Fy = P sin 50 + F cos = 0 Solving the two simultaneous equations for P and F,

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1 ENGR Statics and Mechanics of Materials 1 (161) Homework # Solution Set 1. Summing forces in the y-direction allows one to determine the magnitude of F : Fy 1000 sin sin 37 F sin 45 0 F N Then, summing forces in the x-direction and using the value of F allows one to determine the magnitude of F 1 : Fx F 1 + F cos cos cos 37 0 F N Thus, F N, F 544 N. (a) Noting that the force F exerted on the block by the smooth surface is in the direction normal to the surface, the free-body diagram for the block is y y P x 0 30 x lb The xy- and x y -coordinate systems are for the two alternate solution approaches below. (b) Using the xy-coordinate system, the force equilibrium equations are: Fx P cos 50 F sin 30 0 F Fy P sin 50 + F cos Solving the two simultaneous equations for P and F, P lb, F lb An alternative approach, that avoids having to solve simultaneous equations, is to use the x y -coordinate system. First summing forces in the x -direction, Fx P cos sin 30 0 P lb Then summing forces in the y -direction, Fy F + P sin cos 30 0 F lb Using either approach, the answer is P 53. lb, F 68.4 lb

2 3. The vectors and distances from D to A, B, and C are r DA 10i + 5j 8k r DB 5i 5j 8k r DC 7i j 8k r DA (10) + (5) + ( 8) m r DA (5) + ( 5) + ( 8) m r DA ( 7) + ( ) + ( 8) m It follows that the unit vectors in the directions of the cable forces are e DA r DA r DA 0.774i j k e DB r DB r DB i j k e DC r DC r DC i j k Noting that the net lift of the balloon is 10k (kn), the force equilibrium equation is F 10k + FDA e DA + F DB e DB + F DC e DC 0 where F DA, F DB, and F DC are the forces exerted on the balloon by the three cables. This gives three simultaneous equations to be solved for F DA, F DB, and F DC : 0.774F DA F DB F DC F DA F DB F DC F DA F DB F DC 10 kn Solving these simultaneous equations gives F DA 5.73 kn, F DB 1.48 kn, F DC 7.51 kn

3 4. Note that the weight of the 500 kg mass is (500 kg)(9.81 m/s ) 4905 N, let the tensions in cables A and B be F A and F B, respectively, and consider the free-body diagram below: y F B Fy F B sin F A 45 x F B N Fx F B cos 45 F A 0 F A 4905 N 4905 N The minimum allowable cable cross-sectional area, A min, is given in terms of the cable tension, F, and allowable normal stress, 150 MPa, by σ F A A min F. The minimum allowable cross-sectional area is related to the minimum allowable diameter, d min, by A min π 4 d min. Cable A Cable B π 4 d min A min π 4 d min A min 4905 N N/m m d min m 6.45 mm N N/m m d min m 7.67 mm

4 5. Note: In a problem, like this one, where an allowable normal stress is given without mentioning tension or compression, the allowable normal stress is assumed to be the same in both tension and compression. This may not be entirely clear from the textbook. Use appropriate free-body diagrams, as shown below, to determine the axial force, and subsequently the minimum allowable cross-sectional area, for each pipe section. Segment AB 30 kip 5 kip 15 kip D C B Fx P AB P AB 10 kip P AB A AB P AB 10 kip in 4 kip/in Segment BC P BC C 5 kip 15 kip D Fx P BC P BC 0 kip A BC P BC 0 kip in 4 kip/in Segment CD P CD Fx P CD P CD 15 kip D 15 kip A CD P CD 15 kip 0.65 in 4 kip/in

5 6. Using a free-body diagram of the solid aluminum rod to find the shear force V : V P L Fx P V 0 V P 100 lb The interface area that the shear force is distributed over is Thus, the average shear stress is A (πd)l π(0.75 in)(0.75 in) in τ V A 100 lb 56.6 psi in 7. Note first that the cross-sectional area of the bar is A (0.15 m)t and the angle of inclination of the weld is θ There will be a minimum thickness based on the allowable normal stress and a minimum thickness based on the allowable shear stress. The answer is the larger of the two, so that neither allowable stress is exceeded. Normal Stress Shear Stress Answer: σ n P A cos θ A min P cos θ A min N N/m cos m t min A min 0.15 m m τ n P A sin θ cos θ A min P sin θ cos θ τ all A min N N/m sin 30 cos m t min A min 0.15 m m t min 56.3 mm

6 8. Note first that the angle of inclination of the joint is θ 90 φ and the range of inclinations to be considered is 0 θ 45. The cross-sectional area of the member is A ( in)(3 in) 6 in. Normal Stress: Given the equation for the normal stress on the plane of the joint, σ n P A cos θ the maximum safe load based on the allowable tensile stress in the glue is P σ A cos θ (50 psi)(6 in ) cos θ 300 lb cos θ Shear Stress: Given the equation for the shear stress on the plane of the joint, τ n P sin θ cos θ A the maximum safe load based on the allowable shear stress in the glue is P τ τ alla sin θ cos θ (35 psi)(6 in ) sin θ cos θ 10 lb sin θ cos θ At any given inclination level, θ, the maximum safe load is the smaller of P σ and P τ. Plotting these results as a function of θ for 0 θ 45, P all (lb) 600 P opt P σ P τ θ opt 45 θ (a) The optimum angle is where the two plots cross, P σ P τ, 300 lb cos θ opt 10 lb tan θ opt 10 sin θ opt cos θ opt 300 θ opt 35.0 Thus, φ opt 90 θ opt 55.0 (b) The maximum safe load at this optimum angle is P opt 300 lb cos θ opt 10 lb sin θ opt cos θ opt 447 lb