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Extremely tricky forces question

  1. Mar 12, 2008 #1
    Hi,
    I'm really stuck on this question...

    "A truck dealer wants to suspend a 2.5 tonne truck as shown for advertising. the distance b (which is 17.3m) shows a horizontal distance of the suspension point B from the anchor point A. The sum of the lengths of cables AB and BC is 42m. calculate the tensions in the cables."

    The diagram shows a 40m horizontal distance from anchor point A to anchor point C.
    I can see straight away that hanging point B is 17.3m horizontally from the left (point A) and therefore 22.7m from the right (point C) but also the force verticle would be
    2,500kg*9.81ms^-2 =24,525N in a downward direction (Weight of the truck).

    I know some theory to use where the F=T1sin[tex]\theta[/tex]1+T2sin[tex]\theta[/tex]2

    however to work any other tensions out i require the angles of the cables to verticle but i dont know even 2 sides lol. I know the AB+BC=42 and AC=40

    challenging question :( unless I'm missing something simple and fundimental...

    plz help
    ben
     
  2. jcsd
  3. Mar 12, 2008 #2

    tiny-tim

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    Hi ben! Welcome to PF! :smile:

    This is an ordinary geometry problem.

    You have a triangle with two sides unknown, except that you know the sum of those two sides.

    Let D be the point on AC directly above B.

    Then the two right-angled triangles ABD and CBD have a common side, BD.

    Now use Pythagoras' theorem on each triangle to get two formulas for the length BD, which must be equal, and then solve them. :smile:
     
  4. Mar 12, 2008 #3
    hey

    Thanks for the reply but I don't think you've fully understood I'm not understanding you lol.

    I didn't explain it well I'm sorry but i keep ending up with too many unknowns, because if i split it into two triangles like i was working on originally then it gives me two unknowns in pythag and i only know the combined ABC is 42m so i don't know how to split it to get AB and BC because they're squared in pythag's equation and 42squared isn't the same :(

    can you make sense of that? lol

    where am i going wrong??

    Ben
     
  5. Mar 12, 2008 #4


    i meant you're not fully understanding m8 OR I'm not understanding still lol
     
  6. Mar 12, 2008 #5

    tiny-tim

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    Yes! That's what i thought you meant! :smile:

    When you have a problem like this, pretend you know the length of one of the sides, and call it L.

    Then one triangle gives you BD^2 = 17.3^2 + L^2,
    and the other gives you BD^2 = 22.7^2 + (42 - L)^2;

    so 17.3^2 + L^2 = 22.7^2 + (42 - L)^2,
    which is a quadratic equation for L, and the solution is … ? :smile:
     
  7. Mar 14, 2008 #6
    Hi thats a massive step lol was so stuck but i thought that on the fisrt triangle the pythagorus would be L^2 = BD^2 + 17.3^2 hence BD^2 = L^2 - 17.3 and
    BD^2 = L^2 - 22.7 is that correct? :S because th earea on the hyp is the sum of the areas on the other two sides a^2 + b^2 = c^2 with c as the hypotenuse

    but cheers it still helps so i get an equation on both sides i had before but didn't know what to do with it :smile: so i tried attacking from somewhere else.

    anyways, i worked it through as follows but still get a math error this is a nightmare! lol

    BD^2 = L^2 - 17.3^2
    BD^2 = (42-L)^2 - 22.7 (42-L)^2 = L^2-84L + 1764
    L^2 - 299.29 = L^2 - 84L + 1248.71
    hence ending up with;
    0 = 0L^2 - 84L + 1548

    so in the quadratic equation a = 0 and so I'll be dividing by 0 which is math error...
     
  8. Mar 14, 2008 #7

    tiny-tim

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    oops!

    Hi b3nji3! :smile:

    oops! :redface:

    You're right - I put pluses where I should have put minuses!

    That's what happens when I try to do it in my head without actually drawing a diagram! :redface:

    Sorry!

