PROBLEM SOLUTION

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1 PROBLEM The rectangular plate shown weighs 80 l and is supported three vertical wires. Determine the weight and location of the lightest lock that should e placed on the plate if the tensions in the three wires are to e equal. Free-Bod Diagram: Let W j e the weight of the lock and and z the lock s coordinates. Since tensions in wires are equal, let TA = TB = TC = T Σ M0 = 0: ( r Tj) + ( r Tj) + ( r Tj) + r ( Wj) + ( i+ zk) ( W j ) = 0 A B C G or (75 k) Tj+ (15 k) Tj+ (60i+ 30 k) Tj+ (30i+ 45 k) ( Wj) + ( i+ zk) ( W j ) = 0 or 75Ti 15Ti+ 60Tk 30Ti 30Wk + 45Wi W k + W zi = 0 Equate coefficients of unit vectors to zero: i : 120T + 45W + W z = 0 (1) k : 60T 30W W = 0 (2) Also, Σ F = 0: 3T W W = 0 (3) Eq. (1) + 40 Eq. (3): 5 W + ( z 40) W = 0 (4) Eq. (2) 20 Eq. (3): 10 W ( 20) W = 0 (5) 463

2 PROBLEM (Continued) Solving Eqs. (4) and (5) for W / W and recalling that 0 Eq. (4): Eq. (5): W W W W = z 40 0 = = = 60in., 0 z 90in., Thus, ( W ) min = 0.5W = 0.5(80) = 40 l ( W ) min = 40.0l Making W = 0.5W in Eqs. (4) and (5): 5 W + ( z 40)(0.5 W) = 0 z = 30.0 in. 10 W ( 20)(0.5 W) = 0 = 0in. 464

3 PROBLEM Solve Prolem 4.115, assuming that cale EF is replaced a cale attached at points E and H. PROBLEM The rectangular plate shown weighs 75 l and is held in the position shown hinges at A and B and cale EF. Assuming that the hinge at B does not eert an aial thrust, determine (a) the tension in the cale, () the reactions at A and B. r = ( 8) i = 30i BA / r = (30 4) i+ 20k EA / = 26i+ 20k rga / = i k = 19i+ 10k EH = 30i+ 12j 20k EH = in. EH T T= T = ( 30i+ 12j 20 k) EH Σ M = 0: r T+ r ( 75 j) + r B = 0 A E/ A G/ A B/ A i j k i j k i j k T = B Bz T Coefficient of i: (12)(20) = 0 T = T = 118.8l Coefficient of j: Coefficient of k: ( ) 30Bz = 0 B z = 8.33l (26)(12) B = 0 B = l B= (15.00 l) j (8.33 l) k 487

4 PROBLEM (Continued) Coefficient of i: Coefficient of j: Σ F =0: A+ B+ T (75 l) j = 0 30 A (118.75) = 0 A = l 12 A (118.75) 75 = 0 A = 22.5 l 20 Coefficient of k: Az 8.33 (118.75) = 0 Az = l A = (93.8 l) i+ (22.5 l) j+ (70.8 l) k 488

5 PROBLEM Solve Prolem 4.118, assuming that the earing at D is removed and that the earing at C can eert couples aout aes parallel to the and z aes. PROBLEM The ent rod ABEF is supported earings at C and D and wire AH. Knowing that portion AB of the rod is 250 mm long, determine (a) the tension in wire AH, () the reactions at C and D. Assume that the earing at D does not eert an aial thrust. Free-Bod Diagram: Dimensions in mm Δ ABH is equilateral. r = 50i+ 250j HC / r = 350i+ 250k FC / T= T(sin 30 ) j T(cos30 ) k = T(0.5j k) Σ M = 0: r ( 400 j) + r T + ( M ) j+ ( M ) k = 0 C F/ C H/ C C C z i j k i j k T + ( M ) j+ ( M ) k = C C z Coefficient of i: T = 0 T = N T = 462 N Coefficient of j: 43.3(461.9) + ( M C ) = 0 ( MC) = N mm ( M ) = 20.0 N m C 3 493

6 PROBLEM (Continued) Coefficient of k: (461.9) + ( ) = 0 M C z ( M ) = N mm C z ( M ) = N m C z Σ F = 0: C + T 400j = 0 3 Coefficient of i: C = 0 Coefficient of j: C + 0.5(461.9) 400 = 0 C = N Coefficient of k: C 0.866(461.9) = 0 C = 400 N z z M C = (20.0 N m) j+ (151.5 N m) k C= (169.1 N) j+ (400 N) k 494

7 PROBLEM A lever AB is hinged at C and attached to a control cale at A. If the lever is sujected to a 500-N horizontal force at B, determine (a) the tension in the cale, () the reaction at C. Triangle ACD is isosceles with C = = 120 Thus, DA forms angle of 60 with the horizontal ais. (a) 1 A= D= ( ) = We resolve F AD into components along AB and perpendicular to AB. Σ M = 0: ( F sin 30 )(250 mm) (500 N)(100 mm) = 0 F = 400 N C AD () Σ F = 0: (400 N)cos60 + C 500 N = 0 C =+ 300 N Σ F = 0: (400 N)sin 60 + C = 0 C = N AD C = 458 N

8 PROBLEM Two slots have een cut in plate DEF, and the plate has een placed so that the slots fit two fied, frictionless pins A and B. Knowing that P = 15 l, determine (a) the force each pin eerts on the plate, () the reaction at F. Free-Bod Diagram: (a) Σ F = 0: 15 l Bsin30 = 0 B = 30.0 l 60.0 () Σ M = 0: (30 l)(4 in.) + Bsin 30 (3 in.) + Bcos30 (11in.) F(13 in.) = 0 A 120 l in. + (30 l)sin 30 (3 in.) + (30 l)cos30 (11in.) F(13 in.) = 0 (a) Σ F = 0: A 30 l + Bcos30 F = 0 A 30 l + (30 l)cos l = 0 F = l F = l A = l A = 20.2 l 532

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