1. Problem

2. Draw

Free-body diagram:

3. Knowns and Unknowns

Knowns:

• w = 220 N/m
• OA = 0.5 m
• AC = 0.2 m
• AB = 0.4 m
• L = 1.9 m

Unknowns:

• N_c
• V_c
• M_c

4. Approach

Use equilibrium equations. First solve for reaction forces, then make a cut at C and solve for the internal forces.

5. Analysis

$$w=\frac{F}{L}\\F=wL\\F_R=220N/m\cdot 1.9m\\F_R=418N\\\sum F_X=0=B_X$$

Find reaction forces:

$$\sum M_A=0=B_y(0.4m)-F_R(0.45m)\\(0.4m)B_y=418N(0.45m)\\B_y=\frac{188.1 N\cdot m}{0.4m}\\B_y=470.25N$$

$$\sum F_y=0=-F_R+A_y+B_y\\A_y=F_R-B_y\\A_y=418N-470.25N\\A_y=-52.25N$$

The answer we got for A_y is negative, which means that the arrow should be drawn in the other direction. We will change it for our next sketch.

Make a cut at C:

Now solve for the internal forces:

$$\sum F_x=0\:\:;\:\:N_c=0\\\sum F_y=0=-A_y-V_c-(w\cdot L_A)\\V_c=-52.25N-(220N/m\cdot 0.7m)\\V_c=-206.25 N\\\sum M_c=A_y(0.2m)+M_c+(F_{Rc}\cdot 0.35m)\\M_c=-52.25 N (0.2m)-(220N/m\cdot 0.7m\cdot 0.35m)\\M_c=-64.35N\cdot m$$

Final FBD, showing the arrows in the correct directions:

6. Review

It makes sense that A_y and B_y are in different directions, because the resultant force Fr of the solar panel on the beam is not between A and B. It also makes sense that the moment at C is in the clockwise direction rather than the counterclockwise directions, when you think about the direction of the forces applied to the beam.

1. Problem

2. Draw

Sketch:

Free-body diagram:

3. Knowns and Unknowns

Knowns:

• F1 = 500 N
• F2 = 300 N

Unknowns:

• A_y
• A_x
• C_y
• V_B
• M_B
• N_B

4. Approach

Shear/moment equations, EOM equations

5. Analysis

Solve for reaction forces:

(Ax, Cy)

\begin{aligned}
\sum F_{x}=0=A_{x}=0 \\
\sum M_{A}=0 &=-F_{1} \cdot 2 m-F_{2} \cdot 5.5 m+C_{y} \cdot 8 m \\
C_{y}=&\frac{F_{1} \cdot 2 m+F_{2} \cdot 5.5 m}{8m} \\
C_{y} &=\frac{500 N \cdot 2 m+300 N \cdot 5.5 m}{8 m} \\
C_{y} &=331.25 \mathrm{~N}
\end{aligned}

(Ay)

\begin{aligned}
\sum F_{y}=0 &=A_{y}+C_{y}-F_{1}-F_{2} \\
A_{y} &=F_{1}+F_{2}-C_{y} \\
A_{y} &=500 \mathrm{~N}+300 \mathrm{~N} – 331.25 \mathrm{~N} \\
A_{y} &=468.75 \mathrm{~N}
\end{aligned}

Cut 1: at B

\begin{aligned}
\sum F_{X}=0=A_{X} &+N_{B}=0 \\
& N_{B}=0 \\
\sum F_{y}=0 &=A_{y}-V_{B}-F_{1} \\
V_{B} &=A_{y}-F_{1} \\
V_{B} &=468.75 N – 500 N \\
V_{B}=-31.25 N
\end{aligned}

\begin{aligned}
\sum M_{B}=& 0=-A_{y}(4 m)+F_{1}(2 m)+M_{B} \\
& M_{B}=A_{y}(4 m)-F_{1}(2 m) \\
& M_{B}=468.75 N(4 m)-500 N(2 m) \\
M_{B} &=875 \mathrm{~N} \cdot \mathrm{m}
\end{aligned}

Cut 2: At the point where F₁ is applied

\begin{aligned}
\sum M_{1}=0 &=-A_{y}(2 m)+M_{1}=0 \\
M_{1} &=A_{y}(2 m) \\
M_{1} &=468.75 N(2 m) \\
M_{1} &=937.5 \mathrm{~N} \cdot m
\end{aligned}

Cut 3: At the point where F₂ is applied

\begin{aligned}
\sum M_{2}=0 =-A_{y}(5.5 \mathrm{~m})+F_{1}(3.5 \mathrm{~m})+M_{2} \\
M_{2} =A_{y}(5.5 \mathrm{~m})-F_{1}(3.5 \mathrm{~m}) \\
M_{2} =468.75 \mathrm{~N}(5.5 \mathrm{~m})-500 \mathrm{~N}(3.5 \mathrm{~m}) \\
M_{2} =828.125 \mathrm{~N} \cdot \mathrm{m}
\end{aligned}

Answer: N_B = 0, V_B = -31.25 N, M_B = 875 Nm

6. Review

The reaction forces make sense as they offset the applied forces. The shear/moment diagrams returned to zero so they are correct too. The moment found at B is in the moment diagram, it is smaller than the maximum.