    And you're right - it isn't a quadratic equation after all - the L^2s cancel, and it's a nice simple linear equation, aL -b = 0. :smile:
     
  9. Mar 14, 2008 #8
    yeah i figured it in the end but you helped so much nice one!!

    now i have to do the hard it with summation of forces etc once i get the angles :smile:

    hope i'm better at that despite never being taught it lol
     
  10. Mar 14, 2008 #9
    so you're only new to this place yourself then? (420 posts)?
     
  11. Mar 14, 2008 #10

    tiny-tim

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    :wink: :smile: Just shows how easy it is to make mistakes! :smile: :wink:
     
  12. Mar 14, 2008 #11
    yeah tell me about it lol but yeah cheers still~! gotta work on the forces bit now with the angle i get then move to the next questions. got more difficulties to come lol in the form of adding waves etc and trig identities and working them thru in my head and showing i can do it.

    oh by the way i have "cos74 . cos29 + sin74 . sin29"

    common trig identities means its cos(74-29) hence cos(45) which is 1/[tex]\sqrt{2}[/tex]

    now, dont ask me how, but i happen to know that that also equates to [tex]\sqrt{2}[/tex]/2 which in rms values happens to be a figure that stuck in my head and is 0.7071 now to prove i did it in my head i need to show i know the connection of the recipracol and the half of the square root of two being the same.

    dont suppose you know anything on this???
     
  13. Mar 14, 2008 #12

    tiny-tim

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    You're getting quite good at this! :smile:

    It certainly helps when these formulas get stuck in your head and just pop out exactly when they're needed!

    (btw, you could have typed the "2" inside the tex brackets - tex brackets aren't dangerous magic - just put whatever you like inside them - within reason, of course!
    Alternatively, if you type alt-v, your computer prints √ anyway.)​

    The reason why 1/√2 = [tex]\sqrt{2}/2[/tex]

    is simply because you can multiply both the top and bottom of the fraction by √2; and since √2 times √2 = 2 … :smile:
     
  14. Mar 14, 2008 #13
    you're good! lol but yeah that makes sense duh to me lol im mech elec so sometimes things from one help in others after all the maths is always the same just with different letters

    its just a language tool as i see it and im interested in it but not so good at it. sometimes helps my gf doing a maths degree but i try to avoid asking because it wastes the time we have.

    was only asking if you're new because i figured someone like you would have thousands of posts
     
  15. Mar 14, 2008 #14
    but again thats ace :P

    cheers
     
  16. Mar 14, 2008 #15

    tiny-tim

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    Grandfather, or goldfish?

    Hi ben!

    I've only just joined the forum! :smile:

    (btw, you can click on people's profiles to find out how long they've been here.)

    I'm at home for a few weeks because of an injury, so I decided to mess around on the computer!

    I've been having fun learning all about LaTeX and smilies! :smile:

    I hope your gf is well (erm … is that grandfather, or goldfish? :confused:)
    ™​
     
  17. Mar 16, 2008 #16
    lmao! yeah my girl friend's fine lol she has some books on latex and a cd but im not it the know bout that stuff i dont think its the same as the latex stuff on here :wink:.

    82sin(100pi*.05-pi/4) is what i just typed into goolge and it gave me the answer with its built in calc lmao! i didnt realise it would do so complicated and it even put new brackets where it wanted it... anyways i was doing it to check whether i needed calc in radians... google says i do :smile:

    need to rearrange it for .05

    v1 = Tsin(100pi.t -pi/4)

    im after the t bit to the left

    i got [tex]\frac{\Pi/4+arcsin\frac{V1}{T}}{100\Pi}[/tex]

    is it wrong lol??
     
  18. Mar 16, 2008 #17
    i missed the = t bit on the end lol
     
  19. Mar 16, 2008 #18
    actually it has to be right... it works out at 7.5ms and at 50Hz a wavelength is 1/50 hence 20ms so if that is 2\Pi then it would be lagging at \Pi/4 so thats 0.0025 cause its half of the 5ms or \Pi/2 for the original peak so its 5ms plus 2.5 ms or technically \Pi/4 + \Pi/2 or 3quarters \Pi which is 3\Pi/4 :smile: sorry for the waste of time!
     
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