PROBLEM 2.18 PROBLEM 2.5 Solve Problem 2.3 using trigonometry Two control rods are attached at A to lever AB. Using trigonometry and knowing that the force in the left-hand rod is F1 120 N, determine (a) the required force F2 in the right-hand rod if the resultant R of the forces exerted by the rods on the lever is to be vertical, (b) the corresponding magnitude of R. SOLUTION Problem 2.3: Two forces P and Q are applied as shown at point A of a hook support. Knowing that P 60 N and Q 100 N, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. SOLUTION Using the force triangle and the Laws of Cosines and Sines We have: 60 N 180q 15q 30q J 135q Graphically, by the triangle law Then: F2 # 108 N We measure: = 22085.28 N2 R # 77 N or By trigonometry: Law of Sines 100 N F2 sin D D 90q 28q R sin 38q 62q, E and 120 sin E 180q 62q 38q R2 = (60 N)2 + (100 N)2 – 2 (60 N) (100 N) cos 135° 80q R = 148.6 N 100 N 148.6 N = sin b sin 135∞ sin b = Then: F2 sin 62q R sin 38q FG 100 N IJ sin 135° H 148.6 N K 0.4758 120 N sin 80q E or (a) F2 (b) R 107.6 N W 75.0 N W Then: 28.41q D E 75q D 180q 76.59q R = 148.6 N 76.6° J PROBLEM 2.25 PROBLEM 2.34 While emptying a wheelbarrow, a gardener exerts on each handle AB a force P directed along line CD. Knowing that P must have a 135-N horizontal component, determine (a) the magnitude of the force P, (b) its vertical component. Determine the resultant of the three forces of Problem 2.23. 1.2 m Problem 2.23: Determine the x and y components of each of the forces shown. 1.4 m 2.25 m 900 N 950 N 108 kN SOLUTION 1.5 m 2m SOLUTION The components of the forces were determined in Problem 2.23. P (a) F900 = – (417 N)i + (799.2 N) j Px cos 40q F950 = (494.4 N)i + (812.3 N)j 135 N cos 40q Thus or P (b) Py Px tan 40q F1800 = – (1440 N)i – (1080 N)j R = Rx + Ry 176.2 N W P sin 40q R = – (528.6)i + (531.5)j 135 N tan 40q Now: or Py 113.3 N W tan a = R Ry = 531.5 N a = tan–1 a Rx = –528.6 N and R= 5315 . N 528.6 N 5315 . = 45.16° 528.6 b-528.6 Ng + b5315. Ng 2 2 = 749.6 N R = 750 N 45.2° J PROBLEM 2.43 PROBLEM 2.54 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. 500 mm 300 mm Two cables tied together at C are loaded as shown. Determine the range of values of W for which the tension will not exceed 1050 N in either cable. 750 mm B A 300 mm 400 mm C 15 2.7 kN 8 690 N 400 mm W 525 mm SOLUTION SOLUTION Free-Body Diagram Free-Body Diagram From geometry: The sides of the triangle with hypotenuse CB are in the ratio 8:15:17. The sides of the triangle with hypotenuse CA are in the ratio 3:4:5. Thus: 3 15 15 0: TCA TCB 680 N 5 17 17 6Fx 0 or 2700 N 1 5 TCA TCB 5 17 From the geometry, we calculate the distances: AC = BC = b400 mmg + b300 mmg b500 mmg + b525 mmg 2 2 2 2 6Fy or Hence: (a) (b) 0: = 725 mm 4 8 8 TCA TCB 680 N W 5 17 17 0 or Â Fx = 0: - and (1) and = 500 mm Then, from the Free Body Diagram of point C: or 200 N T BC = Â Fy = 0: 1 2 TCA TCB 5 17 400 525 TAC + TBC = 0 500 725 300 T + 500 T – 2700 N = 0 AC BC 500 725 F H 1 W 4 (2) Then, from Equations (1) and (2) 725 ¥ 4 T AC 525 5 300 500 725 ¥ 4 T T + AC 500 AC 725 525 5 80 N I – 2700 N = 0 K TCB 680 N TCA 25 W 28 17 W 28 Now, with T d 1050 N TCA : TCA TAC = 1982.5 N TAC = 1982.5 N J or TBC = 2190.2 N J and W TCB : TCB or W 1050 N 1176 N 1050 N 609 N 25 W 28 680 N 17 W 28 ? 0 d W d 609 N W PROBLEM 2.61 PROBLEM 2.71 Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension in each cable is 900 N, determine (a) the magnitude of the largest force P which may be applied at C, (b) the corresponding value of D. A load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes 800 N, over the pulley A and supports a load P. Knowing that P determine (a) the tension in cable ACB, (b) the magnitude of load Q. SOLUTION SOLUTION Free-Body Diagram: Pulley C Free-Body Diagram: C Force Triangle = 900 N 900 N 900 N 2E P (a) 0 2303.5 N TACB 47.5q 2 900 N cos 47.5 q 180q 55q 47.5q 77.5q 6Fy (b) 1216 N P Since P ! 0, the solution is correct. D 0: TACB cos 30q cos 50q 800 N cos 50q TACB Hence 180q 85q E (b) 6Fx (a) Force triangle is isoceles with 1216 N W D 77.5q W 0: TACB sin 30q sin 50q 800 N sin 50q Q 2303.5 N sin 30q sin 50q 800 N sin 50q Q or Q 3529.2 N 2.30 kN W 0 0 Q 3.53 kN W

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