# Question on tension

1. Sep 30, 2006

### 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.

2. Sep 30, 2006

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.

3. Sep 30, 2006

### K-Lamb10

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...

4. Sep 30, 2006

### K-Lamb10

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?

5. Sep 30, 2006

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.

6. Sep 30, 2006

### K-Lamb10

ok i dont think im understanding this equation of equilibrium...is it the t1=w?

7. Sep 30, 2006