3.1 Particles in Two-Dimensional Force Systems

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Stanley Norris
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1 3.1 Particles in Two-Dimensional Force Sstems

2 + 3.1 Particles in Two-Dimensional Force Sstems Eample 1, page 1 of 1 1. Determine the tension in cables and lb 1 Free-bod diagram of connection F 2 Equilibrium equations for connection + F = 0: F cos 30 + F cos 60 = 0 F = 0: F sin 30 + F sin lb = 0 F = 60 Solving these equations gives F = 45.0 lb F = 77.9 lb 90 lb

3 3.1 Particles in Two-Dimensional Force Sstems Eample 2, page 1 of 2 2. The spring balance reads 500 N. Determine the tensions in cords and D. Spring balance 1.5 m 2 m 3 m D 2 m 4 m

4 + 3.1 Particles in Two-Dimensional Force Sstems Eample 2, page 2 of 2 1 Free bod diagram of connection F F D Force in spring = tension of 500 N. 2 m +1.5 m = 3.5 m 3 Geometr 3.5 m = tan -1 ( ) = m ( 2 m = tan -1 ) 4 m = m 2 + Equilibrium equations for connection F = 0 F cos + (500 N) cos = 0 (1) 2 m 4 m 4 Substituting and = in Eqs. 1 and 2 and solving gives F = 0: F sin + (500 N) sin F D = 0 (2) F = 901 N F D = 1006 N

5 + 3.1 Particles in Two-Dimensional Force Sstems Eample 3, page 1 of 2 3. The 10-kg block is supported b two identical 1 Free bod diagram of connection springs. The unstretched length of each spring is 500 mm. alculate the spring constant k. F F mm kg Weight = mg = (10 kg)(9.81 m/s 2 ) = 98.1 N 2 + Equilibrium equations for connection : F = 0 F cos 40 + F cos 40 = 0 (1) F = 0: F sin 40 + F sin N = 0 (2)

6 3.1 Particles in Two-Dimensional Force Sstems Eample 3, page 2 of 2 3 Geometr 4 Etension of spring = stretched length unstretched length = mm 500 mm 600 mm = mm = mm sin 40 = mm 5 F = force in spring = k (etension of spring) = k (433.4 mm) (3) 6 Solving Eqs. (1), (2), and (3) gives F = 76.3 N = F k = N/mm = kn/m

7 + 3.1 Particles in Two-Dimensional Force Sstems Eample 4, page 1 of 2 4. Determine the value of angle and the tension in the cables required for equilibrium. 1 Free bod diagram of connection D F lb F lb 60 lb 2 Equilibrium equations for connection : + F = 0: F cos 10 + F cos 20 = 0 (1) F = 0: F sin 10 + F sin lb = 0 (2) Solving Eqs. 1 and 2 gives F = lb F = lb

8 3.1 Particles in Two-Dimensional Force Sstems Eample 4, page 2 of 2 3 Free bod diagram of connection 4 F D ++ Equilibrium equations for connection F = 0: (340.3 lb) cos 20 + F D cos = 0 (5) F = 0: (340.3 lb) sin 20 + F D sin 150 lb = 0 (6) Eqs. 5 and 6 can be solved with a calculator that can handle two simultaneous nonlinear equations. lternativel re-write Eqs. 5 and 6 as F = lb F D cos = (340.3) cos 20 (7) 150 lb F D sin = (340.3) sin (8) Eliminate F D b dividing Eq. 8 b Eq. 7: F D sin (340.3) sin = F D cos (340.3) cos 20 (9) tan 5 Solving Eq. 9 gives = 39.8 Using this result in Eq. 7 gives F D = 416 lb

9 + 3.1 Particles in Two-Dimensional Force Sstems Eample 5, page 1 of 2 5. The collar at weighs 5 lb and can slide freel over the rod. If spring constant k is 2 lb/in., determine the unstretched length of the spring. 1 Free bod diagram of collar D in. N 2 Since the collar can slide freel over the rod, onl a normal force N is present; no friction force (which would be parallel to the rod) is present. 5 lb F spring 3 Equilibrium equations for : + F = 0: N sin + F spring cos = 0 (1) F = 0: N cos + F spring sin lb = 0 (2)

10 3.1 Particles in Two-Dimensional Force Sstems Eample 5, page 2 of 2 4 Geometr: and 6 Geometr: stretched length, 35 D D = 125 stretched length = = = = in. Law of sines: sin 125 sin 35 = 14 in. Solving gives = in Using = 55 and = 35 in Eqs. 1 and 2 and solving simultaneousl gives N = lb and F spring = lb F spring is related to the etension of the spring: F spring = k (stretched length unstretched length) (3) 7 Substituting "stretched length" = = in., k = 2 lb/in., and F spring = lb in Eq. 3 gives Solving gives lb = (2 lb/in.)(9.802 in. unstretched length). unstretched length = 8.28 in.

11 Particles in Two-Dimensional Force Sstems Eample 6, page 1 of 1 6. Two 2-g spheres are suspended b light cords and then given 1 electrostatic charges of opposite sign. alculate the attractive force, F, acting between the spheres. Free bod diagram of sphere 30 F F 30 D F F Weight = mg = (2 g)(1 kg/1000 g)(9.81 m/s 2 ) = kg m/s 2 = N 2 Equilibrium equations for sphere F = 0: F cos 30 + F = 0 (1) F = 0: F sin N = 0 (2) Solving Eqs. 1 and 2 gives F = N = 39.2 mn F = N = 34.0 mn

12 Particles in Two-Dimensional Force Sstems Eample 7, page 1 of 2 7. The figure shows top and side views of a control knob that slides smoothl in a slot cut in a panel and is attached to a spring underneath the panel. Determine the force in the spring and the distance s, for which the knob will be in equilibrium when a 0.5-lb force is applied. The unstretched length of the spring is 4 in., and the weight of the knob is negligible. Top view 0.5 lb 1 F spring Free bod diagram of control knob (Side view) 0.5 lb N 2 Since the knob "slides smoothl", onl a normal force N is present; no friction force is present. s 0.5 lb 3 Equilibrium Equations for control knob : F = 0 F spring cos 0.5 lb = 0 (1) 4 in. k = 5 lb/in. F = 0: F spring sin + N = 0 (2) Side view

13 3.1 Particles in Two-Dimensional Force Sstems Eample 7, page 2 of 2 4 in. 4 Geometr 5 Substituting the "stretched length" epression given in Eq. 4 into the force-etension relationship for a spring gives s F spring = k (stretched length unstretched length) tan = (3) 4 in. s Recall that Eq. 1 is 4 in. = (5 lb/in.) ( 4 in.) (5) sin stretched length of spring = 4 in. = (4) sin F spring cos 0.5 = 0 (Eq. 1 repeated) Substituting the epression for F spring from Eq. 5 into Eq. 1 and multipling through b sin gives 5(4 4 sin ) cos 0.5 sin = 0 (6) Eq. 6 is nonlinear and must be solved numericall b, for eample, using the solver on a calculator. The result is = Using the value = 69.1 in Eqs. 2, 3, and 4 and then solving gives N = 1.31 lb F spring = 1.40 lb s = 1.53 in.

EQUILIBRIUM OF PARTICLES (PROBLEMS)

EQUILIBRIUM OF PARTICLES (PROBLEMS) EQUILIRIUM OF PARICLES (PROLEMS) 1. Determine the force P required to maintain the 2-kg engine in the position for which q=3 o. he diameter of the pulle at is negligible. 2. 4-kg sphere rests on the smooth