# The 30-kg pipe is supported at A

The 30-kg pipe is supported at A by a system of five cords. Determine the force in each cord for equilibrium. Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

Let us draw a free body diagram focusing on ring A. Now, we will write an equation of equilibrium for y-axis forces.

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

$T_{AB}\text{sin}\,(60^0)\,-\,294.3\,=\,0$ N

Solve for $T_{AB}$:

$T_{AB}\,=\,339.8$ N

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

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

$T_{AE}\,-\,339.8\text{cos}\,(60^0)\,=\,0$

(Remember we just found $T_{AB}\,=\,339.8$ N)

Solve for $T_{AE}$:

$T_{AE}\,=\,169.9$ N

We can now focus on ring B and draw a free body diagram. Again, we will write an equation of equilibrium for y-axis forces.

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

$T_{BD}\dfrac{3}{5}\,-\,339.8\text{sin}\,(60^0)\,=\,0$

Solve for $T_{BD}$:

$T_{BD}\,=\,490.4$ N

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

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

$490.4\left(\dfrac{4}{5}\right)\,+\,339.8\text{cos}\,(60^0)\,-\,T_{BC}\,=\,0$

(Remember, we found $T_{BD}\,=\,490.4$ N)

Solve for $T_{BC}$:

$T_{BC}\,=\,562.2$ N

$T_{AB}\,=\,339.8$ N
$T_{AE}\,=\,169.9$ N
$T_{BD}\,=\,490.4$ N
$T_{BC}\,=\,562.2$ N