Vector Mechanics for Engineers: Statics
Vector Mechanics for Engineers: Statics
12th Edition
ISBN: 9781259977268
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 5.2, Problem 5.48P

5.48 and *5.49 Determine by direct integration the centroid of the area shown.

Fig. P5.48

Chapter 5.2, Problem 5.48P, 5.48 and 5.49 Determine by direct integration the centroid of the area shown. Fig. P5.48

Expert Solution & Answer
Check Mark
To determine

The centroid of shaded area in Fig. P5.48 by method of direct integration.

Answer to Problem 5.48P

Centroid is located at (9.27a,3.09a).

Explanation of Solution

Refer the figure P5.48 and figure given below.

Vector Mechanics for Engineers: Statics, Chapter 5.2, Problem 5.48P

Write the equation for curve r.

r=aeθ

Here, y is the radial coordinate, a is an unknown constant, and θ is the polar angle.

In the figure given above, black dot denotes the center of mass of triangular shaped differential area element.

Write the expression for the x-coordinate of center of mass of triangular shaped differential area element.

x¯EL=23rcosθ

Here, x¯EL is the center of mass of element dA.

Rewrite the above relation by substituting aeθ for r .

x¯EL=23aeθcosθ

Write the expression for the y-coordinate of center of mass of triangular shaped differential area element.

y¯EL=23rsinθ

Here, y¯EL is the y-coordinate of center of mass of element dA.

Rewrite the above relation by substituting aeθ for r .

y¯EL=23aeθsinθ

Write the expression to calculate the triangular differential area element.

dA=12(r)(rdθ)

Here, dA is the differential area element and dθ denotes a small change in θ.

Rewrite the above relation by substituting aeθ for r .

dA=12(aeθ)(aeθdθ)=12a2e2θdθ

Write the expression to calculate the total area of shaded region in P5.48.

A=dA

Here, A is the area of shaded region in P5.48.

Rewrite the above equation by substituting 12a2e2θdθ for dA and integrate it from θ=0 to θ=π.

A=0π12a2e2θdθ=12a2[12e2θ]02L=14a2(e2π1)=133.623a2

Calculate x¯ELdA by substituting 12a2e2θdθ for dA and 23aeθcosθ for x¯EL.

x¯ELdA=0π23aeθcosθ(12a2e2θdθ)=a330πe3θcosθdθ (I)

Apply the integration by parts method to solve the above integral. The required formula is given below.

u(θ)v(θ)dθ=u(θ)v(θ)dθ[(du(θ)dθ)v(θ)dθ]

Here, u(θ) and v(θ) are functions of θ.

Substitute e3θ for u(θ) and cosθ for v(θ) and solve integral.

e3θcosθdθ=e3θcosθdθ[(de3θdθ)cosθdθ]=e3θsinθ(3e3θsinθ)=e3θsinθ3[e3θ(cosθ)3e3θ(cosθ)dθ]=e3θsinθ+3e3θcosθ9e3θcosθdθe3θcosθdθ+9e3θcosθdθ=e3θsinθ+3e3θcosθ10e3θcosθdθ=e3θsinθ+3e3θcosθ

Rewrite the above equation in terms of e3θcosθdθ.

e3θcosθdθ=e3θ10(sinθ+3cosθ)

Calculate a330πe3θcosθdθ by using the above result.

a330πe3θcosθdθ=a33[e3θ10(sinθ+3cosθ)]0π=a33[e3π10(sinπ+3cosπ)e010(sin0+3cos0)]=a33[e3π10(3)310]=a330(3e3π3)=1239.26a3

Rewrite equation (I) by substituting the above result.

x¯ELdA=1239.26a3

\

Calculate y¯ELdA by substituting 12a2e2θdθ for dA and 23aeθsinθ for y¯EL.

y¯ELdA=0π23aeθsinθ(12a2e2θdθ)=a330πe3θsinθdθ (II)

Apply the integration by parts method to solve the above integral. The required formula is given below.

u(θ)v(θ)dθ=u(θ)v(θ)dθ[(du(θ)dθ)v(θ)dθ]

Here, u(θ) and v(θ) are functions of θ.

Substitute e3θ for u(θ) and sinθ for v(θ) and solve integral.

e3θsinθdθ=e3θsinθdθ[(de3θdθ)sinθdθ]=e3θcosθ(3e3θ(cosθ))=e3θcosθ+3[e3θsinθ3e3θsinθdθ]=e3θcosθ+3e3θsinθ9e3θsinθdθe3θsinθdθ+9e3θsinθdθ=e3θcosθ+3e3θsinθ10e3θsinθdθ=e3θ(cosθ+3sinθ)

Rewrite the above equation in terms of e3θsinθdθ.

e3θsinθdθ=e3θ10(cosθ+3sinθ)

Calculate a330πe3θsinθdθ by using the above result.

a330πe3θcosθdθ=a33[e3θ10(cosθ+3sinθ)]0π=a33[e3π10(cosπ+3sinπ)e010(cos0+3sin0)]=a33[e3π10+110]=a330(e3π+1)=413.09a3

Rewrite equation (I) by substituting the above result.

y¯ELdA=413.09a3

Write the expression for first moment of whole area about y-axis.

x¯A=x¯ELdA

Here, x¯ is the x-coordinate of center of mass of total area.

Rewrite the above relation by substituting 133.623a2 for A and 1239.26a3 for x¯ELdA.

x¯(133.623a2)=1239.26a3

Rewrite the above relation in terms of x¯.

x¯=1239.26a3133.623a2=9.27a

Write the expression for first moment of whole area about x-axis.

y¯A=y¯ELdA

Here, y¯ is the y-coordinate of center of mass of total area.

Rewrite the above relation in terms of y¯ by substituting 133.623a2 for A and 413.09a3 for y¯ELdA.

y¯=413.09a3133.623a2=3.09a

Therefore, the centroid is located at (9.27a,3.09a).

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Chapter 5 Solutions

Vector Mechanics for Engineers: Statics

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