PROBLEM 8.3. S F = 0: N -(250 lb)cos 30 -(50 lb)sin 30 = SOLUTION
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1 PROLEM 8. Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when θ = and P = 5 lb. ssume equilibrium: S F = : F-(5 lb)sin + (5 lb)cos = S F = : N -(5 lb)cos -(5 lb)sin = F = lb N =+ 4.5 lb Maimum friction force: F m = m N s =.(4.5 lb) = 7.5 lb We note that F> F m. Thus, block moves down ctual friction force: F= F = m N =.(4.5 lb) = 48. lb, F = 48. lb k k Copright McGraw-Hill Education. ll rights reserved. No reproduction or distribution without the prior 8
2 PROLEM 8. The -lb uniform rod is held in the position shown b the force P. Knowing that the coefficient of static friction is. at and, determine the smallest value of P for which equilibrium is maintained. Free-bod diagram S F = : N +.N - lb = N = -. N S M = : N (4 in.) -. N (7.5 in.) -N (7.5 in.) -. N (4 in.) = G.5N - 8.N = ( N ) N = N =.849 lb S M = : (4 in.) + P(7.5 in.) -N (5 in.) -. N (8 in.)= 7.5P + 4 -(.849)( 5) -.(.849)( 8) = P =.955 lb min Copright McGraw-Hill Education. ll rights reserved. No reproduction or distribution without the prior 65
3 PROLEM 5. Locate the centroid of the plane area shown. rea : Rectangle mm b mm rea : Triangle b = 6 mm, h = mm,mm,mm,mm,mm,mm S X S = 5 X (5.6 mm ) = 8.64 mm X = 55.4 mm YS =S 6 Y (5.6 mm ) =.464 mm Y = 9.8 mm Copright McGraw-Hill Education. ll rights reserved. No reproduction or distribution without the prior 56
4 PROLEM 5.8 Locate the centroid of the plane area shown. 6 in. r = 8 in. in.,in,in.,in.,in,in p (8) = ,58-6 = S , Then S in X = = S 948. in S 4, in Y = = S 948. in X =.64 in. Y = 7.46 in. Copright McGraw-Hill Education. ll rights reserved. No reproduction or distribution without the prior 566
5 4 N/m 6 m 6 N/m PROLEM 5.66 For the beam and loading shown, determine (a) the magnitude and location of the resultant of the distributed load, (b) the reactions at the beam supports. () a RI = (4N/m)(6m) = N R II = (6 N/m)(6m) = 48 N + S F = : - =- - R RI RII R = N + 48 N R = 6 N S M = : - X ( 6) =-( ) -4( 48) X =.6 m R = 6 N, X =.6 m (b) Reactions S F = : = S F = : - 6 = = 6 N = 6 N S M = : M -(6 N)(.6 m) = M.6 kn m = Copright McGraw-Hill Education. ll rights reserved. No reproduction or distribution without the prior 68
6 PROLEM 9.6 Determine the moments of inertia of the shaded area shown with respect to the and aes. ssign area to be a rectangle and area to be a semicircle. We have where I = ( I ) -( I ) 6 4 ( I ) = (5 mm)(5 mm) = 65.4 mm p p = (75 mm) + (75 mm) 5 mm = mm 8 ( I ) ( ) I = 65.4 mm mm lso I = ( I) -( I) 6 4 I = 5 mm Then ( I ) = (5 mm)(5 mm) = 6.76 mm 6 4 p ( I ) = (75 mm) =.45 mm I = 6.76 mm -.45 mm 6 4 I = 5. mm Copright McGraw-Hill Education. ll rights reserved. No reproduction or distribution without the prior 466
7 PROLEM 7. force P is applied to a bent rod that is supported b a roller and a pin and bracket. For each of the three cases shown, determine the internal forces at Point J. (a) FD Rod: S M = : ap- a= D P = P S F = : V - = P V = FD J: S F = : F = P S MJ = : M- a = ap M = (b) FD Rod: a æ ö S MD = : ap- = ç è5 ø P = Copright McGraw-Hill Education. ll rights reserved. No reproduction or distribution without the prior 46
8 FD J: PROLEM 7. (Continued) 4P S F = : - V = 8P 5 V = P S F = : - F = F = P 5 8aP M = (c) FD Rod: æ4ö æö S MD = : ap- a - a = ç è 5 ø çè 5 ø 45P S F = : V - æ ö = ç è54 ø 5P S F = : - F = 54 45P S MJ = : M- a æ ö = ç è54 ø 5P = 4 V = M = P 7 P F = 4 ap 7 Copright McGraw-Hill Education. ll rights reserved. No reproduction or distribution without the prior 47
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