kN, 2 kN, 3.4 kN and 1.8 kN are acting on the same point, as shown in FIGURE 7.1 below. Determine, by means of calculations, the magnitude... show full transcript

Worked Solution & Example Answer:Four pulling forces of 1.2 kN, 2 kN, 3.4 kN and 1.8 kN are acting on the same point, as shown in FIGURE 7.1 below - NSC Mechanical Technology Automotive - Question 7 - 2017 - Paper 1

Step 1

Calculate Horizontal and Vertical Components

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Answer

To find the resultant force, we need to determine the horizontal (HC) and vertical (VC) components of each force:

For 1.2kN1.2 \, kN at 4040^\circ:

  • HC = 1.2cos(40)1.2 \cdot \cos(40^\circ)
  • VC = 1.2sin(40)1.2 \cdot \sin(40^\circ)

For 2kN2 \, kN at 5050^\circ:

  • HC = 2cos(50)2 \cdot \cos(50^\circ)
  • VC = 2sin(50)2 \cdot \sin(50^\circ)

For 3.4kN3.4 \, kN at 00^\circ:

  • HC = 3.4cos(0)3.4 \cdot \cos(0^\circ)
  • VC = 3.4sin(0)3.4 \cdot \sin(0^\circ)

For 1.8kN1.8 \, kN at 7070^\circ:

  • HC = 1.8cos(70)1.8 \cdot \cos(70^\circ)
  • VC = 1.8sin(70)1.8 \cdot \sin(70^\circ)

Calculating these gives:

  • HC = 1.20.766020.6428+3.41.80.3420=4.39kN1.2 \cdot 0.7660 - 2 \cdot 0.6428 + 3.4 - 1.8 \cdot 0.3420 = 4.39 \, kN
  • VC = 1.20.6428+20.7660+01.80.9397=0.61kN1.2 \cdot 0.6428 + 2 \cdot 0.7660 + 0 - 1.8 \cdot 0.9397 = 0.61 \, kN

Step 2

Determine the Resultant

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Answer

The magnitude of the resultant force (R) can be calculated using the Pythagorean theorem:

R=HC2+VC2R = \sqrt{HC^2 + VC^2}
Substituting the values:
R=(4.39)2+(0.61)2=4.43kNR = \sqrt{(4.39)^2 + (0.61)^2} = 4.43 \, kN

To find the direction (angle θ\theta) with respect to the horizontal:

tan(θ)=VCHC\tan(\theta) = \frac{VC}{HC}
Substituting gives: θ=tan1(0.614.39)=7.91\theta = \tan^{-1}\left(\frac{0.61}{4.39}\right) = 7.91^\circ
Thus, the resultant direction is 7.917.91^\circ north of east.

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