# HW14 - L1 PROBLEM 5.35 Determine by direct integration the...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: L1 PROBLEM 5.35 Determine by direct integration the centroid of the area shown. SOLUTION First note that symmetry implies f : 0 4 For the element (EL) shown y : R0039, x : Rsinﬂ dx : Recs 6 d (9 dA = ydx : R200529d6 Hence a = éRZ (211 Sin 2a) A: jdAzzﬁchoszﬁdﬁlez 9+5m26 2 4 0 a Jig/ﬂ?! = 2f§cosﬂ(R2 0052 8516') = Riécoszasing + gsinﬁ) 0 R3 = ?(cosla sinar + 25ina) 3 Rw(coszasina + 25mg) But FA : pad/4 so p : 3 2 £2 (2a + sinZa) cosza + 2 or i : ZRsinat L——) 3 (20: + sm 2a) _ i 2 Allematively, } = gRsina 23 Sin :1 a + SH] 0! 4 PROBLEM 5.49 From the solution to Problem 5.]: A = 240 1:19, 3M 2 5.601'11..l7area = 6.60 in. (Area) From the solution to Problem 5.21: L = 72 imjm = 4.67m, Pm = 6.67 in. Applying the theorems of Pappus-Guldinus, we have (a) Rotation about the x axis: A. = erl’iinel. = 21r(6.67in.)(72in.) = 301141113 A = 3020 in2 1 , = zzrmA = 2x(6.60in.)[240in2) = 9952.6 in3 (b) Rotation aboutx = 16 in.: A,=16 : 222(16 — XML = 2::[06 i 4.67)in.](72in.) = 5125.5 in2 A”,6 = sum2 4 Fm 5 2::(16 — Xaml)A = 2n'[(16 — 5.60)in.](240in2) =15 682.8:‘n3 r;=J6 :1563x103in34 PROBLEM 5.53 A 15-mm-diameter hole is drilled in a piece of 20-mm-thick steel; the hole is then countersunk as shown. Determine the volume of steel removed during the countersinking process. SOLUTION The required volume can be generated by rotating the area shown about the y axis. Applying the second theorem of Pappus-Guldinus, we have V=27rfA=2n[(§+7,5jmm:|x|:%x5mmx5mm] or V = 720mm3< ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern