# Determine the position x and the tension developed

Cable ABC has a length of 5 m. Determine the position x and the tension developed in ABC required for equilibrium of the 100-kg sack. Neglect the size of the pulley at B.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

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Note that we will assume the pulley to be frictionless. This means the tension across the rope at any given point is going to be the same. Knowing this, we can draw a free body diagram like so:

Let us write an equation of equilibrium for the x-axis forces.

$\rightarrow ^+\sum \text{F}_\text{x}\,=\,0$

$T\text{cos}\theta\,-\,T\text{cos}\phi\,=\,0$

(Factor out T and divide both sides of the equation by T)

$\text{cos}\theta\,=\,\text{cos}\phi$

Therefore,

$\theta\,=\,\phi$

Now, we will write an equation of equilibrium for the y-axis forces.

$+\uparrow \sum \text{F}_\text{y}\,=\,0$

$2T\text{sin}\theta\,-\,981\,=\,0$ (eq.1)

(remember, we established that $\theta\,=\,\phi$)

Now, let us draw a diagram of the dimensions given to us like so:

Using trigonometry, we can write the following:

$\dfrac{3.5\,-\,x}{\text{cos}\theta}\,+\,\dfrac{x}{\text{cos}\theta}\,=\,5$

(remember, we know the total length of the rope is 5m)

Isolate for $\theta$:

$\theta\,=\,\text{cos}^{-1}\left(\dfrac{3.5}{5}\right)$

$\theta\,=\,45.6^0$

With the angle found, we can write the following using our diagram:

$x\text{tan}\,(45.6^0)\,+\,0.75\,=\,(3.5\,-\,x)\text{tan}\,(45.6^0)$

(solve for x)

$x\,=\,1.38$ m

Going back to eq.1 which was our equilibrium equation written for the y-axis forces, we can substitute the $\theta$ value we found.

$2T\text{sin}\theta\,-\,981\,=\,0$

$2T\text{sin}\,(45.6^0)\,=\,981$

(solve for T)

$T\,=\,686.5$ N

x = 1.38 m

T = 686.5 N