Step 1

We are given the following data:

The magnitude of the force is $F = 800\;{\rm{N}}$.

The magnitude of the uniform loading is $W = 200\;{\rm{N/m}}$.

The value of angle is $\theta = 60^\circ $.

We are asked to determine the resultant force at pins $A$, $B$ and $C$ on the three member frame.

 
Step 2

Let us consider a points on the vertical member of the frame is $E$ and a point $D$ on the horizontal axis perpendicular below to the point $C$.

The distance between the point $A$ and $D$ is ${x_1}$.

The distance between the point $B$ and $E$ is $BE = 2\;{\rm{m}}$.

The distance between the point $C$ and $E$ is $CE = 2\;{\rm{m}}$.

The distance between the point $C$ and $D$ is $CD = 2\;{\rm{m}}$.

The equation to calculate the resultant force of the uniform distributed loading is given by,

\[{F_1} = \left( W \right) \times \left( {\frac{{CD}}{{\sin \theta }}} \right)\]

Here, ${F_1}$ represent the resultant point force of the uniform distributed load.

Substitute all the known values in the above formula.

\[\begin{array}{c} {F_1} = \left( {200\;{\rm{N/m}}} \right)\left( {\frac{{\left( {2\;{\rm{m}}} \right)}}{{\sin 60^\circ }}} \right)\\ = \left( {200\;{\rm{N/m}}} \right) \times \left( {2.31\;{\rm{m}}} \right)\\ = 462\;{\rm{N}} \end{array}\]
 
Step 3

The formula to calculate the distance of the resultant point force from the point $A$ is given by,

\[{y_1} = \left( {\frac{{CD}}{{2\sin \theta }}} \right)\]

Here, ${y_1}$ represent the distance of the resultant point force from the point $A$.

Substitute all the known values in the above formula.

\[\begin{array}{c} {y_1} = \frac{{\left( {2\;{\rm{m}}} \right)}}{{\left( {2\sin 60^\circ } \right)}}\\ \approx 1.16\;{\rm{m}} \end{array}\]

The distance between the point $A$ and the point of the effective point force is same as the distance between the point of effective point force and the point $C$. So,

\[{y_2} = 1.16\;{\rm{m}}\]
 
Step 4

The equation to calculate the distance between the point $A$ and $D$ is given by,

\[{x_1} = \frac{{CD}}{{\tan \theta }}\]

Substitute all the known values in the above formula.

\[\begin{array}{c} {x_1} = \frac{{\left( {2\;{\rm{m}}} \right)}}{{\tan 60^\circ }}\\ \approx 1.16\;{\rm{m}} \end{array}\]
 
Step 5

We will draw a free body diagram of the part $AC$ of the frame.

Here, ${A_y}$ and ${C_{\rm{y}}}$ are the vertical reaction force, and ${A_{\rm{x}}}$ and ${C_{\rm{x}}}$ are the horizontal force.

We will take the sum of the moments about point $A$ equal to zero to determine the equation between the force ${C_{\rm{x}}}$. and ${C_{\rm{y}}}$.

\[\begin{array}{c} \sum {{M_{\rm{A}}}} = 0\\ \left[ \begin{array}{l} \left( {{C_{\rm{y}}}} \right)\left( {{x_1}} \right) + {C_{\rm{x}}}\left( {CD} \right)\\ - \left( {{F_1}} \right)\left( {{y_1}} \right) \end{array} \right] = 0 \end{array}\]

Substitute the known value in the above equation.

\[\begin{array}{c} \left[ \begin{array}{l} \left( {{C_{\rm{y}}}} \right)\left( {1.16\;{\rm{m}}} \right) + \left( {{C_{\rm{x}}}} \right)\left( {2\;{\rm{m}}} \right)\\ - \left( {462\;{\rm{N}}} \right)\left( {1.16\;{\rm{m}}} \right) \end{array} \right] = 0\\ \left[ {\left( {{C_{\rm{y}}}} \right)\left( {1.16\;{\rm{m}}} \right)} \right] = \left[ \begin{array}{l} \left( {535.92\;{\rm{N}} \cdot {\rm{m}}} \right)\\ - \left( {{C_{\rm{x}}}} \right)\left( {2\;{\rm{m}}} \right) \end{array} \right]\\ {C_{\rm{y}}} = \left[ \begin{array}{l} \left( {462\;{\rm{N}}} \right)\\ - \left( {1.724} \right){C_{\rm{x}}} \end{array} \right] \end{array}\]
 
Step 6

We will draw the free-body diagram of the member $BC$ of the frame.

