Determine the components of the support reactions at the fixed support A on the cantilevered beam.

### Solution:

Show me the final answerâ†“

Our first step is to draw a free body diagram like so:

Remember that since the beam is attached at a fixed support, a moment is also created at A.

Let us now write an equilibrium equation for the x-axis forces.

\rightarrow^+ \Sigma F_x=0;

4\text{cos}\,(30^0)-A_x=0

A_x=3.464 kN

4\text{cos}\,(30^0)-A_x=0

A_x=3.464 kN

Next, we can write one for the y-axis forces.

\uparrow^+ \Sigma F_y=0;

A_y-6-4\text{sin}\,(30^0)=0

A_y=8 kN

A_y-6-4\text{sin}\,(30^0)=0

A_y=8 kN

Lastly, we can write a moment equation at point A.

\circlearrowleft^+\Sigma M_A=0;

M_A-6(1.5)-(4\text{sin}\,(30^0))(1.5+1.5+1.5\text{cos}\,(30^0))-(4\text{cos}\,(30^0))(1.5\text{sin}\,(30^0))=0

M_A=20.196 \,\text{kN}\cdot\text{m}

M_A-6(1.5)-(4\text{sin}\,(30^0))(1.5+1.5+1.5\text{cos}\,(30^0))-(4\text{cos}\,(30^0))(1.5\text{sin}\,(30^0))=0

M_A=20.196 \,\text{kN}\cdot\text{m}

### Final Answer:

A_x=3.46 kN

A_y=8 kN

M_A=20.2 \,\text{kN}\cdot\text{m}

A_y=8 kN

M_A=20.2 \,\text{kN}\cdot\text{m}