# Determine the tension developed in each cord 6

Determine the tension developed in each cord required for equilibrium of the 20-kg lamp.

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

#### Solution:

We will start off by drawing a free body diagram focusing on ring D.

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

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

$F_{DE}\text{sin}\,(30^0)\,-\,196.2\,=\,0$

(Solve for $F_{DE}$)

$F_{DE}\,=\,392.4$ N

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

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

$F_{DE}\text{cos}\,(30^0)\,-\,F_{DC}\,=\,0$

(Substitute the value of $F_{DE}$ we found)

$392.4\text{cos}\,(30^0)\,-\,F_{DC}\,=\,0$

(solve for $F_{DC}$)

$F_{DC}\,=\,339.8$ N

We can now draw a free body diagram focusing on ring C.

Again, we will write our equations of equilibrium.

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

$339.8\,-\,F_{CA}\dfrac{3}{5}\,-\,F_{CB}\text{cos}\,(45^0)\,=\,0$ (eq.1)

(Remember that we already found $F_{DC}\,=\,339.8$ N)

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

$F_{CA}\dfrac{4}{5}\,-\,F_{CB}\text{sin}\,(45^0)\,=\,0$ (eq.2)

Let us solve for $F_{CA}$ and $F_{CB}$

Isolate for $F_{CA}$ in eq.2.

$F_{CA}\,=\,\dfrac{F_{CB}\text{sin}\,(45^0)}{\dfrac{4}{5}}$

(Simplify)

$F_{CA}\,=\,0.88F_{CB}$ (eq.3)

Substitute this value into eq.1.

$339.8\,-\,0.88F_{CB}\dfrac{3}{5}\,-\,F_{CB}\text{cos}\,(45^0)\,=\,0$

(solve for $F_{CB}$)

$F_{CB}\,=\,275.1$ N

Substitute this value back into eq.3 to figure out $F_{CA}$.

$F_{CA}\,=\,(0.88)(275.1)$

$F_{CA}\,=\,242.1$ N

$F_{DE}\,=\,392.4$ N

$F_{DC}\,=\,339.8$ N

$F_{CB}\,=\,275.1$ N

$F_{CA}\,=\,242.1$ N