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Die meganiese werkswinkel benodig 'n hidrouliese pers - NSC Mechanical Technology Automotive - Question 9 - 2017 - Paper 1

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Die meganiese werkswinkel benodig 'n hidrouliese pers. Die diameter van Suier B is 180 mm en beweeg 12 mm op. Die krag wat op Suier A toegespas word is 550 N. Suier ... show full transcript

Worked Solution & Example Answer:Die meganiese werkswinkel benodig 'n hidrouliese pers - NSC Mechanical Technology Automotive - Question 9 - 2017 - Paper 1

Step 1

Die diameter van Suier A

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Answer

First, we need to determine the volume of Suier B using the formula for the area and stroke length:

VB=ABimesSBV_B = A_B imes S_B

Where:

  • AB=π(DB2)2=π(180mm2)2=25446.9mm2A_B = \pi \left(\frac{D_B}{2}\right)^2 = \pi \left(\frac{180 \, mm}{2} \right)^2 = 25446.9 \, mm^2
  • Stroke length, SB=12mm=0.012mS_B = 12 \, mm = 0.012 \, m

Thus,

VB=25.4469×103m2×0.012m=0.3054×103m3V_B = 25.4469 \times 10^{-3} \, m^2 \times 0.012 \, m = 0.3054 \times 10^{-3} \, m^3

Next, we can find the diameter of Suier A, given the volume is the same:

AA=VASA=0.305×1030.06A_A = \frac{V_A}{S_A} = \frac{0.305 \times 10^{-3}}{0.06}

Solving for DAD_A:

DA=4×AAπ800mmD_A = \sqrt{\frac{4 \times A_A}{\pi}} \approx 800 \, mm

Step 2

Die krag wat deur Suier A uitgeoefen word

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Answer

To find the force exerted by Suier A, we apply the formula:

PA=FAAAP_A = \frac{F_A}{A_A}

Given that the pressure PAP_A is equal to PBP_B, we also know:

PB=FBAB=108.268×103PaP_B = \frac{F_B}{A_B} = 108.268 \times 10^3 \, Pa

Using the area calculated for Suier A:

FA=PA×AA=108.268×103Pa×0.055×102=2.76kNF_A = P_A \times A_A = 108.268 \times 10^3 \, Pa \times 0.055 \times 10^{-2} = 2.76 \, kN

Step 3

Die druk wat op Suier B uitgeoefen word

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Answer

The pressure exerted on Suier B can be calculated using the area from above:

FB=PB×AMF_B = P_B \times A_M

Where:

  • AM=AB=108.268×103×25.45×103A_M = A_B = 108.268 \times 10^3 \times 25.45 \times 10^{-3}

Thus,

FB=2755.42N=2.76kNF_B = 2755.42 \, N = 2.76 \, kN

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