HW4_Solutions

# HW4_Solutions - PROBLEM 3.41 Slider P can move along rod...

This preview shows pages 1–7. Sign up to view the full content.

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

View Full Document

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

View Full Document

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: PROBLEM 3.41 Slider P can move along rod 0A. An elastic cord PC is attached to the slider and to the vertical member BC. Knowing that the distance from O to P is 0.12 m and the tension in the cord is 30 N, determine (a) the angle between the elastic cord and the rod 0A, (27) the projection on 0/1 of the force exerted by cord PC at point P. Dimensions in mm SOLUTION 2 (a) By deﬁnition km, AFC = (I) (1) cos 8 where km = (0.24)2 + (0.24)2 + (0.12)2 n1 3 3 3 Knowing thatlrfwl = LOA : 0.36 m and that P is located 0.12 m from 0, it follows that the coordinates of P are % the coordinates of A. P(0.08 m, 0.08 m, —0.040 m) (0.10 m)i + (0.22 m)j + (0.28 m)k Then KPC : 7 '7 0 ./(0.10)“ + (0.22)‘ + (0.28)' m = 0.27037i + 0.59481j + 0.75703}; 2. 2 . 1 . . - 1 -. 31 + :1 — :k (0.270371 + 0.59481] + 0.7370311) = 0056 J J and a = cos—1(032445) = 71.068° or 6 = 711" 4 (b) (Zack/4 = TPC c0519 2 (30 N)cos71.068° (7%)“ = 9.7334N or (3am = 9.73 N 4 PROBLEM 3.45 The 0.732 x 1.2-rnlid ABCD of a storage bin is hinged along side AB and is held open by looping cord DEC over a frictionless hook at E. If the tension in the cord is 54 N, determine the moment about each of the coordinate axes of the force exerted by the cord at D. I 3601:1511 SOLUTION r\ First note 2 = (0.732)2 — (0.132)2 m l D {1732777 among : 0.720m T A l—e— 2 ————>‘ Then do}; = (0.360)2 + (0.720)2 + (0.720)2 m =l.081’n and rm) = (0.360 m)i + (0.720 m) j — (0.720 m)k Have TDE = g—Iii(rE,D) a ﬂ(0.360: + 0.720j — 0.720k) _ 1.08 = (18.0 Nji + (36.0 N)j — (36.0 N)k NOW MA = rD/A X TDE where rDM = (0.132 n1)j + (0.720 m)k i j k Then MA = 0 0.132 0.720 N-m 18.0 36.0 M360 PROBLEM 3.45 CONTINUED MA = {[(0.132)(—36.0) — (0.720)(36.o)]i + [(0.720)(18.0) — 0] j +[0 — (0.132)(18.0)]k} N-m MA = m(30.7 N.m)i + (12.96 N-m)j — (2.38 N-m)k ‘. Mx = —30.7N'm, My : 12.96 N-m, M2 = —2.38N-m { PROBLEM 3.47 A fence consists of wooden posts and a steel cable fastened to each post and anchored in the ground at A and D. Knowing that the sum of the moments about the z axis of the forces exerted by the cable on the posts at B and C is —66 N- In, determine the magnitude TCD when TR4 = 56 N. SOLUTION ”‘ = (24 N)i — (16 N)j + (43 N)k TCD = KCDTCD 3.0m CD = grog (2i — j — 2k) —(66 N-m) = k . {(1 m)j x [(24 N)i ~— (16 N)j + (48 N)k]} + k -{(1m)jx[%Tco(2i"j‘ 2k)]} or —66 = —24 — §TCD 3 TCD = E(66 — 24) N or TCD = 63.0N 4 PROBLEM 3.51 A force P is applied to the lever of an arbor press. Knowing that P lies in a plane parallel to the yz plane and that MI = 230 lb-in., M = —2001b-in., and M2 = ~35 lboin., determine the magnitude of P y and the values of gﬁ and 19. SOLUTION Mx = (Pcos¢)[(9 in.)sin6] — [Psin¢)[(9 in.)cosa] (1) My = —(Pcos¢)(5 in.) (2) M: = —(Psin ¢)(5 in.) (3) Equation (3). M2 __ -(PSin (I5) (5) Equation (2). My _ —(P cos ¢)(5) —35 or tan = 45 —200 = 0.175 925 = 99262" or 4b 2 993° 4 Substituting gﬁ into Equation (2) —200 lb-in. = —(Pcos9.9262°)(5 in.) P = 40.608113 or P = 40.6 lb ‘ Then, from Equation (1) 230113.111. = [(40.608 1b)cos9.9262°] [(9 in.) sin a] — [(40.608 1b)sin9.9262°][(9 in.)cos9] or 0.98503 sinB — 01723800056 = 0.62932 Solving numerically, 6 = 489° 4 PROBLEM 3.57 A rectangular tetrahedron has Six edges of length a. A force P is directed as shown along edge BC. Determine the moment of P about edge 0A. (on)x HaVe MOA = ADA ' (r00 P) O I i C x where [0933 From triangle OBC (a4) = E I 2 o a 1 a B (0A)z = (0A)xtan30 : ii] = m 2 Since (0/02 : (0A): + (0A): + (0A2) 2 2 or = + (0A): + 2 2 (0A) : a2 —i—£— =aJ§ V 4 12 3 Then 1' —3i+a\F'+ik {#0 _ 2 3] 2J3: l [2 1 and l : ii+ w'+—k 0A 2 P = iBCP‘ [asin30°)i — (acos30°)k : ———————(P) : 123(1 — ﬁk) 1'00 2 PROBLEM 3.57 CONTINUED ...
View Full Document

## This note was uploaded on 10/09/2008 for the course COE 2001 taught by Professor Valle during the Fall '08 term at Georgia Institute of Technology.

### Page1 / 7

HW4_Solutions - PROBLEM 3.41 Slider P can move along rod...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online