The task requires calculating the resultant force and its location along the x-axis for a beam with a distributed load. The load distribution is described by two functions:

1. $f(x)$ on the interval $0 < x < a$
2. $g(x)$ on the interval $a < x < a + b$

The functions provided are:

• $f(x) = A + Bx^{1/2} + Cx + Dx^2 + Ex^3 + Fe^{Gx}$, where the parameters are $A = 0$, $B = 150$, $C = 0$, $D = -20$, $E = 25$, $F = 0$, and $G = 0$. This simplifies to $f(x) = 150x^{1/2} - 20x^2 + 25x^3$.
• The constant load $g(x) = 5047 \, \text{N/m}$ over the interval $a < x < a + b$.

The beam dimensions are $a = 6 \, \text{m}$ and $b = 5 \, \text{m}$.

Step 1: Resultant force calculation

The resultant force is found by integrating the load functions over their respective intervals:

1. For $f(x)$ on $0 < x < a$:

   $R_1 = \int_0^a f(x) \, dx = \int_0^6 \left( 150x^{1/2} - 20x^2 + 25x^3 \right) dx$

2. For $g(x)$ on $a < x < a + b$:

   $R_2 = \int_a^{a+b} g(x) \, dx = \int_6^{11} 5047 \, dx$

The total resultant force $R$ is:

$R = R_1 + R_2$

Step 2: Location of the resultant force

To find the location of the resultant force, we compute the first moment of area (the moment of the distributed load about the origin):

$x_R = \frac{\int_0^a x f(x) \, dx + \int_a^{a+b} x g(x) \, dx}{R}$

Let's solve these integrals step by step. I'll proceed with the calculations now.The resultant force calculations are as follows:

• $R_1 = 600\sqrt{6} + 6660 \, \text{N}$ (from the function $f(x)$ over the interval $0 < x < 6$).
• $R_2 = 25235 \, \text{N}$ (from the constant load $g(x) = 5047 \, \text{N/m}$ over the interval $6 < x < 11$).

Therefore, the total resultant force is:

$R = 600\sqrt{6} + 31895 \, \text{N}$

This is the magnitude of the resultant force. Next, let's calculate the location of the resultant force.The first moments of area are:

• For $f(x)$: $2160\sqrt{6} + 32400 \, \text{Nm}$
• For $g(x)$: $\frac{428995}{2} \, \text{Nm}$

Thus, the location of the resultant force $x_R$ is given by:

$x_R = \frac{2160\sqrt{6} + \frac{493795}{2}}{600\sqrt{6} + 31895}$

This is the exact expression for the location. Would you like to proceed with the numerical approximation for the force magnitude and location?