Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 13P from Chapter 9 from Hibbeler's Engineering Mechanics.
Problem 13P
Chapter:
Problem:
Locate the centroid x of the area.
Step-by-Step Solution
Step 1
Step 2
Step 3
We are given the function $y = 4 - \frac{{{x^2}}}{{16}}$.
We are asked to find the centroid $\bar x$.
Step 2
The free body diagram of the area is shown below:
To find the $x$ coordinate we will use the relation,
\[\bar x = \frac{{\int\limits_A {xdA} }}{{\int\limits_A {dA} }}\]...... (1)To find the area of the strip from the above diagram we will use the relation,
\[dA = \left( {4 - y} \right)dx\]On plugging the values in the above relation, we get,
\[dA = \left( {4 - \left( {4 - \frac{{{x^2}}}{{16}}} \right)} \right)dx\]Step 3
On plugging the values in the equation (1), we get,
\[\begin{array}{l} \bar x = \frac{{\int_0^8 {x\left( {4 - \left( {4 - \frac{{{x^2}}}{{16}}} \right)} \right)dx} }}{{\int_0^4 {\left( {4 - \left( {4 - \frac{{{x^2}}}{{16}}} \right)} \right)dx} }}\\ \bar x = \frac{{\int_0^8 {\left( {\frac{{{x^3}}}{{16}}} \right)dx} }}{{\int_0^8 {\left( {\frac{{{x^2}}}{{16}}} \right)dx} }}\\ \bar x = \frac{{\left( {\frac{{{x^4}}}{{64}}} \right)_0^8}}{{\left( {\frac{{{x^3}}}{{48}}} \right)_0^8}} \end{array}\]On further solving the above equation, we get,
\[\begin{array}{l} \bar x = \frac{{\left( {\frac{{{{\left( 8 \right)}^4}}}{{64}} - \frac{{{{\left( 0 \right)}^4}}}{{64}}} \right)}}{{\left( {\frac{{{{\left( 8 \right)}^3}}}{{48}} - \frac{{{{\left( 0 \right)}^3}}}{{48}}} \right)}}\\ \bar x = 6.003\;{\rm{m}} \end{array}\]