VECTOR MECHANICS FOR ENGINEERS: STATICS
VECTOR MECHANICS FOR ENGINEERS: STATICS
12th Edition
ISBN: 9781260912814
Author: BEER
Publisher: MCG
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Chapter 5.2, Problem 5.44P

5.43 and 5.44

Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b.

Chapter 5.2, Problem 5.44P, 5.43 and 5.44 Determine by direct integration the centroid of the area shown. Express your answer in

Fig. P5.44

Expert Solution & Answer
Check Mark
To determine

The centroid of shaded area in Fig. P5.44 by method of direct integration.

Answer to Problem 5.44P

Centroid is located at (a,1735b).

Explanation of Solution

Refer the figure P5.44 and figure given below.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 5.2, Problem 5.44P

Write the equation for curve y.

y=kx2 (I)

Here, y is the y co-ordinate, x is the x-coordinate, and k is an unknown constant.

Consider the point (a,2b).

Here, a is a length on x-axis and b  is a length parallel to y axis.

Rewrite equation (I) by substituting a for x and 2b for y.

2b=ka2

Rewrite the above equation in terms of k.

k=2ba2

Rewrite equation (I) by substituting the above relation for k.

y=2ba2x2

Divide the shaded region in P5.44 into two parts for the purpose of integration. Region 0xa and   ax2a.

Consider the region 0xa in P5.44.

Consider a rectangular differential area element in the region. Write the expression for the x-coordinate of center of mass of differential area element.

x¯EL1=x

Here, x¯EL1 is the center of mass of differential area element in region 0xa.

Write the expression for the y-coordinate of center of mass of differential area element in region 0xa in P5.43.

y¯EL1=y2

Here, y¯EL1 is the y-coordinate of differential area element

Rewrite the above relation by substituting 2ba2x2 for y in the above equation.

y¯EL1=12(2ba2x2)=ba2x2 (II)

Write the expression to calculate the differential area element in 0xa.

dA1=ydx

Here, dA1 is the differential area element in 0xa and dx denotes a small change in x.

Rewrite the above relation by substituting 2ba2x2 for y in the above equation.

dA1=2ba2x2dx (III)

Consider the region ax2a in in P5.44.

Write the expression for the x-coordinate of center of mass of differential area element in region ax2a.

x¯EL2=x

Here, x¯EL2 is the center of mass of differential area element in region ax2a

Write the expression for the y-coordinate of center of mass of differential area element in region ax2a.

y¯EL2=y22 (IV)

Here, y¯EL2 is the y-coordinate of center of mass of differential area element in region ax2a and y2 is a straight line in region ax2a.

Calculate the slope of y2 using points (a,b) and (2a,0).

m=0b2aa=ba

Here, m is the slope of y2.

Write the equation of y2 using slope-point form of equation for straight line. Use the point (2a,0) for finding the equation.

y20=m(x2a)

Rewrite the above equation by substituting ba for m to find y2.

y2=ba(x2a)=b(2xa)

Rewrite equation (IV) by substituting b(2xa) for y2.

y¯EL2=b2(2xa) (V)

Write the expression for dA2.

dA2=y2dx

Rewrite the above relation by substituting b(2xa)  for y2.

dA2=b(2xa)dx (VI)

Write the equation to calculate the total area of shaded region in P5.44.

A=0adA1+a2adA2

Here, A is the area shaded region in P5.44.

Rewrite the above equation by substituting equation (III) and (VI).

A=0a2ba2x2dx+02ab(2xa)dx=2ba2[x33]0a+b[a2(2xa)2]02a=76ab

Write the expression for x¯ELdA.

x¯ELdA=0ax¯EL1dA1+a2ax¯EL2dA2=0AxdA1+a2axdA2

Rewrite the above equation by substituting equation (III) and (VI).

x¯ELdA=0ax2ba2x2dx+02axb(2xa)dx=2ba2[x44]0a+b[x2(x33a)2]a2a=16a2b+b[((2a)2a2)+13a((2a)2a3)]=76a2b

Write the expression for y¯ELdA.

y¯ELdA=0ay¯EL1dA1+a2ay¯EL2dA2

Rewrite the above equation by substituting equations (II), (III), (V) and (VI).

y¯ELdA=0aba2x22ba2x2dx+a2ab2(2xa)b(2xa)dx=2b2a4[x55]0a+b22[a3(2xa)3]a2a=1730ab2

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 76ab for A and 76a2b for x¯ELdA.

x¯(76ab)=76a2b

Rewrite the above relation in terms of x¯.

x¯=76a2b76ab=a

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 76ab for A and 1730ab2 for y¯ELdA.

y¯=1730ab276ab=1735b

Therefore, the centroid is located at (a,1735b).

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Chapter 5 Solutions

VECTOR MECHANICS FOR ENGINEERS: STATICS

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