Determine the magnitude of the reactions on the beam at A and B. Neglect the thickness of 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 like so:
We can now write a moment equation at point A to determine force B_y:
\circlearrowright^+ M_A=0;
600(4)+400\text{cos}\,(15^0)(12)-B_y(12)=0
B_y=586.37 N
600(4)+400\text{cos}\,(15^0)(12)-B_y(12)=0
B_y=586.37 N
Now, we can write an equilibrium equation for the x-axis forces.
\rightarrow^+\Sigma F_x=0;
A_x-400\text{sin}\,(15^0)=0
A_x=103.53 N
A_x-400\text{sin}\,(15^0)=0
A_x=103.53 N
Lastly, we can write an equilibrium equation for the y-axis forces.
\uparrow^+ \Sigma F_y=0;
A_y+B_y-600-400\text{cos}\,(15^0)=0
A_y+B_y-600-400\text{cos}\,(15^0)=0
(Remember we already found B_y=586.37 N)
A_y=400 N
Now that we know A_x and A_y, we can find the magnitude. Remember, magnitude is:
F_A=\sqrt{(A_x)^2+(A_y)^2}
F_A=\sqrt{(103.53)^2+(400)^2}
F_A=413.18 N
F_A=\sqrt{(103.53)^2+(400)^2}
F_A=413.18 N
Final Answer:
F_A=413.18 N
F_B=B_y=586.37 N
How can you answer if the question doesn’t not have an angle…..how can you find the magnitude
What angle is not given?