{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW_4 - SOLUTION(a(b PROBLEM 2.71 A horizontal circular...

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: SOLUTION (a) (b) PROBLEM 2.71 A horizontal circular plate is suspended as shown from three wires which are attached to a support at D and form 30° angles with the vertical. Knowing that the x component of the force exerted by wire AD on the plate is 220.6 N, determine (a) the tension in wire AD, (b) the angles 19x, By, and 62 that the force exerted at A forms with the coordinate axes. @536“ E vs wine Fx = F sin30°sin50° = 220.6 N (Given) A D F - “r—*—.‘—" \ sm3 0° sm50° ,. ~ 220'6N = 575.95 N . cosﬁx = i = = 0.3830 F 575.95 0x = 67.5°4 Fy = Fcos30° = 498.79N F 0030}, = —y = 498'79 = 0.86605 F 575.95 Q, = 300° 4 F2 = —Fsin30°cos50° = —(575.95 N)sin30°cos50° = —185.107 N c0502 = E— = = —0.32139 F 575.95 Oz = 108.7° 4 PROBLEM 2.85 A force F of magnitude 400 N acts at the origin of a coordinate system. Knowing that 6x = 28.5°, Fy = —80 N, and F1 > 0, determine (a) the components Fx and F2, (b) the angles ﬂy and 02. SOLUTION (a) Have Fx = Fcosﬂx = (400 N)cos28.5° Fx = 351.5 N 4 Then: ﬂ=ﬁ+ﬁ+ﬁ So: (400 N)2 = (352.5 N)2 + (—80 N)2 + F} Hence: F2 = + (400 N)2 — (351.5 N)2 — (—80 N)2 Fz = 173.3 N4 (19) cosﬁy = % = 2T? = —0.20 0y =101.5° 4 c050 = g = = 0.43325 (92 = 64.3° { 2 F 400 PROBLEM 2.39 n . A 150W Plateismsupportedpy three cables as shown. Knowing that the tension in cable 4:3 is. 204 1b,:detem1ine the components Of the force exerted on thepIatg gt 3, . V ' SOLUTION 37 e (32 my. .+ (4,8 huh: (36 in.)k (32 in)? 13(48'in32 2+ (33\$,m.)? .= 68 ‘ 1 __... BA 2043-11) 3’ ' 3" BA 68m; j ,— (36 in.)k‘]':»; 1333(961b)i§+;(14421b)j;1—-(108x1b)k g .e .» I t ., Fx =+96.0 lb, Fy =+144.o lb, F, =—108.01b< ...
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