Question

2500 lb 5500 lb 4 7 5 σmx = 0 σv = 0 σmr₂ = 0 = 16r₁-(2500×12)-(5500×5) r₁ = 2,500 / 16

Explanation:

Step1: Analyze moment - equilibrium equation

We have the moment - equilibrium equation about point $R_2$: $\sum M_{R_2}=0 = 16R_1-(2500\times12)-(5500\times5)$. The left - hand side represents the sum of the moments about $R_2$. The first term $16R_1$ is the moment due to the reaction force $R_1$ (with a moment arm of 16 feet), the second term is the moment due to the 2500 - lb force (with a moment arm of 12 feet) and the third term is the moment due to the 5500 - lb force (with a moment arm of 5 feet).

Step2: Solve for $R_1$

First, expand the right - hand side of the equation:
\[

$$\begin{align*} 0&=16R_1-(2500\times12)-(5500\times5)\\ 0&=16R_1 - 30000-27500\\ 0&=16R_1 - 57500\\ 16R_1&=57500\\ R_1&=\frac{57500}{16}=3593.75\text{ lb} \end{align*}$$

\]
Also, from $\sum V = 0$, we have $R_1 + R_2=2500 + 5500$. Substitute $R_1 = 3593.75$ into this equation:
\[

$$\begin{align*} 3593.75+R_2&=8000\\ R_2&=8000 - 3593.75=4406.25\text{ lb} \end{align*}$$

\]

Answer:

$R_1 = 3593.75\text{ lb}$, $R_2 = 4406.25\text{ lb}$