Step 1

We are given the moment developed at point A is $M = 1500\;{\rm{N}} \cdot {\rm{m}}$.

We are asked the smallest force F that must be applied along the rope in order to cause the curved rod to fail at the support.

 
Step 2

The following is the diagram of the system:

We have the radius of curved rod is $R = 4\;{\rm{m}}$.

 
Step 3

The position vector of component AB is,

\[{R_{AB}} = B - A\]
 
Step 4

Substitute the values in the above expression.

\begin{array}{l} {R_{AB}} = \left( {4\sin 45^\circ i + 0j + 4\cos 45^\circ k} \right){\rm{ m}} - \left( {0i + 0j + 4k} \right){\rm{ m}}\\ {R_{AB}} = \left( {2.8i + 0j - 1.2k} \right)\;{\rm{m}} \end{array}
 
Step 5

The position vector of component BC is,

\[{R_{BC}} = C - B\]
 
Step 6

Substitute the values in the above expression.

\begin{array}{l} {R_{BC}} = \left( {6i + 6j + 0k} \right){\rm{ m}} - \left( {4\sin 45^\circ i + 0j + 4\cos 45^\circ k} \right){\rm{ m}}\\ {R_{BC}} = \left( {3.2i + 6j - 2.8k} \right){\rm{ m}} \end{array}
 
Step 7

The Cartesian vector form of force is,

\[F = F \cdot \frac{{{R_{BC}}}}{{\left| {{R_{BC}}} \right|}}\]
 
Step 8

Substitute the values in the above expression.

\begin{array}{l} F = F \cdot \frac{{\left( {3.2i + 6j - 2.8k} \right){\rm{ m}}}}{{\sqrt {{{\left( {3.2} \right)}^2} + {{\left( 6 \right)}^2} + {{\left( { - 2.8} \right)}^2}} \;{\rm{m}}}}\\ F = \left( {0.435Fi + 0.816Fj - 0.380Fk} \right) \end{array}
 
Step 9

The moment of the force about point A is,

\[M = F \times {R_{AB}}\]
 
Step 10

Substitute the values in the above expression.

\[\begin{array}{l} M = \left( {0.435Fi + 0.816Fj - 0.380Fk} \right) \times \left( {2.8i + 0j - 1.2k} \right)\;{\rm{m}}\\ M = \left| {\begin{array}{*{20}{c}} i&j&k\\ {0.435F}&{0.816F}&{ - 0.380F}\\ {2.8}&0&{ - 1.2} \end{array}} \right|\\ M = \left\{ \begin{array}{l} i\left[ {0.816F \times \left( { - 1.2} \right) - \left( { - 0.380F} \right) \times \left( 0 \right)} \right] - j\left[ {0.435F \times \left( { - 1.2} \right) - \left( { - 0.380F} \right) \times \left( {2.8} \right)} \right]\\ + k\left[ {0.435F \times \left( 0 \right) - \left( {0.816F} \right) \times \left( {2.8} \right)} \right] \end{array} \right\}\\ M = \left( { - 0.98Fi - 0.54Fj - 2.28Fk} \right)\;{\rm{m}} \end{array}\]
 
Step 11

The magnitude of the moment of the force about point A is,

\[M = \left| M \right|\]
 
Step 12

Substitute the values in the above expression.

\begin{array}{c} 1500\;{\rm{N}} \cdot {\rm{m}} = \sqrt {{{\left( { - 0.98F} \right)}^2} + {{\left( { - 0.54F} \right)}^2} + {{\left( { - 2.28F} \right)}^2}} \;{\rm{m}}\\ 1500\;{\rm{N}} \cdot {\rm{m}} = 2.54F\;{\rm{m}}\\ F = 590.6\;{\rm{N}} \end{array}