Use single integration (use a thin, long vertical slice as your integration element) to determine the x and y coordinates of the centroid of the plane region shown. y (mm) xy = 3175 x (mm) 102 mm 25 mm 127 mm

Use single integration (use a thin, long vertical slice as your integration element) to determine the x and y coordinates of the centroid of the plane region shown. y (mm) xy = 3175 x (mm) 102 mm 25 mm 127 mm

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter9: Moments And Products Of Inertia Of Areas

Section: Chapter Questions

Problem 9.34P: Use integration to verify the formula given in Table 9.2 for Ixy of a half parabolic complement.

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Concept explainers

Free Body Diagram

The system of forces acting on a body tends to either rotate it or make the body undergo motion. In general, when an external agent exerts a force on a body, the body undergoes translational motion and rotational motion. The body has six degrees of freedom under the influence of that force. Here, three degrees of freedom are from the translational motion while the other three are from the rotation motion. However, sometimes the bodies are not allowed to undergo any motion and are fixed, under such conditions the body is said to be constrained.

Question

Use single integration (use a thin, long vertical slice
as your integration element) to determine the x and y coordinates of the
centroid of the plane region shown.
y (mm)
xy = 3175
x (mm)
102 mm
25 mm
127 mm

Expert Solution

Step 1 Area of the whole region.

XY = 3175

Y = 3175 / X

Taking a vertical strip of area dA,

dA = Y . dx

dA = [ 3175 / X ] dx

Area = $\int d A$

$= \int_{25}^{120} \frac{3175}{X} d x$

$= 3175 [\ln (120) - \ln (25)]$

Area $= 4980.355 m m^{2}$

Step 2 Y coordinate of the centroid.

The centroid coordinates are given by,

$\bar{y} = \frac{\int Y d A}{A} \bar{x} = \frac{\int X d A}{A}$

For y coordinate,

$\bar{y} = \frac{\int Y d A}{A}$

$\bar{y} = \frac{\int_{25}^{120} \frac{3175}{2 X} [\frac{3175}{X} d x]}{A}$

$= \frac{\frac{{(3175)}^{2}}{2} {[\frac{1}{X}]}_{25}^{120}}{A}$

$= \frac{- 5040312.5 [\frac{1}{120} - \frac{1}{25}]}{A} = \frac{159609.8958}{4980.335}$

$\bar{y} = 32.04 m m$

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