Here, ${B_{\rm{x}}}$ is the horizontal force and ${B_{\rm{y}}}$ is the vertical force at point $B$ of the frame.

We will take the sum of the moments about point $B$ equal to zero to determine the equation between the force ${C_{\rm{x}}}$. and ${C_{\rm{y}}}$.

\[\begin{array}{c} \sum {{M_{\rm{B}}}} = 0\\ \left[ {\left( {{C_{\rm{y}}}} \right)\left( {CE} \right) + \left( F \right)\left( {BE} \right) - \left( {{C_{\rm{x}}}} \right)\left( {BE} \right)} \right] = 0 \end{array}\]

Substitute the known value in the above equation.

\[\begin{array}{c} \left[ \begin{array}{l} \left\{ {\left( {462\;{\rm{N}}} \right) - \left( {1.724} \right){C_{\rm{x}}}} \right\}\left( {2\;{\rm{m}}} \right)\\ - \left( {{C_{\rm{x}}}} \right)\left( {2\;{\rm{m}}} \right)\\ + \left( {800\;{\rm{N}}} \right)\left( {2\;{\rm{m}}} \right) \end{array} \right] = 0\\ \left[ \begin{array}{l} \left( {2524} \right) - \left( {3.448} \right)\left( {{C_{\rm{x}}}} \right)\\ - \left( {{C_{\rm{x}}}} \right)\left( 2 \right) \end{array} \right]\;{\rm{N}} \cdot {\rm{m}} = - \left( {1600\;{\rm{N}} \cdot {\rm{m}}} \right)\\ \left[ {\left( { - 5.448} \right){C_{\rm{x}}}} \right] = \left( { - 4124\;{\rm{N}}} \right)\\ {C_{\rm{x}}} = 461.88\;{\rm{N}} \end{array}\]
 
Step 7

Substitute the value ${C_{\rm{x}}} = 461.88\;{\rm{N}}$ in the equation ${C_{\rm{y}}} = \left[ {\left( {462\;{\rm{N}}} \right) - \left( {1.724} \right){C_{\rm{x}}}} \right]$ to obtain the value of the force ${C_{\rm{y}}}$

\[\begin{array}{c} {C_{\rm{y}}} = \left[ \begin{array}{l} \left( {462\;{\rm{N}}} \right)\\ - \left( {1.724} \right)\left( {461.88\;{\rm{N}}} \right) \end{array} \right]\\ = \left[ {\left( {462\;{\rm{N}}} \right) - \left( {796.28\;{\rm{N}}} \right)} \right]\\ = - 338.12\;{\rm{N}} \end{array}\]
 
Step 8

For the member $AC$ ,

We will take the sum of the force with $x$ components equal to zero to determine the horizontal force ${A_{\rm{x}}}$.

\[\begin{array}{c} \sum {{F_{\rm{x}}}} = 0\\ \left[ {{A_{\rm{x}}} + \left( {{F_1}} \right)\left( {\sin \theta } \right) - {C_{\rm{x}}}} \right] = 0 \end{array}\]

Substitute all the known values in the above formula.

\[\begin{array}{c} \left[ {{A_{\rm{x}}} + \left( {462\;{\rm{N}}} \right)\left( {\sin 60^\circ } \right) - \left( {461.88\;{\rm{N}}} \right)} \right] = 0\\ \left[ {{A_{\rm{x}}} - \left( {61.8\;{\rm{N}}} \right)} \right] = 0\\ {A_{\rm{x}}} = 61.8\;{\rm{N}} \end{array}\]
 
Step 9

We will take the sum of the force with $y$ components equal to zero to determine the vertical force ${A_{\rm{y}}}$.

\[\begin{array}{c} \sum {{F_{\rm{y}}}} = 0\\ \left[ {{A_{\rm{y}}} + \left( {{C_{\rm{y}}}} \right) - \left( {{F_1}\cos \theta } \right)} \right] = 0 \end{array}\]

