# Determine the tension developed in each wire

Determine the tension developed in each wire used to support the 50-kg chandelier. Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

We will first draw a free body diagram around ring D. We can now write our equations of equilibrium.

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

$T_{DC}\text{cos}\,(30^0)\,-\,T_{DB}\text{cos}\,(45^0)\,=\,0$ (eq.1)

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

$T_{DC}\text{sin}\,(30^0)\,+\,T_{DB}\text{sin}\,(45^0)\,-\,490.5\,=\,0$ (eq.2)

Now we can solve for $T_{DC}$ and $T_{DB}$ by first isolating for $T_{DC}$ in eq.1.

$T_{DC}\,=\,\dfrac{T_{DB}\text{cos}\,(45^0)}{\text{cos}\,(30^0)}$

(Simplify)

$T_{DC}\,=\,0.816T_{DB}$ (eq.3)

Substitute this value into eq.2 to figure out $T_{DC}$.

$0.816T_{DB}\text{sin}\,(30^0)\,+\,T_{DB}\text{sin}\,(45^0)\,-\,490.5\,=\,0$

(Solve for $T_{DB}$)

$T_{DB}\,=\,439.9$ N

Substitute this value back into eq.3 to figure out $T_{DC}$.

$T_{DC}\,=\,(0.816)(439.9)$

$T_{DC}\,=\,358.9$ N

Let us now switch our attention to ring B by drawing a free body diagram. (Remember we found $T_{DB}\,=\,439.9$ N)

Again, we will write our equations of equilibrium, however, this time, we will write our equation of equilibrium for y-axis forces first. Why? It will give us a direct answer to $T_{BA}$.

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

$T_{BA}\text{sin}\,(30^0)\,-\,439.9\text{sin}\,(45^0)\,=\,0$

(Solve for $T_{BA}$)

$T_{BA}\,=\,622.1$ N

Now, we can write our equation of equilibrium for the x-axis forces.

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

$439.9\text{cos}\,(45^0)\,+\,T_{BC}\,-\,622.1\text{cos}\,(30^0)\,=\,0$

(Remember we just found $T_{BA}\,=\,622.1$ N)

Solve for $T_{BC}$

$T_{BC}\,=\,227.7$ N

$T_{DB}\,=\,439.9$ N
$T_{DC}\,=\,358.9$ N
$T_{BA}\,=\,622.1$ N
$T_{BC}\,=\,227.7$ N