Determine the horizontal and vertical components of reaction at the pin A and the reaction of the rocker B on the beam.
Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.
Solution:
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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
