Find Tension in Cords | Figure (a) & (b) | Homework Help

[Heat]

Homework Statement

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).

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: http://img399.imageshack.us/img399/7552/98301308qg2.jpg

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

 

[Doc Al]

Sum of Fx =

Acos30 = Ax
Bcos45 = Bx

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.

Sum of Fy =

Asin30 = Ay
Bsin45 = By
w = Cy

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

 

[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. :(

 

[Doc Al]

so it would be:

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

Good.

now, the A's cancel out,

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...

 

[Heat]

ok so it would be

Bcos45/cos30 + B sin 45 -w = 0

.81649658B + .707106781B - w = 0

1.523603361B = w

B = w/ 1.523603361

..

 

[Doc Al]

ok so it would be

Bcos45/cos30 + B sin 45 -w = 0

You forgot the sin30 in the first term.

 

[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

 

[Doc Al]

Looks OK.

 

[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

instead of this

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

I get .8164962w

good enough?

 

[Doc Al]

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

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

Calculate 1/1.115355071

 

[Heat]

:O

wow, forgot about that 1 in the numerator.

thank you for your help. :)