Physics 218: Tension in Cords with Suspended Object Weight w

[K-Lamb10]

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.

 

[radou]

The three forces in the strings have to be in equilibrium; \vec{a}+\vec{b}+\vec{w}=\vec{0}. 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.

 

[K-Lamb10]

im sorry, i still cannot get it, i mean, I am plugging in these numbers to my formulas and it just won't come out, and the only way i can get the answer, is the wrong way because I am subtracting an x component from a y component, and i don't think you can do that...unless I am wrong and you can in which i think i may have it...

 

[K-Lamb10]

ok I am wrong...im using the equations in the book, and it does not work out to match the answer, and I am 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?

 

[radou]

im sorry, i still cannot get it, i mean, I am plugging in these numbers to my formulas and it just won't come out, and the only way i can get the answer, is the wrong way because I am subtracting an x component from a y component, and i don't think you can do that...unless I am wrong and you can in which i think i may have it...

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.

 

[K-Lamb10]

ok i don't think I am understanding this equation of equilibrium...is it the t1=w?

 

[radou]

ok i don't think I am understanding this equation of equilibrium...is it the t1=w?

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.

 

[K-Lamb10]

ok, i actually understand what ur saying, i haven't tryed it to see if it works, but thanks a bunch for taking time outta ur day to help me, it means a lot to me man..thanx