Step 1

We are given the mass of skid-steer loader ${M_1} = 1.18\;{\rm{Mg}}$, the mass of the bucket with stone as ${M_2} = 30{\rm{0}}\;{\rm{kg}}$, the length $AB$ as $AB = 0.75\;{\rm{m}}$, and the length $DE$ as $DE = 1.25\;{\rm{m}}$.

We are asked to determine the reaction force at the wheel $A$ and $B$, and the force in the hydraulic cylinder $CD$ and at the pin $E$.

 
Step 2

We will draw a free body diagram of the skid-steer loader.

Here, ${N_A}$ is the reaction force at point $A$, ${N_B}$ is the reaction force at point $B$, and $g = 9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$ is the acceleration due to gravity.

The horizontal length ${G_2}A$ is ${G_2}{A_x} = 1.5\;{\rm{m}}$.

 
Step 3

We will take the moment about point $A$ to find the reaction force at point $B$.

\[\begin{array}{c} \sum {{M_A}} = 0\\ {N_B} \times AB + {M_2}g \times {G_2}A - {M_1}g \times AC = 0\\ {N_B} \times AB + {M_2}g \times {G_2}A = {M_1}g \times AC \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} \left[ \begin{array}{l} {N_B} \times \left( {0.75\;{\rm{m}}} \right) + \\ \left( {30{\rm{0}}\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {1.5\;{\rm{m}}} \right) \end{array} \right] = \left( {1.18\;{\rm{Mg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {0.6\;{\rm{m}}} \right)\\ {N_B} \times \left( {0.75\;{\rm{m}}} \right) = \left[ \begin{array}{l} \left( {1.18\;{\rm{Mg}} \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{\rm{3}}}\;{\rm{kg}}}}{{{\rm{1}}\;{\rm{Mg}}}}} \right)} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {0.6\;{\rm{m}}} \right) - \\ \left( {30{\rm{0}}\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {1.5\;{\rm{m}}} \right) \end{array} \right]\\ {N_B} = \frac{{\left[ {\left( {6945.48\;{\rm{kg}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{s}}^{\rm{2}}}} \right) - \left( {4414.5\;{\rm{kg}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{s}}^{\rm{2}}}} \right)} \right]}}{{\left( {0.75\;{\rm{m}}} \right)}} \times \left( {\frac{{{\rm{1}}\;{\rm{N}}}}{{{\rm{1}}\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}}}} \right)\\ = 3374.64\;{\rm{N}} \end{array}\]
 
Step 4

We will resolve the force in vertical direction to find the reaction force at point $A$.

\[\begin{array}{c} \sum {{F_y}} = 0\\ {N_A} + {N_B} - {M_1}g - {M_2}g = 0\\ {N_A} = \left( {{M_1} + {M_2}} \right)g - {N_B} \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} {N_A} = \left( {1.18\;{\rm{Mg}} + 30{\rm{0}}\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) - 3374.64\;{\rm{N}}\\ {N_A} = \left( {1.18\;{\rm{Mg}} \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{\rm{3}}}\;{\rm{kg}}}}{{{\rm{1}}\;{\rm{Mg}}}}} \right) + 30{\rm{0}}\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {\frac{{{\rm{1}}\;{\rm{N}}}}{{{\rm{1}}\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}}}} \right) - 3374.64\;{\rm{N}}\\ = 11144.16\;{\rm{N}} \end{array}\]
 
Step 5

We will draw a free body diagram of the bucket.

Here, ${F_{CD}}$ is the force in the member $CD$, ${E_y}$ is the reaction force at point $E$, and ${E_x}$ is the reaction force at point $E$.

The length ${G_2}D$ is ${G_2}D = 1.5\;{\rm{m}}$.

 
Step 6

We will take the moment about point $E$ to find the reaction force $CD$.

\[\begin{array}{c} \sum {{M_E}} = 0\\ {M_2}g \times \left( {{G_2}D + DE} \right) - {F_{CD}}\sin 30^\circ \times DE = 0\\ {F_{CD}}\sin 30^\circ \times DE = {M_2}g \times \left( {{G_2}D + DE} \right) \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} {F_{CD}}\sin 30^\circ \times \left( {{\rm{1}}{\rm{.25}}\;{\rm{m}}} \right) = \left( {30{\rm{0}}\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {{\rm{1}}{\rm{.5}}\;{\rm{m}} + {\rm{1}}{\rm{.25}}\;{\rm{m}}} \right)\\ {F_{CD}} = \frac{{\left( {30{\rm{0}}\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {{\rm{1}}{\rm{.5}}\;{\rm{m}} + {\rm{1}}{\rm{.25}}\;{\rm{m}}} \right)}}{{\sin 30^\circ \times \left( {{\rm{1}}{\rm{.25}}\;{\rm{m}}} \right)}} \times \left( {\frac{{{\rm{1}}\;{\rm{N}}}}{{{\rm{1}}\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}}}} \right)\\ = 12949.2\;{\rm{N}} \end{array}\]
 
Step 7

There are two hydraulic cylinder placed on each side of the loader.

We will find the force in each cylinder.

\[{F_{CD}}' = \frac{{{F_{CD}}}}{2}\]

Substitute the given value in the above equation.

\[\begin{array}{c} {F_{CD}}' = \frac{{12949.2\;{\rm{N}}}}{2}\\ = 6474.6\;{\rm{N}} \end{array}\]
 
Step 8

We will resolve the force in vertical direction to find the reaction force ${E_y}$.

\[\begin{array}{c} \sum {{F_y}} = 0\\ {F_{CD}}\sin 30^\circ - {E_y} - {M_2}g = 0\\ {E_y} = {F_{CD}}\sin 30^\circ - {M_2}g \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} {E_y} = \left( {12949.2\;{\rm{N}}} \right)\sin 30^\circ - \left( {30{\rm{0}}\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\\ = \left( {12949.2\;{\rm{N}}} \right)\sin 30^\circ - \left( {30{\rm{0}}\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {\frac{{{\rm{1}}\;{\rm{N}}}}{{{\rm{1}}\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}}}} \right)\\ = 3531.6\;{\rm{N}} \end{array}\]
 
Step 9

We will resolve the force in vertical direction to find the reaction force ${E_y}$.

\[\begin{array}{c} \sum {{F_x}} = 0\\ {E_x} = {F_{CD}}\sin 30^\circ \end{array}\]

Substitute the given value in the above equation.

\[\begin{array}{c} {E_x} = \left( {12949.2\;{\rm{N}}} \right) \times \cos 30^\circ \\ \approx 11214.3\;{\rm{N}} \end{array}\]
 
Step 10

We will find the force acting at point $E$.

\[{F_E} = \sqrt {{{\left( {{E_x}} \right)}^2} + {{\left( {{E_y}} \right)}^2}} \]

Substitute the given value in the above equation.

\[\begin{array}{c} {F_E} = \sqrt {{{\left( {11214.3\;{\rm{N}}} \right)}^2} + {{\left( {3531.6\;{\rm{N}}} \right)}^2}} \\ \approx 11757.2\;{\rm{N}} \end{array}\]
 
Step 11

There are two members on each side of the loader.

We will find the force in each member.

\[{F_E}' = \frac{{{F_E}}}{2}\]

Substitute the given value in the above equation.

\[\begin{array}{c} {F_E}' = \frac{{11757.2\;{\rm{N}}}}{2}\\ = 5878.6\;{\rm{N}} \end{array}\]