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 5.43P

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

Fig. P5.43

Expert Solution & Answer
Check Mark
To determine

The centroid of shaded area in Fig. P5.43 by method of direct integration.

Answer to Problem 5.43P

Centroid is located at (17130a,1126b).

Explanation of Solution

Refer the figure P5.43 and figure given below.

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 5.2, Problem 5.43P

Write the equation for curve x.

x=ky2 (I)

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

Consider the point (a,b).

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 b for y1.

a=kb2

Rewrite the above equation in terms of k

k=ba2

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

x=ba2y2

Rewrite the above equation in terms of y.

y=bax

Divide the shaded region in P5.43 into two parts for the purpose of integration. Region 0xa2 and   a2xa.

Consider the region 0xa2 in P5.43.

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 0xa2

Write the expression for the y-coordinate of center of mass of differential area element in region 0xa2 in P5.43.

y¯EL1=y2

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

Rewrite the above relation by substituting bax for y in the above equation.

y¯EL1=b2ax (II)

Write the expression to calculate the differential area element in 0xa2.

dA1=ydx

Here, dA1 is the differential area element in 0xa2 and dx denotes a small change in x.

Rewrite the above relation by substituting bax for y in the above equation.

dA1=baxdx (III)

Consider the region a2xa in in P5.43.

Write the expression for the x-coordinate of center of mass of differential area element in region a2xa.

x¯EL2=x

Here, x¯EL2 is the center of mass of differential area element in region a2xa

Write the expression for the y-coordinate of center of mass of differential area element in region a2xa.

y¯EL2=12(y+y1) (IV)

Here, y¯EL is the y-coordinate of center of mass of differential area element in region a2xa and y1 is the straight line in a2xa.

Calculate the slope of y1 using points (a2,0) and (a,b2).

m=b20aa2=ba

Here, m is the slope of y1.

Write the equation of y1 using slope-point form of equation for straight line. Use the point (a2,0) for finding the equation.

y10=m(xa2)

Rewrite the above equation by substituting ba for m to find

y1=ba(xa2)

Rewrite equation (IV) by substituting the above relation for y1 and bax for y.

y¯EL2=12(bax+ba(xa2))=b2(xa+xa12) (V)

Write the expression for dA2.

dA2=(yy1)dx

Rewrite the above relation by substituting bax for y and ba(xa2) for y1.

dA2=(baxba(xa2))dx=b(xaxa+12)dx (VI)

Write the equation to calculate the total area of shaded region in P5.43.

A=0a/2dA1+a/2adA2

Here, A is the area shaded region in P5.43.

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

A=0a/2baxdx+a/2ab(xaxa+12)dx=ba[23x3/2]0a/2+b[23x3/2ax22a+x2]a/2a=2b3a[(a2)3/2+a3/2(a2)3/2]+b[12a(a2(a2)2)+12(aa2)]=1324ab

Write the expression for x¯ELdA.

x¯ELdA=0a/2x¯EL1dA1+a/2ax¯EL2dA2=0a/2xdA1+a/2axdA2

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

x¯ELdA=0a/2xbaxdx+a/2axb(xaxa+12)dx=ba[25x5/2]0a/2+b[25x5/2ax33a+x44]a/2a=2b5a[(a2)5/2+a5/2(a2)5/2]+b[13a(a3(a2)3)+14(a2(a2)2)]=71249a2b

Write the expression for y¯ELdA.

y¯ELdA=0a/2y¯EL1dA1+a/2ay¯EL2dA2

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

y¯ELdA=0a/2b2axbaxdx+a/2ab2(xa+xa12)b(xaxa+12)dx=b22a[12x2]0a/2+b22[x22a13a(xa12)3]a/2a=b4a[(a2)2+a2(a2)2]b26a(a212)3=1148ab2

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 1324ab for A and 71249a2b for x¯ELdA.

x¯(1324ab)=71249a2b

Rewrite the above relation in terms of x¯.

x¯=(71249a2b)(1324ab)=17130a

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 1324ab for A and 1148ab2 for y¯ELdA.

y¯=(1148ab2)(1324ab)=1126b

Therefore, the centroid is located at (17130a,1126b).

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Chapter 5 Solutions

VECTOR MECHANICS FOR ENGINEERS: STATICS

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