Determine the horizontal and vertical components of reaction at the pin A and the reaction of the rocker B on the beam.

**Solution:**

Show me the final answerâ†“

Let us first draw a free body diagram showing all the forces affecting the beam. Remember that since the beam is in equilibrium, all forces added must equal 0.

We can first solve for N_B by writing a moment equation at point A.

\circlearrowleft^+ \Sigma M_A=0;

N_B\text{cos}\,(30^0)(8)-4(6)=0

N_B=3.464kN

N_B\text{cos}\,(30^0)(8)-4(6)=0

N_B=3.464kN

Now, we can write an equation along the x-axis like so:

\rightarrow^+\Sigma F_x=0;

A_x-3.464\text{sin}\,(30^0)=0

A_x=1.732 kN

A_x-3.464\text{sin}\,(30^0)=0

A_x=1.732 kN

Next, we can write another equation along the y-axis:

\uparrow^+\Sigma F_y=0;

A_y+3.464\text{cos}\,(30^0)-4=0

A_y=1 kN

A_y+3.464\text{cos}\,(30^0)-4=0

A_y=1 kN

**Final Answers:**

A_x=1.73 kN

A_y=1.00 kN

N_B=3.46kN

A_y=1.00 kN

N_B=3.46kN

why you redirected the angle 30 like that? i don’t understand how the angle become like that. kindly explain

It’s a roller at B, which means the force will be perpendicular to the ground, so if you draw it out, you will see that since it’s perpendicular, the angle will be 30 degrees from the y-axis (vertical line).