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FIGURE 8.1 below shows a system of four forces acting on the same point - NSC Mechanical Technology Fitting and Machining - Question 8 - 2022 - Paper 1

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FIGURE 8.1 below shows a system of four forces acting on the same point. Calculate the magnitude and direction of the resultant for this system of forces. HINT: Dra... show full transcript

Worked Solution & Example Answer:FIGURE 8.1 below shows a system of four forces acting on the same point - NSC Mechanical Technology Fitting and Machining - Question 8 - 2022 - Paper 1

Step 1

Calculate the horizontal components

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Answer

To find the horizontal components, we can use:

HC=50cos(30)85+75cos(70)\sum H_C = 50 \cos(30^\circ) - 85 + 75 \cos(70^\circ)

Calculating each term gives:

  • 50cos(30)=43.30 N50 \cos(30^\circ) = 43.30 \ N
  • 75cos(70)=25.65 N75 \cos(70^\circ) = 25.65 \ N

Thus, we get: HC=43.3085+25.65=16.05 N\sum H_C = 43.30 - 85 + 25.65 = -16.05 \ N

Step 2

Calculate the vertical components

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Answer

For the vertical components, we use:

VC=50sin(30)+2575sin(70)\sum V_C = 50 \sin(30^\circ) + 25 - 75 \sin(70^\circ)

Calculating the terms:

  • 50sin(30)=25 N50 \sin(30^\circ) = 25 \ N
  • 75sin(70)=70.48 N75 \sin(70^\circ) = 70.48 \ N

Therefore, we have: VC=2575sin(70)=2570.48=45.48 N\sum V_C = 25 - 75 \sin(70^\circ) = 25 - 70.48 = -45.48 \ N

Step 3

Determine the magnitude and direction of the resultant

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Answer

Now, we find the resultant force using:

R=(VC)2+(HC)2R = \sqrt{(\sum V_C)^2 + (\sum H_C)^2}

Substituting the values gives: R=(70.48)2+(16.05)2=5225.03372.28 NR = \sqrt{(-70.48)^2 + (-16.05)^2} = \sqrt{5225.033} \approx 72.28 \ N

To find the angle, we use:

tan(θ)=VCHC    θ=tan1(70.4816.05)77.17\tan(\theta) = \frac{\sum V_C}{\sum H_C} \implies \theta = \tan^{-1}\left(\frac{-70.48}{-16.05}\right) \approx 77.17^\circ

This angle can be expressed as 77101277^\circ 10' 12' West from South.

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