# Find the tension help

1. Oct 31, 2007

### Heat

1. The problem statement, all variables and given/known data

Find the tension in each cord in the figure, if the weight of the suspended object is w.

1)Find the tension of the cord A in the figure (a)
2)Find the tension of the cord B in the figure (a).
3)Find the tension of the cord C in the figure (a).
4)Find the tension of the cord A in the figure (b).
5)Find the tension of the cord B in the figure (b).
6)Find the tension of the cord C in the figure (b).

3. The attempt at a solution

I got the solutions, but I want to learn how to do this problem.

What I have done is broken down the tension into x and y components.

as so:

if they are in equilibrium then the total force is 0.

ok I now understand part c and f, because the only tension acting on that string is the weight of the object pulling down.

update:

I progressed some more (hopefully what I did is in the right track):

Sum of Fx =

Acos30 = Ax
Bcos45 = Bx

Sum of Fy =

Asin30 = Ay
Bsin45 = By
w = Cy

Solutions: 1).73205w 2) .896575w 3) w 4) 2.732w 5) 3.346w 6) w

Last edited: Oct 31, 2007
2. Oct 31, 2007

### Staff: Mentor

Signs matter. I'd say:
Ax = -Acos30 (because it points to the left)
Bx = +Bcos45

Sum of Fx = 0, so:
Ax + Bx = 0
-Acos30 + Bcos45 = 0

That's one equation.

Correct the signs and write the vertical equilibrium equation. That's your second equation. You can solve them together to get A and B.

3. Oct 31, 2007

### Heat

so it would be:

Sum of Fx::: -Acos30 + Bcos45 = 0
Sum of Fy::: Asin30 + Bsin45 -w = 0

now, the A's cancel out, and it would be 2Bcos45sin45-w=0....b=w? can't be. :(

4. Oct 31, 2007

### Staff: Mentor

Good.

What do you mean they cancel out?

Write one variable in terms of the other (from one equation) and substitute (into the other equation).

-Acos30 + Bcos45 = 0
Acos30 = Bcos45
A = Bcos45/cos30

etc...

5. Oct 31, 2007

### Heat

ok so it would be

Bcos45/cos30 + B sin 45 -w = 0

.81649658B + .707106781B - w = 0

1.523603361B = w

B = w/ 1.523603361

..

6. Oct 31, 2007

### Staff: Mentor

You forgot the sin30 in the first term.

7. Oct 31, 2007

### Heat

B(cos45/cos30)(sin30) + B(sin 45) -w = 0

B (.81649658)(.5) + B (.707106781) -w

.40824829B + .707106781B - w = 0

1.115355071B = w

B = w/1.115355071

8. Oct 31, 2007

### Staff: Mentor

Looks OK.

9. Oct 31, 2007

### Heat

but the answer for the tension of B is "T_B =.896575w"

but if I try doing it like this:

((Bcos45)/cos30)(sin30) + B(sin 45) -w = 0

B(cos45/cos30)(sin30) + B(sin 45) -w = 0

I get .8164962w

good enough?

10. Oct 31, 2007

### Staff: Mentor

Realize that B/1.115355071 = (1/1.115355071)B

Calculate 1/1.115355071

11. Oct 31, 2007

### Heat

:O

wow, forgot about that 1 in the numerator.

thank you for your help. :)