Step 1

We are given that the force applied at handle is $F = 350\;{\rm{N}}$ and the angle of tilt of handle is $\theta = 30^\circ $.

We are asked to calculate the clamping force at point A.

We have the horizontal distance between point A and C is ${d_1} = 235\;{\rm{mm}}$.

We have the horizontal distance between point C and B is ${d_2} = 70\;{\rm{mm}}$.

We have the distance between point B and point of force is ${d_3} = 275\;{\rm{mm}}$.

We have the vertical distance between point C and B is ${h_1} = 30\;{\rm{mm}}$.

We have the vertical distance between point B and E is ${h_2} = 30\;{\rm{mm}}$.

 
Step 2

The free body diagram of the handle is shown as:

 
Step 3

Applying the moment of force equation about point C:

\[{F_{BE}}\cos \theta {d_2} - {F_{BE}}\sin \theta {h_1} - F\cos \theta \left( {{d_3}\cos \theta + {d_1}} \right) - F\sin \theta \left( {{d_3}\cos \theta } \right) = 0\]
 
Step 4

Substitute the known values in the equation:

\[\begin{array}{c} \left\{ \begin{array}{l} {F_{BE}}\cos 30^\circ \left( {70\;{\rm{mm}}} \right) - {F_{BE}}\sin 30^\circ \left( {30\;{\rm{mm}}} \right) - \\ \left( {350\;{\rm{N}}} \right)\cos 30^\circ \left( {\left( {275\;{\rm{mm}}} \right)\cos 30^\circ + 70\;{\rm{mm}}} \right) - \\ \left( {350\;{\rm{N}}} \right)\sin 30^\circ \left( {\left( {275\;{\rm{mm}}} \right)\sin 30^\circ } \right) \end{array} \right\} = 0\\ \left\{ \begin{array}{l} {F_{BE}}\left( {0.866} \right)\left( {70\;{\rm{mm}}} \right) - {F_{BE}}\left( {0.5} \right)\left( {30\;{\rm{mm}}} \right) - \\ \left( {350\;{\rm{N}}} \right)\left( {0.866} \right)\left( {\left( {275\;{\rm{mm}}} \right)\left( {0.866} \right) + 70\;{\rm{mm}}} \right) - \\ \left( {350\;{\rm{N}}} \right)\left( {0.5} \right)\left( {\left( {275\;{\rm{mm}}} \right)\left( {0.5} \right)} \right) \end{array} \right\} = 0\\ {F_{BE}}\left( {45.62\;{\rm{mm}}} \right) = 93400.3\;{\rm{N}} \cdot {\rm{mm}} + 24602.5\;{\rm{N}} \cdot {\rm{mm}}\\ {F_{BE}} = 2574.8\;{\rm{N}} \end{array}\]
 
Step 5

Applying the equilibrium force of equation along x-axis:

\[\begin{array}{c} \sum {F_x} = 0\\ {C_x} + F\sin \theta - {F_{BE}}\sin \theta = 0 \end{array}\]
 
Step 6

Substitute the known values in the equation:

\[\begin{array}{c} {C_x} + \left( {350} \right)\sin 30^\circ - \left( {2574.8\;{\rm{N}}} \right)\sin 30^\circ = 0\\ {C_x} = \left( {2574.8\;{\rm{N}}} \right)\left( {0.5} \right) - \left( {350\;{\rm{N}}} \right)\left( {0.5} \right)\\ = 1287.4\;{\rm{N}} - 175\;{\rm{N}}\\ = 1112.4\;{\rm{N}} \end{array}\]
 
Step 7

The free body diagram of the member CD is shown as:

 
Step 8

Applying the moment of force equation about point D:

\[{C_x}\left( {{h_1} + {h_2}} \right) - {F_C}{d_1} = 0\]
 
Step 9

Substitute the known values in the equation:

\[\begin{array}{c} \left( {1112.4\;{\rm{N}}} \right)\left( {30\;{\rm{mm}} + {\rm{30}}\;{\rm{mm}}} \right) - {F_C}\left( {235\;{\rm{mm}}} \right) = 0\\ {F_C}\left( {235\;{\rm{mm}}} \right) = \left( {1112.4\;{\rm{N}}} \right)\left( {60\;{\rm{mm}}} \right)\\ {F_C} = \frac{{\left( {1112.4\;{\rm{N}}} \right)\left( {60\;{\rm{mm}}} \right)}}{{\left( {235\;{\rm{mm}}} \right)}}\\ = 284\;{\rm{N}} \end{array}\]