Step 1

We are given the maximum tension of each cord as ${T_{BC}} = {T_{BD}} = {T_{BA}} = {T_{HA}} = {T_{AE}} = 500\;{\rm{N}}$.

We are asked to determine the largest mass of pipe.

 
Step 2

The free-body diagram of the forces acting at point H can be drawn as:

Here, ${T_{HA}}$ is the tension in the cord HA and W is the weight of the pipe.

To find the force in cord HA, we need to apply equilibrium condition along the vertical direction as:

\[\begin{array}{c} \sum {F_y} = 0\\ {T_{HA}} - W = 0\\ {T_{HA}} = W \end{array}\]

To find the force in the cord AB and AE, the free-body diagram of the forces acting at point A can be drawn as:

Here, ${T_{AE}}$ is the tension in the rope AE and ${T_{AB}}$ is the tension in the rope AB.

On applying the equilibrium condition along y-axis, we get:

\[\begin{array}{c} \sum {F_y} = 0\\ {T_{AB}}\sin 60^\circ - W = 0\\ {T_{AB}} = 1.154W \end{array}\]

Similarly, on applying the equilibrium condition along x-axis, we get:

\[\begin{array}{c} \sum {F_x} = 0\\ {T_{AE}} - {T_{AB}}\cos 60^\circ = 0\\ {T_{AE}} = \left( {1.154W} \right)\cos 60^\circ \\ {T_{AE}} = 0.577W \end{array}\]
 
Step 3

To find the force in the cord BD and BC, the free-body diagram of the forces acting at point B can be drawn as:

Here, ${T_{BD}}$ is the tension in the rope BD and ${T_{BC}}$ is the tension in the rope BC.

On applying the equilibrium condition along the vertical direction, we get:

\[\begin{array}{c} \sum {F_y} = 0\\ {T_{BD}}\left( {\frac{3}{5}} \right) - {T_{AB}}\sin 60^\circ = 0\\ {T_{BD}}\left( {\frac{3}{5}} \right) = \left( {1.154W} \right) \times \sin 60^\circ \\ {T_{BD}} = 1.66W \end{array}\]

Similarly, on applying the equilibrium condition along the horizontal direction, we get:

\[\begin{array}{c} \sum {F_x} = 0\\ {T_{BD}}\left( {\frac{4}{5}} \right) + {T_{AB}}\cos 60^\circ - {T_{BC}} = 0\\ \left( {1.66W} \right) \times \left( {\frac{4}{5}} \right) + \left( {1.154W} \right)\cos 60^\circ - {T_{BC}} = 0\\ {T_{BC}} = 1.905W \end{array}\]

It is given that,

\[{T_{BC}} = 500\;{\rm{N}}\]

Substitute the value in the above relation, we get:

\[\begin{array}{c} \left( {500\;{\rm{N}}} \right) = 1.905W\\ W = 262.46\;{\rm{N}} \end{array}\]

Hence, the mass of the pipe can be calculated as:

\[\begin{array}{c} W = mg\\ \left( {262.46\;{\rm{N}}} \right) = m \times \left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)\\ m = 26.7\;{\rm{kg}} \end{array}\]