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:
Show me the final answer↓
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).