Recent content by DevonZA

  1. DevonZA

    Hydraulic jack - Pascal's principle

    A1= pi/4 * d2 = pi/4 * (0.15)2 = 0.0177m2 A2 = pi/4 * d2 = pi/4 * (0.05)2 = 0.00196m2 Ratio = 0.0177/0.00196 = 9 Therefore ratio is 9:1
  2. DevonZA

    Hydraulic jack - Pascal's principle

    I only want to confirm that my answers are correct. In 1.2 I believe they want to know the distance moved on the side of the mass when 10mm pushed down on the small piston side
  3. DevonZA

    Hydraulic jack - Pascal's principle

    Homework Statement Homework Equations P1=P2 P=F/A F=PA F1/A1=F2/A2 F2 = F1(A2/A1) F2 = W(A2/A1) W=mg W=F*d The Attempt at a Solution A1= pi/4 * d2 = pi/4 * (0.15)2 = 0.0177m2 A2 = pi/4 * d2 = pi/4 * (0.05)2 = 0.00196m2 P1=P2 P=F/A F=PA F1/A1=F2/A2 F2 = F1(A2/A1) F2 = W(A2/A1)...
  4. DevonZA

    Centre of gravity, moment of inertia and radius of gyration

    The formulae given for a rectangular triangle: about aa axis I=1/18md^2 about bb axis (through G) I=1/18mh^2 because we want the inertia about xx axis I used =1/18md^2 where d=distance to xx axis=200mm
  5. DevonZA

    Centre of gravity, moment of inertia and radius of gyration

    Okay so then X=m1x1-m2x2/m1-m2 = 8*(500)-1.2*(600)/8-1.2 = 482.35mm Is this correct? What about my answers for 1.2 and 1.3?
  6. DevonZA

    Centre of gravity, moment of inertia and radius of gyration

    Meaning they should be positive? If they were below the X axis then they would be negative? In other words if the XX axis were at the top of the rectangle.
  7. DevonZA

    Centre of gravity, moment of inertia and radius of gyration

    That should be 500, from the centre of the rectangle to the XX axis is 500mm. I believe the centre of the triangle should be 2/3*600 = 400mm Then added to the 200mm from the start of the triangle to the XX axis should give 400+200=600mm. I put a negative in front of the 400 and 1000/3 as they...
  8. DevonZA

    Centre of gravity, moment of inertia and radius of gyration

    Homework Statement Homework Equations Centre of gravity: X=m1x1-m2x2/m1-m2 MOI rectangle: 1/3ml^2 MOI triangle: 1/18md^2 Radius of gyration: Ixx=mk^2 The Attempt at a Solution Mass of body 1: b*l*p = 0.8*1*10=8kg Mass of body 2: 1/2b*h*p = 1/2(0.4)*0.6*10=1.2kg1.1 X=m1x1-m2x2/m1-m2...
  9. DevonZA

    How Do You Calculate the Inside Diameter and Mass Savings for Hollow Shafts?

    Homework Statement A solid and hollow shaft of the same material must transmit the same maximum torque. The diameter of the solid shaft is 203mm and the outside diameter of the hollow shaft is to be 216mm. a) Determine the inside diameter of the hollow shaft. (answer = 138.7mm) b) What...
  10. DevonZA

    Change of internal energy of an ideal gas

    Thank you let me give that a try.
  11. DevonZA

    Change of internal energy of an ideal gas

    Is this because (a-Ru) is a constant? eg. ∫1 dx = x
  12. DevonZA

    Change of internal energy of an ideal gas

    Homework Statement Homework Equations Δu = ∫ [(a-Ru)+bT+cT^2+dT^3]dT The Attempt at a Solution The answer of 6447kJ/kmol is given but I am struggling to get to this answer after integrating the above formula and inserting the given values. Firstly would the integral of...
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