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Help needed on statics, and also to check my answers

  1. Sep 22, 2005 #1
    I have this assignment to hand in and ive been having problems remembering how to do these problems. Im not sure if my answers are correct. I have no idea how to do the third one if anyone can walk me through the solution.

    My answers are:
    #1: -59.8
    #2: Tension=14715N Mechanical Advantage=4, 3678.74
    #3: I have no idea how to solve
    #4: Tension on C= 1.84Kips Tension on B= 1.74kips

    If i have any of these wrong could you please should me the way. Im sorry for giving so many problems. Im thankful for any help or direction that you can give.

    The problems are on the link below
    [img=http://img303.imageshack.us/img303/7843/problems9hm.th.png]
     
    Last edited: Sep 22, 2005
  2. jcsd
  3. Sep 22, 2005 #2
    Too small give me a bigger image! YIKES
     
  4. Sep 22, 2005 #3
    sorry, i just noticed and fixed it. lol
     
  5. Sep 22, 2005 #4
    ahhhhhhhh my eyes! thanks
     
  6. Sep 22, 2005 #5
    For Tension in number two i get 3,678.75, Just cut the ropes, then there is 4T going up and 1500*9.81 going down, and solve for t, very simple.
    The mechanical advantage you provide is 3,678.75 to lift up 1500*9.81 or 14715/3678.75 = 4 times. But im not sure how that is defined, so i may be wrong. A mechanical advantage of 4k just doesnt make sense to me, sorry. Ah, I read what you wrote, you have good numbers but assigned them wrong. 14,715 is the weight of the block no the tension, that 3k value is the tension, and 4 is right for the advantage.
     
    Last edited: Sep 22, 2005
  7. Sep 22, 2005 #6
    Sorry im slow lol. I did T=1500*9.81 and i got 14715.00N. What did you do to get 3678.75. Thanks
     
  8. Sep 22, 2005 #7
    Number 3 is simple as well. For a square, the force acts at the center along the x axis, so the total force is F= 5kn/m * 3m = 15kN acting at x=1.5 m.

    Now for the triangle, the force acts 1/3 from the height of the trangle, or 2/3 of x from the other direciton. so the total force is again the area, so its F = (5kn/m* 3m )/2
    (the area of that triangle). and it acts a distance 2/3 rds away, (BUT add in the distance for the rectangle too!) so its [(2/3) *3 ] +3, the +3 is because of the distance due to the rectangle! Now you have two forces at two distances, and you have to add them, and find the resultant distance this force acts, by summing the moments about the pin. (But first find the reaction of the smooth roller!) I Hope that helps you. I wont do it for you though.
     
  9. Sep 22, 2005 #8
    NO! Weight = 1500*9.81, thatsthe WEIGHT of the block. The tension is what holds it up. Look, just cut the ropes so that its 4 ropes magically pointing up. Do you see what i mean?
    ....T........T........T........T
    .../\...... /\......./\...... /\
    __|_____|_____|_____|
    |.................W ...........|
    |..................|............|
    |.................\/............|
    ------------------------

    So 4T acts UP, and the Weight W, acts DOWN.

    Now find what T is, its 1/4 the total weight, obviously.
     
    Last edited: Sep 22, 2005
  10. Sep 22, 2005 #9
    do you get what im saying?
     
  11. Sep 22, 2005 #10
    Thanks alot, its slowly coming back to me. I hate the summer time it always ruins my memory. Thanks again for your help
     
  12. Sep 22, 2005 #11
    Yeah, but do you understand what im trying to show you? If its not clear let me know.
     
  13. Sep 22, 2005 #12
    Only thing im trying to figure out is where you got the 2/3s from the triangle one. How do i decide the point?
     
  14. Sep 22, 2005 #13
    You have to go back and read about centroids of various areas. This is the location that works when you find the centrioid, or geometric center of a shape based on a mathematical definition. If you have a right triangle, the centroid is ALWAYS 1/3 away form the base so graphically:
    .................................../..|
    ................................./....|
    .............................../......|
    ............................./ .......|
    .......................... /......... |
    ........................../.......... |
    ......................../.............|
    ......................./..............|
    .................... /................|
    ..................../ -----------|
    ........................./\
    ........................ |<-1/3->|

    So that distance of the centroid is 1/3 from the side with the |'s, or 2/3s away from the side with the /, get it?
     
    Last edited: Sep 22, 2005
  15. Sep 22, 2005 #14
    Thank you, now i understand
     
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