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Question on tension

  1. Sep 30, 2006 #1
    This is from the textbook for physics 218 :young and freedman

    #8 the question is what is the tension tention in each cord in a figure if the weight of the suspended object is w. The pic is of a lantern hanging on two cords connected together from the ceiling(both with a different angle form the ceiling, 30 and 45 degrees), those cords are labeld a and b, then cord c is attached to the lantern going down. I cannot figure this thing out, my textbook is of no help...i need to find tension of a and b in terms of w.
  2. jcsd
  3. Sep 30, 2006 #2


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    The three forces in the strings have to be in equilibrium; [tex]\vec{a}+\vec{b}+\vec{w}=\vec{0}[/tex]. I suggest you express this in a graphical way first. Then, solve the vector equation. Or, if you are not too familiar with vectors, express the equation of equilibrium for both x and y directions in order to obtain the magnitudes a and b of the forces in the strings.
  4. Sep 30, 2006 #3
    im sorry, i still cannot get it, i mean, im plugging in these numbers to my formulas and it just wont come out, and the only way i can get the answer, is the wrong way because im subtracting an x component from a y component, and i dont think you can do that...unless im wrong and you can in which i think i may have it...
  5. Sep 30, 2006 #4
    ok im wrong...im using the equations in the book, and it does not work out to match the answer, and im sorry but i do not really understand what you mean by express the equation of equilibrium for both x and y, do you mean like the sqrt of cos(x)^2+sin(45)^2 would equal my magnitude?
  6. Sep 30, 2006 #5


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    You can not subtract x and y components. Find the 'bottom' angles first (the angles at the point where the strings intersect), write an equation of equilibrium for the x direction, and then for the y direction separately. You'll have a system of two equations with two unknowns, from which you'll obtain a and b.
  7. Sep 30, 2006 #6
    ok i dont think im understanding this equation of equilibrium...is it the t1=w?
  8. Sep 30, 2006 #7


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    Ok, let's first write the equation of equilibrium for the x-direction: -a*sinA + b*sinB = 0, where A is the angle between the string 'a' and the 'vertical line', and B is the angle between the string 'b' and the 'vertical line'. Both angles are obtained easily from the triangles which consist of the strings, the ceiling, and the imagined 'vertical line'. So, to get back to our equation, we get a = b*(sinB/sinA) from it (*). So, we expressed the tension a in term of b. Further on, let's write the equation of equilibrium for the y direction: a*cosA + b*cosB - w = 0, which implies b = (w + a*cosA)/cosB. Now, all you have to do is plug the value of a we got out of the first equation (*) into this equation, and you directly obtain the tension b. After that, plug b into equation (*), and get a. I hope it's more clear now.

    By the way, it looks like this: http://upload.wikimedia.org/wikibooks/en/1/1f/Fhsst_forces6.png , where T1 = a and T2 = b and the angles are different, if I remember well.
    Last edited: Sep 30, 2006
  9. Sep 30, 2006 #8
    ok, i actually understand what ur saying, i havent tryed it to see if it works, but thanx a bunch for taking time outta ur day to help me, it means a lot to me man..thanx
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