Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 24P from Chapter 9 from Hibbeler's Engineering Mechanics.
Problem 24P
Step-by-Step Solution
We are given the equation of the curve is $y = - \frac{h}{{{a^2}}}{x^2} + h$.
We are asked the centroid $\bar y$ of the shaded area.
Step 2
Consider a small element parallel to the y-axis of thickness $dx$ and height of y as shown below:
We have the area of small elemental segment is $dA = ydx$.
We have the horizontal centroid of the small segment is located at distance $\tilde x = x$ from y-axis and the vertical centroid of the small segment is located at distance $\tilde y = \frac{y}{2}$ from x-axis.
Step 3
The centroid along the y-axis of the shaded area is,
\[\bar y = \frac{{\int\limits_A {\tilde ydA} }}{{\int\limits_A {dA} }}\]Step 4
Substitute the values in the above expression.
\[\begin{array}{l} \bar y = \frac{{\int\limits_A {\left( {\frac{y}{2}} \right)ydx} }}{{\int\limits_A {ydx} }}\\ \bar y = \frac{{\int\limits_0^a {\frac{1}{2}\left( { - \frac{h}{{{a^2}}}{x^2} + h} \right)\left( { - \frac{h}{{{a^2}}}{x^2} + h} \right)dx} }}{{\int\limits_0^a {\left( { - \frac{h}{{{a^2}}}{x^2} + h} \right)dx} }}\\ \bar y = \frac{{\int\limits_0^a {\frac{1}{2}\left( {\frac{{{h^2}}}{{{a^4}}}{x^4} - \frac{{2{h^2}}}{{{a^2}}}{x^2} + {h^2}} \right)dx} }}{{\int\limits_0^a {\left( { - \frac{h}{{{a^2}}}{x^2} + h} \right)dx} }}\\ \bar y = \frac{{\frac{1}{2}\left[ {\frac{{{h^2}}}{{5{a^4}}}{x^5} - \frac{{2{h^2}}}{{3{a^2}}}{x^3} + {h^2}x} \right]_0^a}}{{\left[ { - \frac{h}{{{a^2}}}\left( {\frac{{{x^3}}}{3}} \right) + hx} \right]_0^a}} \end{array}\]Step 5
On further solving:
\[\begin{array}{l} \bar y = \frac{{\frac{1}{2}\left[ {\frac{{{h^2}}}{{5{a^4}}}{a^5} - \frac{{2{h^2}}}{{3{a^2}}}{a^3} + {h^2}a - 0} \right]}}{{\left[ { - \frac{h}{{{a^2}}}\left( {\frac{{{a^3}}}{3}} \right) + ha - 0} \right]}}\\ \bar y = \frac{2}{5}h \end{array}\]