Step 1

We are given the length of cable $ABC$ as ${l_{ABC}} = 5\;{\rm{m}}$, the sack of mass $m = 100\;{\rm{kg}}$, and the height of $C$ from $A$ is $h = 0.75\;{\rm{m}}$.

We are asked to determine position $x$.

We are asked to determine the tension in $ABC$.

 
Step 2

We first draw a free body diagram of the all the force acting at point $A$.

Figure-(1)

 
Step 3

First find the weight of the sack using the following relation.

\[W = mg\]

Substitute the known values in the above expression.

\[\begin{array}{c} W = \left( {100\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\\ = 981\;{\rm{N}} \end{array}\]
 
Step 4

We add all the force acting along the $x$-axis in the equilibrium condition at point $A$.

\[ - {F_{ABC}}\cos \theta + {F_{ABC}}\cos \phi = 0\]

Substitute the known values in the above expression.

\[\begin{array}{c} {F_{ABC}}\cos \theta = {F_{ABC}}\cos \phi \\ \cos \theta = \cos \phi \\ \theta = \phi \end{array}\]
 
Step 5

Now find the length of $AB$ using the following relation.

\[\cos \theta = \frac{x}{{{l_{AB}}}}\]

Rearrange the relation for length of $AB$.

\[{l_{AB}} = \frac{x}{{\cos \theta }}\]
 
Step 6

Now find the length of $BC$ using the following relation.

\[\cos \theta = \frac{{AC - x}}{{{l_{BC}}}}\]

Substitute the known values in the above expression.

\[\begin{array}{c} \cos \theta = \frac{{3.5\;{\rm{m}} - x}}{{{l_{BC}}}}\\ {l_{BC}} = \frac{{3.5\;{\rm{m}} - x}}{{\cos \theta }} \end{array}\]
 
Step 7

To find the length of $ABC$ we will use the following relation.

\[{l_{ABC}} = {l_{AB}} + {l_{BC}}\]

Substitute the known values in the above expression.

\[\begin{array}{c} 5\;{\rm{m}} = \frac{x}{{\cos \theta }} + \frac{{3.5\;{\rm{m}} - x}}{{\cos \theta }}\\ \cos \theta = \frac{{3.5\;{\rm{m}}}}{{5\;{\rm{m}}}}\\ \theta = 45.57^\circ \end{array}\]
 
Step 6

We add all the force acting along the $y$-axis in the equilibrium condition at point $A$.

\[{F_{ABC}}\sin \theta + {F_{ABC}}\sin \phi - W = 0\]

Substitute the known values in the above expression.

\[\begin{array}{c} {F_{ABC}}\sin 45.57^\circ + {F_{ABC}}\sin 45.57^\circ - 981\;{\rm{N}} = 0\\ 1.43{F_{ABC}} = 981\;{\rm{N}}\\ {F_{ABC}} = 686.01\;{\rm{N}} \end{array}\]
 
Step 7

Now find the vertical distance between $B$ and $C$ using the following relation.

\[a = AB\sin \theta + h\]

Substitute the known values in the above expression.

\[a = \left( {\frac{x}{{\cos \theta }}} \right)\sin 45.57^\circ + 0.75\;{\rm{m}}\]

Substitute the known values in the above expression.

\[\begin{array}{c} a = \left( {\frac{x}{{\cos 45.57^\circ }}} \right)\sin 45.57^\circ + 0.75\;{\rm{m}}\\ a = 1.02x + 0.75\;{\rm{m}} \end{array}\]
 
Step 8

Now find the vertical distance between $B$ and $C$ using the following relation.

\[a = \left( {AC - x} \right)\tan \theta \]

Substitute the known values in the above expression.

\[\begin{array}{c} 1.02x + 0.75\;{\rm{m}} = \left( {3.5\;{\rm{m}} - x} \right)\tan 45.57^\circ \\ 1.02x + 0.75\;{\rm{m}} = 3.57\;{\rm{m}} - 1.02x\\ 2.04x = 2.82\;{\rm{m}}\\ x = 1.38\;{\rm{m}} \end{array}\]