Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

melmel

  • one year ago

solve the pink one

  • This Question is Closed
  1. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

  2. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Blue: Denote tension in cord as T. Vertical reaction on bar equals weight of bar, assuming no friction anywhere. Take moments about B. Solve for T.

  3. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how?

  4. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can you show to me

  5. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    "How" meaning to take moments?

  6. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes but i'm confuse >.<

  7. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    roller gve vertical force to the wall so therefore

  8. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1375615728179:dw|

  9. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1375615816627:dw|

  10. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    also

  11. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I can give an example: |dw:1375616654183:dw| A ladder 5 m long leans agains a smooth wall at B and stands on a rough floor at A at a distance of 3 m from the wall as shown above. The mass of the ladder is 10 kg. Since we don't know the friction on the floor, we can take moments about A so the friction does not come in the equation. Let reaction at B = R Mass at the middle (C) = m Take moments about A, since the ladder is in equilibrium, sum of moments = 0. -R*4 + mg*(3/2)=0 Note: moment = force * distance, clockwise is positive. Solve for R: R=(3mg/2)/4, or =3mg/8 N.

  12. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1375617070605:dw|

  13. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes u are right

  14. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    So you're good for both problems?

  15. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no

  16. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lets continue master

  17. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1375616151580:dw|

  18. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Keep going, you're on the right track. I have an appointment, so have a little time left.

  19. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes, what about the diagram?

  20. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what diagram?

  21. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    In the case of the ladder, there is a horizontal frictional force, F.|dw:1375617403359:dw|

  22. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    But taking moments at A will ignore that.

  23. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the roller does not give horizontal force master this will give only a horizontal force if the roller

  24. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is parallel to the wall

  25. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1375616451865:dw|

  26. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1375617764714:dw| Would that help?

  27. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes u are correct tension is there now can you teach me to take moment

  28. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    where can i put my moment

  29. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    basically i dont know where can i put the moment >.<

  30. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1375617910870:dw| Moments tend to turn the object (bar) that you have isolated as a "free-body". Moment means the product of a force times the distance about a point. The distance must be the perpendicular distance, i.e. the shortest possible distance between the point and the direction of the force. Moments that turn the (free) body in a clockwise direction is positive, anti-clockwise is negative. So far so good?

  31. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes i want to use the moment clock wise

  32. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so therefore?

  33. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    When you take moments about a point (say B), we assume that B does not move but free to turn. When the (free) body is in equilibrium, the sum of moments about any point is zero. So take moments about B: Ra*(3sin(60) - T(2cos(60)) = 0 Since Rb passes through point B, the distance is zero, so the product (moment) is also zero. Now you can solve for T in terms of Ra. (You know from the sum of forces in the vertical direction that Ra=mg).

  34. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes i familiar that Ra=mg

  35. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Are you able to solve for T now? By the way, in the future you would get more attention posting mechanics problems under Physics or Engineering. I enjoy solving mechanics problems.

  36. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wow

  37. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i become your fan :)

  38. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    T=1273.06 ?

  39. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so im right?

  40. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yep, that's what I got (1274.356). You can solve the blue problem the same way. In any case, I've got to go. If you need more help, you may want to post the question again under physics or engineering, whichever you're onto.

  41. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thats all for the blue one?

  42. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i need also to solve the red one >.< dont go :)

  43. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    By the way, the minor difference is because I use g=9.81, when most people use 9.8. Yes, that's all for the blue one. Sorry, I meant you can solve the red one the same way.

  44. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Can you draw the FBD for the red one?

  45. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  46. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1375618026746:dw|

  47. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    let me put the Rb and Ra and correct me if im wrong

  48. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1375618200791:dw|

  49. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Here you know (again) that T=W=mg. |dw:1375619330484:dw|

  50. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    and that Ra=Rb by equilbrium of horizontal and vertical forces, right?

  51. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    where did tou get 10 and 5 the distance of the assume box is 12m

  52. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i think yes their are equal but im not sure

  53. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh, I was sloppy. It should have been 4 and 8 to make 12. The answer would have been the same. (you don't need to know that! :)

  54. melmel
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1375618365908:dw|