Recent content by user12323567

  1. U

    Engineering Why was the thickness of the beam included in calculating this moment?

    Hi all, Why was the thickness of the beam considered when moment was being calculated and why was the horizontal component of the 600 N force used in calculating the moment? That part is unclear to me. Please explain it. Am I looking for the logic behind using the horizontal comp. and the...
  2. U

    Engineering Using method of sections to solve for truss systems

    If you consider the horizontal component of Fhc, you have a vertical perpendicular distance of 1. This would be much easier for me to understand if you explained why we have to use a distance of 3 meters when Fhc is not 3 metres away from A, please explain this.
  3. U

    Engineering Using method of sections to solve for truss systems

    Since Fhc is at an angle, do we consider both its components when we are looking at the moment about point A? i.e. Fhc(1.5)sin(gamma) - Fhc(1)cos(gamma)? Gamma is the angle of Fhc
  4. U

    Engineering Using method of sections to solve for truss systems

    Yes, you take 2/sqrt(13) and multiply it by 3 to get said 1.66 m, I am asking why you are multiplying by 3 instead of 1 (the vertical perpendicular distance between A and Fhc)
  5. U

    Engineering Using method of sections to solve for truss systems

    Taking the moments about points A and C, why is a distance of 3 meters used in Fhc and Fhg? Isn’t it supposed to be 1.5 meters for both? I acknowledge that Fhc and Fhg have x and y components, but were we considering the horizontal or vertical component (or both components) of either Fhc and Fhg...
  6. U

    What does "Low strength but good strength-to-weight ratio" mean?

    Good strength-to-weight-ratio
  7. U

    Engineering Elastic Deformation of an Axially Loaded Member

    So, are you saying that the length of the supporting rod, AB would now change to 749.20 mm?
  8. U

    Engineering Elastic Deformation of an Axially Loaded Member

    Sum of forces in the y-direction = 0 and downwards is +ve P + Fab,y = 0 P + Fab (4/5) = 0 Fab = -1.25P ẟ = FL/AE -> ẟab = FabLab/AabE ẟab = (-1.25P*.75)/(pi*(.01)^2*(200*10^3)) = -0.0149P After this step, I am uncertain of how I can relate the vertical elongation with AB's elongation to find...
  9. U

    Engineering Path integrals in scalar fields when the path is not provided

    Disregard my request for assistance. I have solved the problem.
  10. U

    Engineering Path integrals in scalar fields when the path is not provided

    I meant to say that the path is not given explicitly as a parametrized form.
  11. U

    Engineering Path integrals in scalar fields when the path is not provided

    I cannot seem to start answering the question as a result of the path not being provided. How do I solve this when the path is not provided? See picture below
  12. U

    Engineering Power Triangle and Power factor

    Thank you, that helped!
  13. U

    Engineering Power Triangle and Power factor

    But how do I use that information to obtain what they are asking for?
  14. U

    Engineering Power Triangle and Power factor

    Induction motors: pf = 0.8 triangle: θ = arccos(0.8) = 36.86 degrees Pa = 500/0.8 = 625 VA Pr = sqrt(625^2-500^2) = 375 VAR Synchronous motors: pf = 0.707 triangle: θ = arccos(0.707) = 45 degrees Pa = 500/0.707 = 707 VA Pr = sqrt(707^2-500^2) = 499.85 VAR I am uncertain of how I can...
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    Engineering Can Fatigue Failure of Material be Demonstrated at Home Using Everyday Items?

    Thank you Scott! These are great ideas:)