ENTROIDS AND CENTER OF GRAVITY: The following problems are to locate centroids of areas and lines. PROBLEM 2 ONLY PROVIDE THE SAME PROCESS OF SOLUTION IN THE GIVEN EXAMPLE ? THANK YOU. Problems 1 and 2. Locate the centroid of the shaded arear 100 mmx50 mm-180 mm1.2.100 mm50 mm-200 mm140 mm140 mmX 2. Locate the centroid of the shaded area shown.Divide the area into regular geometric shapes. Makesure the centroids of the geometric shapes can beidentified.6″a, in².(1/2)(12)(6)=36(12)(6)=72|-(1/2) (12) (6)=-36A=72PartTotalΣαχx =Ax = 5"Σανy =Ay = 6"=360724327212"The area can be divided into three partsas shown, with the unshaded triangle asa negative area since it is not part of theshaded area.x, iny, in8463ax, in³(36)(4)=144(72)(6) = 432(-36) (6)=-216360ay, in³(36) (8)=288(72)(3)=216(-36) (2)-72432623. Locate the centroid of the shaded area in the figure,created by cutting a semicircle of diameter r from aquarter circle of radius r.a, in².x, iny, inPartQuartercircleπr²4rax, in³πr²/4r4r3π3π4 3TT.Semicircle4rπr (4rvбл28 6πTotal3r-³12Σαχx =Ax=0.636ry =y = 0.349rTT (+)²2πr²8A =0.25r³0.393r²ay_0.137³= 0.393r²πr²8= 0.393r-²= 0.257-³7.312ay, in³πr² 4r4 3πпре80.13713πr-3164 mi5. Locate the centroid of the built-up section shown in the figure. Refer to Table 6-6.2on page 199 of your textbook for the properties of the elements.8" x 6" x 1"8" x 6" x 1"16" x 1"5" x 3" x 1/2"5" x 3" x 1/2"12"-20.7 lb channelSolution:Since the cross section is symmetrical with respect to a vertical axis, the centroid of thecross section lies on the axis of symmetry. Therefore, only the location of the centroid withrespect to a horizontal axis will be determined.With the base of the cross section as the reference:8" x 6" x 1"8" x 6" x 1"16" x 1"5" x 3" x 1/2"5" x 3" x 1/2"Part2-8"x6"x1"16" x 1"2-5"x3"x1/2"12"-20.7 lb channela, in².2(13)=26(16)(1)=162(3.75)=7.56.03A=55.53ΣαγTotal524.80655.53y =Ay = 9.45" above the base of the cross sectionLOTININ12"-20.7 lb channelRSITYDIFINALy, in16.28 -1.65 14.638 +0.28 8.280.75 +0.28 = 1.030.7016" +0.28" 16.28"ay, in³380.38132.487.7254.221524.806

Question

ENTROIDS AND CENTER OF GRAVITY: The following problems are to locate centroids of areas and lines. PROBLEM 2 ONLY PROVIDE THE SAME PROCESS OF SOLUTION IN THE GIVEN EXAMPLE ? THANK YOU. Problems 1 and 2. Locate the centroid of the shaded arear 100 mmx50 mm-180 mm1.2.100 mm50 mm-200 mm140 mm140 mmX 2. Locate the centroid of the shaded area shown.Divide the area into regular geometric shapes. Makesure the centroids of the geometric shapes can beidentified.6″a, in².(1/2)(12)(6)=36(12)(6)=72|-(1/2) (12) (6)=-36A=72PartTotalΣαχx =Ax = 5&#34;Σανy =Ay = 6&#34;=360724327212&#34;The area can be divided into three partsas shown, with the unshaded triangle asa negative area since it is not part of theshaded area.x, iny, in8463ax, in³(36)(4)=144(72)(6) = 432(-36) (6)=-216360ay, in³(36) (8)=288(72)(3)=216(-36) (2)-72432623. Locate the centroid of the shaded area in the figure,created by cutting a semicircle of diameter r from aquarter circle of radius r.a, in².x, iny, inPartQuartercircleπr²4rax, in³πr²/4r4r3π3π4 3TT.Semicircle4rπr (4rvбл28 6πTotal3r-³12Σαχx =Ax=0.636ry =y = 0.349rTT (+)²2πr²8A =0.25r³0.393r²ay_0.137³= 0.393r²πr²8= 0.393r-²= 0.257-³7.312ay, in³πr² 4r4 3πпре80.13713πr-3164 mi5. Locate the centroid of the built-up section shown in the figure. Refer to Table 6-6.2on page 199 of your textbook for the properties of the elements.8&#34; x 6&#34; x 1&#34;8&#34; x 6&#34; x 1&#34;16&#34; x 1&#34;5&#34; x 3&#34; x 1/2&#34;5&#34; x 3&#34; x 1/2&#34;12&#34;-20.7 lb channelSolution:Since the cross section is symmetrical with respect to a vertical axis, the centroid of thecross section lies on the axis of symmetry. Therefore, only the location of the centroid withrespect to a horizontal axis will be determined.With the base of the cross section as the reference:8&#34; x 6&#34; x 1&#34;8&#34; x 6&#34; x 1&#34;16&#34; x 1&#34;5&#34; x 3&#34; x 1/2&#34;5&#34; x 3&#34; x 1/2&#34;Part2-8&#34;x6&#34;x1&#34;16&#34; x 1&#34;2-5&#34;x3&#34;x1/2&#34;12&#34;-20.7 lb channela, in².2(13)=26(16)(1)=162(3.75)=7.56.03A=55.53ΣαγTotal524.80655.53y =Ay = 9.45&#34; above the base of the cross sectionLOTININ12&#34;-20.7 lb channelRSITYDIFINALy, in16.28 -1.65 14.638 +0.28 8.280.75 +0.28 = 1.030.7016&#34; +0.28&#34; 16.28&#34;ay, in³380.38132.487.7254.221524.806

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Problems 1 and 2. Locate the centroid of the shaded area y r = 100 mm x 50 mm -180 mm 1. 2. 100 mm 50 mm —200 mm 140 mm 140 mm X