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Homework Help: Frictionless ladder question

  1. Apr 17, 2010 #1
    1. The problem statement, all variables and given/known data


    [PLAIN]http://img72.imageshack.us/img72/6052/jdff.jpg [Broken]

    3. The attempt at a solution

    For part ii
    i found the centre of mass of the ladder by
    80x2/80+25 = 1.52m up the ladder
    then
    force of the wall minus the tension
    Fw - T = 0
    Fw = (9.8 x (25x0.51 + 80x0.67)) / 2.83 (height of the ladder against the wall)
    = 229.76 N
    so thats the force on the wall and the tension opposing it
    would that be the answer for the tension in the rope?
    or would i have to consider the tension doing more work?
    like the total force downwards in the y direction?
    Mg + mg = 1029 N
    or is that just the force the ground is giving?
    and im also not sure how to incorporate the fact that the rope is abit higher than ground level
    thanks for help in advanced
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 17, 2010 #2

    ehild

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    Would you please show the forces in the picture?

    ehild
     
  4. Apr 17, 2010 #3
    [PLAIN]http://img44.imageshack.us/img44/5007/fhdf.jpg [Broken]
    the angle is 19.5 aswell
    Fg- force of ground
    Fw force wall
    T tension
    mg gravitational force
    i think theres a Mg force aswell of the person
    I'm having trouble thinking if the height of the rope matters
    like half of me thinks it does
    but the other half thinks it doesnt
    but if it doesnt matter, would that answer be right?
     
    Last edited by a moderator: May 4, 2017
  5. Apr 17, 2010 #4

    ehild

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    You have to include the torque of the rope, too.

    ehild
     
  6. Apr 17, 2010 #5
    ohh ok sorry i forgot
    tz = 0
    but what about for part two?
    does that look alright to you?
    would the tension on the rope change if its height increases?
    when the weight of hte rope doesnt matter?
     
  7. Apr 17, 2010 #6

    ehild

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    What is tz=0? The tension of the rope is T, its torque is T*its height, anticlockwise. This term is missing from the torque equation

    "Fw = (9.8 x (25x0.51 + 80x0.67)) / 2.83 (height of the ladder against the wall)"

    ehild
     
  8. Apr 17, 2010 #7
    oh ok
    the previous line looks like
    i made the pivot at the bottom of the ladder
    -h(Fw) + g(x1m1 + x2m2) + 0 T + 0 Fg = 0
    then i rearagned for Fw
     
  9. Apr 17, 2010 #8

    ehild

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    Why do you multiply the tension of the rope with 0? The arm is 0.5 times cosine of the angle between the ladder and the wall.

    ehild
     
    Last edited: Jun 29, 2010
  10. Apr 17, 2010 #9
    Oh right, so the pivot 0.5 up the ladder where the string is attached
    I was just following an example in my book
    it doesnt have a ladder with a string though
    it says Tnet,z = 0 (torque)
    and times those things by 0
    damnn
    i thought i had this question down
    so is the procedure similar just with the upper length of the ladder?
     
  11. Apr 17, 2010 #10

    ehild

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    yes...

    ehild
     
  12. Apr 17, 2010 #11
    i get 183N
    by pluging in the new x co-ordinante for the person
    9.8(25*0.51 + 80x0.5)/2.83
    im pretty positive im on the wrong track
    the centre of mass would stay the same right?
    and we would still consider the height to be 2.83? or do i have to find the height without the 0.5m of the ladder?
     
  13. Apr 17, 2010 #12

    ehild

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    No, you do not have a new x coordinate. You do not need the centre of mass of the whole system. The centre of mass of the ladder is in the middle. You have forces, points where they act and level arms. Calculate the torque of all forces with respect to the point where the ladder stands. The sum must be zero.

    ehild
     
    Last edited: Jun 29, 2010
  14. Apr 17, 2010 #13
    ohh ok
    so
    Fw + T + Fg + (2/3 x 80 x 9.8) + (0.5 x 25 x 9.8) = 0
    so would all the forces be times'd by H
    and now do we apply that zero thing like they did in my book?
     
  15. Apr 17, 2010 #14

    ehild

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    This equation is entirely wrong. What is torque?

    ehild
     
  16. Apr 19, 2010 #15
    (H)Fw + 1/2mg + 2/3 Mg + hT = 0
    is that one right?
    sorry for the late reply i polished off all of the other questions
    514.5 - 2.83 Fw - 0.47T = 0

    does Fw - T= 0?
    then could we find one of them and solve
    Fw = T

    514.5 - 2.83T - 0.47T = 0
    3.3T = 514.5
    T = 156N = Fw ?
     
