Assistance Needed in Calculating Cable Tension

**Lance2497** · September 25th 2008, 06:00 PM

I'm having so much trouble with this and it's due in 3 days.

Here is what it looks like;

Question:
For W=175 pounds, computer tension force in the cables (both sides) for L=11 and X=(1,2,3,4,5). Which X position creates the greatest tension? Where is the best place to hang the weight?

I'm not asking for a solution, I just wonder if anyone can give me an equation I can use.

**skeeter** · September 25th 2008, 07:56 PM

the system is in equilibrium ... $\sum F = 0$

let $\theta$ = angle $T_1$ makes with the horizontal
$\phi$ = angle $T_2$ makes with the horizontal

$T_1 \cos{\theta} = T_2 \cos{\phi}$

$T_1 \sin{\theta} + T_2 \sin{\phi} = 175$

you'll have to determine the values of the trig ratios from your given dimensions.

**Lance2497** · September 25th 2008, 08:39 PM

Originally Posted by skeeter

the system is in equilibrium ... $\sum F = 0$

let $\theta$ = angle $T_1$ makes with the horizontal
$\phi$ = angle $T_2$ makes with the horizontal

$T_1 \cos{\theta} = T_2 \cos{\phi}$

$T_1 \sin{\theta} + T_2 \sin{\phi} = 175$

you'll have to determine the values of the trig ratios from your given dimensions.

How did you get 175 as the angle measure?

I apologize for the error, but T1 and T2 refer to the suspended cable pieces, not angles.

**skeeter** · September 26th 2008, 06:08 PM

I'm even more sorry that you have no idea what you are solving for ...

$T_1$ and $T_2$ refer to tensions (forces) in the cable, or didn't you know that?

175 lbs is a force, not an angle.

**Lance2497** · September 28th 2008, 05:25 PM

I appreciate your clarity on my confusion, as well as you snarky remarks. Truly makes everything go much smoother when one in need of assistance is given grief for his lack of understanding.

Thank you

**juanfe_zodiac** · February 16th 2009, 08:23 PM

Originally Posted by skeeter

the system is in equilibrium ... $\sum F = 0$

let $\theta$ = angle $T_1$ makes with the horizontal
$\phi$ = angle $T_2$ makes with the horizontal

$T_1 \cos{\theta} = T_2 \cos{\phi}$

$T_1 \sin{\theta} + T_2 \sin{\phi} = 175$

you'll have to determine the values of the trig ratios from your given dimensions.

i dont understain why theta 1 and theta 2 are equal??? they dont look to be equal...looking at the horizontal line they dont look the same...could you please explain me that??? (how are the angles equal

**skeeter** · February 17th 2009, 11:31 AM

I did not say the angles were equal.

**juanfe_zodiac** · February 17th 2009, 06:51 PM

Originally Posted by skeeter

I did not say the angles were equal.

can you explain me what did you do there?

**HallsofIvy** · February 18th 2009, 09:14 AM

He didn't use the angles, he used the fact that the triangle formed by the horizontal and vertical forces has the same angles and is similar to the triangle formed by the roof, vertical, and cable. The horizontal force on each side, divided by the Tension in the rope on that side, is equal to the horizontal distance (x on the left, 10-x on the right) divided by the length of the hypotenuse ( $\sqrt{x^2+ H^2}$ on the left, $\sqrt{(10-x)^2+ H^2}$ on the right). Those must be equal in order that the weight must not move from side to side. That was the first of the two equations skeeter gave.

The vertical force on each side, divided by the tension in the rope on that side, is equal to the vertical distance, H, divided by the length of the hypotenuse ( $\sqrt{x^2+ H^2}$ on the left, $\sqrt{(10-x)^2+ H^2}$ on the right). The total of those two must be the weight itself. That is the second equation skeeter gave.

**juanfe_zodiac** · February 18th 2009, 07:31 PM

Originally Posted by HallsofIvy

He didn't use the angles, he used the fact that the triangle formed by the horizontal and vertical forces has the same angles and is similar to the triangle formed by the roof, vertical, and cable. The horizontal force on each side, divided by the Tension in the rope on that side, is equal to the horizontal distance (x on the left, 10-x on the right) divided by the length of the hypotenuse ( $\sqrt{x^2+ H^2}$ on the left, $\sqrt{(10-x)^2+ H^2}$ on the right). Those must be equal in order that the weight must not move from side to side. That was the first of the two equations skeeter gave.

The vertical force on each side, divided by the tension in the rope on that side, is equal to the vertical distance, H, divided by the length of the hypotenuse ( $\sqrt{x^2+ H^2}$ on the left, $\sqrt{(10-x)^2+ H^2}$ on the right). The total of those two must be the weight itself. That is the second equation skeeter gave.

i agree that
$T_1 \sin{\theta} = T_2 \sin{\phi}$

but why is
$T_1 \sin{\theta} + T_2 \sin{\phi} = 175$ is like saying that H+H=175 and i think that just 1 H is=175

**skeeter** · February 19th 2009, 06:03 AM

Originally Posted by juanfe_zodiac

i agree that
$T_1 \sin{\theta} = T_2 \sin{\phi}$

the above equation is incorrect

but why is
$T_1 \sin{\theta} + T_2 \sin{\phi} = 175$ is like saying that H+H=175 and i think that just 1 H is=175

the horizontal components of $T_1$ and $T_2$ are equal ...

$T_1 \cos{\theta} = T_2 \cos{\phi}$

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