Question

w = 12,000 lb
w = 1,000 lb/ft
w = 12,000 lb
r1
r2
σmr2 = 0
0 = 12r1 - (12,000× ) - (5000× )
r1 = σv = 0 (r1 + r2) = (12,000 + 5000)

Explanation:

Step1: Calculate moments about $R_2$

Taking moments about $R_2$, we know that the sum of moments $\sum M_{R_2}=0$. The distance from $R_1$ to $R_2$ is 12 feet. The 12000 - lb load is at the mid - point (6 feet from $R_1$) and the 5000 - lb load is 8 feet from $R_1$. So the moment equation is $0 = 12R_1-(12000\times6)-(5000\times8)$.

Step2: Solve the moment equation for $R_1$

First, calculate the right - hand side of the equation: $(12000\times6)+(5000\times8)=72000 + 40000=112000$. Then, from $12R_1=112000$, we get $R_1=\frac{112000}{12}=\frac{28000}{3}\approx9333.33$ lb.

Step3: Use vertical equilibrium to find $R_2$

Since the sum of vertical forces $\sum V = 0$, we have $R_1+R_2=12000 + 5000$. Substitute $R_1=\frac{28000}{3}$ into the equation: $\frac{28000}{3}+R_2=17000$. Then $R_2=17000-\frac{28000}{3}=\frac{51000 - 28000}{3}=\frac{23000}{3}\approx7666.67$ lb.

Answer:

$R_1=\frac{28000}{3}\text{ lb}\approx9333.33\text{ lb}$, $R_2=\frac{23000}{3}\text{ lb}\approx7666.67\text{ lb}$