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

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The discussion focuses on calculating the tension in three cords supporting a weight in two different figures. The user attempts to break down the tension into x and y components, applying equilibrium conditions to derive equations for the forces. They progress through calculations for tensions A, B, and C, while correcting signs and ensuring proper relationships between the variables. Despite initial confusion regarding the values, they refine their approach and arrive at a clearer understanding of the equations needed to solve for the tensions. The conversation emphasizes the importance of careful algebraic manipulation and the correct application of trigonometric functions in solving the problem.
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Homework Statement



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

YF-05-44.jpg


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
 
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Heat said:
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.
 
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. :(
 
Heat said:
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...
 
ok so it would be

Bcos45/cos30 + B sin 45 -w = 0

.81649658B + .707106781B - w = 0

1.523603361B = w

B = w/ 1.523603361

..
 
Heat said:
ok so it would be

Bcos45/cos30 + B sin 45 -w = 0
You forgot the sin30 in the first term.
 
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
 
Looks OK.
 
but the answer for the tension of B is "T_B =.896575w" O_o

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?
 
  • #10
Heat said:
but the answer for the tension of B is "T_B =.896575w" O_o
Realize that B/1.115355071 = (1/1.115355071)B

Calculate 1/1.115355071
 
  • #11
:O

wow, forgot about that 1 in the numerator.

thank you for your help. :)
 
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