Substitute all the known values in the above formula.

\[\begin{array}{c} \left[ {{A_{\rm{y}}} + \left( { - 338.12\;{\rm{N}}} \right) - \left( {462\;{\rm{N}}} \right)\left( {\cos 60^\circ } \right)} \right] = 0\\ \left[ {{A_{\rm{y}}} - \left( {569.1\;{\rm{N}}} \right)} \right] = 0\\ {A_{\rm{y}}} \approx \;569\;{\rm{N}} \end{array}\]
 
Step 10

For the member $BC$,

We will take the sum of the forces along the $x$ axis equal to zero to determine the couple ${B_{\rm{x}}}$.

\[\begin{array}{c} \sum {{F_{\rm{x}}}} = 0\\ \left[ {{B_{\rm{x}}} + {C_{\rm{x}}} - F} \right] = 0 \end{array}\]

Substitute the known value in the above equation.

\[\begin{array}{c} \left[ {{B_{\rm{x}}} + \left( {461.8\;{\rm{N}}} \right) - \left( {800\;{\rm{N}}} \right)} \right] = 0\\ \left[ {\left( {{B_{\rm{x}}}} \right) - \left( {338.2\;{\rm{N}}} \right)} \right] = 0\\ {B_{\rm{x}}} = 338.2\;{\rm{N}} \end{array}\]
 
Step 11

We will take the sum of the forces in $y$ direction to zero to determine the vertical force ${B_{\rm{y}}}$.

\[\begin{array}{c} \sum {{F_{\rm{y}}}} = 0\\ \left[ { - {B_{\rm{y}}} - \left( {{C_{\rm{y}}}} \right)} \right] = 0 \end{array}\]

Substitute the known value in the above equation.

\[\begin{array}{c} \left[ { - {B_{\rm{y}}} - \left( { - 338.12\;{\rm{N}}} \right)} \right] = 0\\ \left[ { - {B_{\rm{y}}} + \left( {338.12\;{\rm{N}}} \right)} \right] = 0\\ {B_{\rm{y}}} = 338.12\;{\rm{N}} \end{array}\]
 
Step 12

The formula to calculate the resultant reaction at point $C$ is given by,

\[{F_{\rm{C}}} = \sqrt {{{\left( {{C_{\rm{x}}}} \right)}^2} + {{\left( {{C_{\rm{y}}}} \right)}^2}} \]

Substitute all the known values in the above formula.

\[\begin{array}{c} {F_{\rm{C}}} = \sqrt {{{\left( {461.8\;{\rm{N}}} \right)}^2} + {{\left( { - 338.12\;{\rm{N}}} \right)}^2}} \\ = \left( {\sqrt {327584.374\;} } \right)\;{\rm{N}}\\ \approx 572\;{\rm{N}} \end{array}\]
 
Step 13

The formula to calculate the resultant reaction at point $A$ is given by,

\[{F_{\rm{A}}} = \sqrt {{{\left( {{A_{\rm{x}}}} \right)}^2} + {{\left( {{A_{\rm{y}}}} \right)}^2}} \]

Substitute all the known values in the above formula.

\[\begin{array}{c} {F_{\rm{A}}} = \sqrt {{{\left( {61.8\;{\rm{N}}} \right)}^2} + {{\left( {569.1\;{\rm{N}}} \right)}^2}} \\ = \left( {\sqrt {327694.05\;} } \right)\;{\rm{N}}\\ \approx 572\;{\rm{N}} \end{array}\]
 
Step 14

The formula to calculate the resultant reaction at point $B$ is given by,

\[{F_{\rm{B}}} = \sqrt {{{\left( {{B_{\rm{x}}}} \right)}^2} + {{\left( {{B_{\rm{y}}}} \right)}^2}} \]

Substitute all the known values in the above formula.

\[\begin{array}{c} {F_{\rm{B}}} = \sqrt {{{\left( {338.12\;{\rm{N}}} \right)}^2} + {{\left( {338.12\;{\rm{N}}} \right)}^2}} \\ = \left( {\sqrt {228650.268\;} } \right)\;{\rm{N}}\\ \approx 478\;{\rm{N}} \end{array}\]