Step 1

We are given the weight of the uniform beam as $w = 50\,{\rm{lb}}$, the weight of the block as ${w_b} = 100\,{\rm{lb}}$, the coefficient of static friction between the beam and the block & the rope and peg as ${\mu _s} = 0.4$, the length of the beam as $l = 10\,{\rm{ft}}$, and the height of the cord s $h = 1\,{\rm{ft}}$.

We are asked to determine the maximum distance that the block can be placed from $A$ and still remain in equilibrium.

 
Step 2

Assuming that the block will not tip,

The free body diagram of the block can be drawn as shown below.

Here, $x$ is the distance of the normal force from the central axis of the block.

 
Step 3

Find the normal force at the block by balancing force along $y$-axis using the following relation.

\[N - {w_b} = 0\]

On substituting the values in the above equation we get,

\[\begin{array}{c} N - 100\,{\rm{lb}} = 0\\ N = 100\,{\rm{lb}} \end{array}\]
 
Step 4

Find the friction force between the block and the beam using the following relation.

\[F = {\mu _s}N\]

On substituting the known values in the above equation we get,

\[\begin{array}{c} F = \left( {0.4} \right)\left( {100\,{\rm{lb}}} \right)\\ = 40\,{\rm{lb}} \end{array}\]
 
Step 5

Find the tension in the rope by balancing force along $x$-axis using the following relation.

\[T - F = 0\]

On substituting the known values in the above equation we get,

\[\begin{array}{c} T - 40\,{\rm{lb}} = 0\\ T = 40\,{\rm{lb}} \end{array}\]
 
Step 6

Find the distance $x$ by taking moment about $O$ using the following relation.

\[N \times x - T \times h = 0\]

On substituting the known values in the above equation we get,

\[\begin{array}{c} \left( {100\,{\rm{lb}}} \right) \times x - \left( {40\,{\rm{lb}}} \right) \times \left( {1\,{\rm{ft}}} \right) = 0\\ x = 0.4\,{\rm{ft}} \end{array}\]
 
Step 7

The free body diagram of the peg can be drawn as shown below.

 
Step 8

Find the tension in the rope on the beam side using the following relation.

\[T' = T{e^{{\mu _s}\theta }}\]

Here, the angle of contact of the rope to the peg is $\theta $ having value as $\theta = 90^\circ $.

On substituting the known values in the above equation we get,

\[\begin{array}{c} T' = \left( {40\,{\rm{lb}}} \right){e^{\left( {0.4} \right)\left( {90^\circ \times \left( {\frac{\pi }{{180^\circ }}} \right)} \right)}}\\ = \left( {40\,{\rm{lb}}} \right){e^{\left( {0.4} \right)\left( {\frac{\pi }{2}} \right)}}\\ = \left( {40\,{\rm{lb}}} \right)\left( {1.8745} \right)\\ = 75\,{\rm{lb}} \end{array}\]
 
Step 9

The free body diagram of the beam can be drawn as shown below.

 
Step 10

Find the distance $d$ by taking moment about $A$ using the following relation.

\[T' \times l - w \times \left( {\frac{l}{2}} \right) - N \times \left( {d + x} \right) = 0\]

On substituting the known values in the above equation we get,

\[\begin{array}{c} \left( {75\,{\rm{lb}}} \right) \times \left( {10\,{\rm{ft}}} \right) - \left( {50\,{\rm{lb}}} \right) \times \left( {\frac{{10\,{\rm{ft}}}}{2}} \right) - \left( {100\,{\rm{lb}}} \right) \times \left( {d + 0.4\,{\rm{ft}}} \right) = 0\\ \left( {75\,{\rm{lb}} \cdot {\rm{ft}}} \right) - \left( {25\,{\rm{lb}} \cdot {\rm{ft}}} \right) - \left( {10\,{\rm{lb}}} \right) \times \left( {d + 0.4\,{\rm{ft}}} \right) = 0\\ d + 0.4\,{\rm{ft}} = 5\,{\rm{ft}}\\ d = 4.6\,{\rm{ft}} \end{array}\]