    Last edited: Apr 19, 2010
  17. Apr 19, 2010 #16

    ehild

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    (H)Fw should have "-" sign, as Fw would rotate the ladder clockwise while the other three forces would rotate anti-clockwise.

    - (H)Fw + 1/2mg + 2/3 Mg + hT = 0,

    but where did the 541.5 come from?

    Yes,the magnitude of Fw is the same as the magnitude of the tension in the rope, Fw=T. But the torque of Fw is negative, that of the tension is positive.

    ehild
     
  18. Apr 19, 2010 #17
    oh right, sorry

    i got 541.5 just by adding 1/2mg + 2/3 Mg
    which i may have done wrong haha one sec
    (53.33+12.5)g= 645.1
    ahh i mixed up the 1/2 and 2/3
    does that look better?
    then the same procedure so like

    -2.83Fw + 645 + 0.47T = 0
    Fw = T
    645 = 2.36T
    T = 273.3N
     
  19. Apr 19, 2010 #18

    ehild

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    It looks all right at last.


    ehild
     
  20. Apr 19, 2010 #19
    thanks loads =D
    sorry about being a hassle at the start i wasnt thinking properly
     
  21. Apr 20, 2010 #20
    hey ehild sorry to be a pain again
    i full understood it last night
    but im just thinking over it again
    so we made the rope the pivot point
    so should it it be different
    cause i dont get why its 2/3 and 1/2,
    and then why Fw is times H would it be times 3?
    i understood the first time because they're = to the x components of the distance from where the laddy touches the ground
    how come it isnt, say if L = length of the ladder = 3
    L2/3 and L/2
    but if we take the pivot at 0.5ms up the ladder
    would everything change?
    like the distance of everything relative to the pivot?
    so like i got the guy is 1.5m from the pivot which is just L/2 then the centre of mass would be the same as the original or would it change?
    and would T just be 0 because its 0 from the pivot?
    and would there be a new height?
    i did it all again and this is what it looked like
    1/2Mg + 1/2 mg = 2.35T = 219N
    am i just over thinking it?
    im just confused about the pivot and why we use 2/3 and 1/2 and then HFw
    sorry for being a pain
     
  22. Apr 20, 2010 #21

    ehild

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    I took the bottom of the ladder as the pivot point. This makes things easy as we do not need to calculate with the normal force from the ground.

    The torque is force times level arm (the distance of the "attack line" of force from the pivot. See pic.


    I will show you a picture where the pivot is the point where the rope is attached to the ladder in the next post.

    ehild
     
    Last edited: Jun 29, 2010
  23. Apr 20, 2010 #22
    oh thanks, i was just rewriting my post because it was all jibber jabber
    so if we took the pivot at the rope we would ignore T because itll be 0T and have to find Fground right?
    just another question
    i had my head around it last night, but somone at uni questioned it today
    how come for the masses of the person and the ladder, we use the x components from the pivot
    like
    in my head it makes sense
    and i dont want to question it haha, but then i think why isnt it how far he is up the ladder?
    is it somthing like, thats where the force is acting relative to where the pivot is?
    how come the y component isnt considered for the masses? or just like 2m and 1.5 ms
     
  24. Apr 20, 2010 #23

    ehild

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    Yes, if the end of the rope is the pivot point then its torque is zero. And we have to take the torque of the normal force into account.

    As for your next question, remember, the torque is the turning effect of a force. If you push a door, it will not turn if you push in the direction of the pivot. It is easiest to turn if your force is perpendicular to the door and as far from the pivot as possible.

    The picture shows a level. Fa is perpendicular to the level, and its attack point is at distance a from the pivot. The level arm is a, the torque is Ta=a*Fa, and clockwise.

    Fc is in-line with the level. Its line of attack crosses the pivot point. The torque is zero.

    Fb makes an angle with the level. You get the level arm by drawing a straight line thorough the pivot which makes a 90° angle with the force Fb. The length of the straight line between the pivot point and the line of force is the level arm. Tb=b*Fb, and it is anti-clockwise.

    Or you can calculate the torque of Fb in an other way. The in-line component of Fb does not rotate the level, its torque is zero. Only that component of Fb counts which is normal to the level (Fn). So Fb=c*Fn



    ehild
     
    Last edited: Jun 29, 2010
  25. Apr 20, 2010 #24
    thanks ehild that helped alot =]
     
  26. Apr 20, 2010 #25

    ehild

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    I do not know how much do you learnt about vectors, but the torque is a vector product of the position vector of the "attack point" of the force with the force vector.

    [tex]T = \vec r \times \vec F[/tex]

    The magnitude of the torque is T = r F sin(alpha) where alpha is the angle between the force and the position vector r.

    ehild
     
    Last edited: Jun 29, 2010
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