To find the position of the center of gravity of a rod of length 3 m with a linear density that varies as \(\lambda = 2 + x\), we can follow these steps:
### Step-by-Step Solution:
1. **Define the Variables**:
- Let the length of the rod be \(L = 3\) m.
- The linear density of the rod is given by \(\lambda(x) = 2 + x\), where \(x\) is the position along the rod from \(0\) to \(3\) m.
2. **Express the Mass Element**:
- The mass element \(dm\) can be expressed in terms of the linear density:
\[
dm = \lambda(x) \, dx = (2 + x) \, dx
\]
3. **Set Up the Integral for Center of Gravity**:
- The position of the center of gravity \(r_{cg}\) is given by the formula:
\[
r_{cg} = \frac{\int_0^L x \, dm}{\int_0^L dm}
\]
- Substitute \(dm\) into the integrals:
\[
r_{cg} = \frac{\int_0^3 x (2 + x) \, dx}{\int_0^3 (2 + x) \, dx}
\]
4. **Calculate the Numerator**:
- First, calculate the integral in the numerator:
\[
\int_0^3 x(2 + x) \, dx = \int_0^3 (2x + x^2) \, dx
\]
- This can be split into two integrals:
\[
= \int_0^3 2x \, dx + \int_0^3 x^2 \, dx
\]
- Calculate each integral:
\[
\int_0^3 2x \, dx = 2 \left[\frac{x^2}{2}\right]_0^3 = 2 \cdot \frac{9}{2} = 9
\]
\[
\int_0^3 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^3 = \frac{27}{3} = 9
\]
- Thus, the numerator becomes:
\[
\int_0^3 x(2 + x) \, dx = 9 + 9 = 18
\]
5. **Calculate the Denominator**:
- Now calculate the integral in the denominator:
\[
\int_0^3 (2 + x) \, dx = \int_0^3 2 \, dx + \int_0^3 x \, dx
\]
- Calculate each integral:
\[
\int_0^3 2 \, dx = 2 \cdot 3 = 6
\]
\[
\int_0^3 x \, dx = \left[\frac{x^2}{2}\right]_0^3 = \frac{9}{2}
\]
- Thus, the denominator becomes:
\[
\int_0^3 (2 + x) \, dx = 6 + \frac{9}{2} = \frac{12}{2} + \frac{9}{2} = \frac{21}{2}
\]
6. **Calculate the Center of Gravity**:
- Now substitute the values into the formula for \(r_{cg}\):
\[
r_{cg} = \frac{18}{\frac{21}{2}} = 18 \cdot \frac{2}{21} = \frac{36}{21} = \frac{12}{7} \text{ m}
\]
### Final Answer:
The position of the center of gravity of the rod is \(\frac{12}{7}\) m.
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