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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?

Laske oheinen palkin jakaantuneen kuormituksen resultantti ja sen sijainti x-akselin suunnassa, kun välillä 0 < x < a jakaantunut kuorma (N/m) on funktion f(x) ja välillä a < x < (a+b) funktion g(x) mukainen. Kuormitusfunktion f(x) parametrit ovat A = 0, B = 150, C = 0, D = -20, E = 25, F = 0, G = 0 ja kuormitus g(x) = 5047.0 N/m. Palkin mitat ovat a = 6 m ja b = 5 m.

Learn how to calculate the resultant force and its location on a beam with a distributed load described by a function over two intervals. The task involves integral calculus and applies to statically determinate beams.

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