Math Problem Statement

A beam of length 10 meters supports a load that decreases from 1200 N/m at the left end to 200 N/m at the right end. The weight of the load is given as a function of length, x: w = 10(120 - x^2). Find the weight of the load and the x-coordinate of the centroid of the load.

Solution

The problem describes a beam of length 10 meters that supports a load with a variable weight along its length. The load at the left end is 1200 N/m and decreases to 200 N/m at the right end. The weight per unit length is given as a function of xx, where xx is the distance from the left end of the beam:

w(x)=10(120x2)N/mw(x) = 10(120 - x^2) \, \text{N/m}

We are tasked with solving two parts:

  1. Find the total weight of the load: To find the total weight of the load, we need to integrate the weight function w(x)w(x) over the length of the beam (from x=0x = 0 to x=10x = 10).

  2. Find the x-coordinate of the centroid of the load: The centroid xˉ\bar{x} is the point where the total load can be considered to act. It is calculated by the formula:

xˉ=010xw(x)dx010w(x)dx\bar{x} = \frac{\int_0^{10} x \cdot w(x) \, dx}{\int_0^{10} w(x) \, dx}

Let me calculate these values for you.

Step 1: Total Weight of the Load

To find the total weight, we integrate the weight function over the length of the beam:

Total Weight=010w(x)dx=01010(120x2)dx\text{Total Weight} = \int_0^{10} w(x) \, dx = \int_0^{10} 10(120 - x^2) \, dx

Step 2: x-Coordinate of the Centroid

To find the centroid, we calculate:

xˉ=010xw(x)dx010w(x)dx\bar{x} = \frac{\int_0^{10} x \cdot w(x) \, dx}{\int_0^{10} w(x) \, dx}

Let me compute these integrals for you. sympy as sp

Define the variable

x = sp.symbols('x')

Define the weight function

w = 10 * (120 - x**2)

Step 1: Total weight of the load

total_weight = sp.integrate(w, (x, 0, 10))

Step 2: x-coordinate of the centroid

centroid_numerator = sp.integrate(x * w, (x, 0, 10)) centroid_denominator = total_weight

centroid = centroid_numerator / centroid_denominator

total_weight, centroidThe results of the calculations are:

  1. Total weight of the load: The total weight of the load is 260003N\frac{26000}{3} \, \text{N}, which is approximately 8666.67 N.

  2. x-coordinate of the centroid: The x-coordinate of the centroid is 10526m\frac{105}{26} \, \text{m}, which is approximately 4.04 meters from the left end of the beam.

Would you like further details or clarification on these calculations?

Here are some related questions you might find useful:

  1. How do changes in the load distribution affect the centroid?
  2. What is the physical meaning of the centroid of a load distribution?
  3. How would the calculations change if the beam had a different length?
  4. How can you calculate the bending moment for a distributed load?
  5. What are the key factors to consider when designing a beam to withstand a varying load?

Tip: When calculating centroids for distributed loads, breaking the load into small segments and integrating over those intervals can help solve complex load distributions.

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Math Problem Analysis

Mathematical Concepts

Integration
Centroids of Distributed Loads
Definite Integrals

Formulas

w(x) = 10(120 - x^2)
Total Weight = \( \int_0^{10} w(x) \, dx \)
Centroid x-coordinate = \( \frac{\int_0^{10} x \cdot w(x) \, dx}{\int_0^{10} w(x) \, dx} \)

Theorems

Properties of definite integrals
Centroid of a distributed load

Suitable Grade Level

Undergraduate Engineering or Physics