situation 3. replace the loading system by an equivalent resultant force and specify where the resultant’s…

Explanation:

Step1: Calculate resultant force

The resultant force $F_R$ is the sum of all the forces. Here, $F_1 = 500$ lb, $F_2=250$ lb, $F_3 = 500$ lb. So, $F_R=F_1 + F_2+F_3=500 + 250+500=1250$ lb.

Step2: Calculate the moment about point O

Let the distances of the forces from point O be $x_1$, $x_2$, $x_3$. Assume the distances are $x_1 = a$, $x_2=b$, $x_3 = c$ (not given in the problem - but for general moment - calculation). The moment about O, $M_O=F_1x_1+F_2x_2 + F_3x_3=500a + 250b+500c$.

Step3: Find the location of the resultant force

We know that $M_O=F_Rx$, where $x$ is the distance of the line - of - action of the resultant force from O. So, $x=\frac{M_O}{F_R}=\frac{500a + 250b+500c}{1250}$.

Answer:

Resultant force $F_R = 1250$ lb. The location of the line - of - action of the resultant force from O is $x=\frac{500a + 250b+500c}{1250}$, where $a$, $b$, $c$ are the distances of the 500 - lb, 250 - lb, and 500 - lb forces respectively from O.