• Rc Hibbeler Solutions 13th Free
• Engineering Mechanics Statics 14th Edition Solution Manual Pdf
• Hibbeler Statics 13th Edition Solutions Pdf

Engineering mechanics statics 14th edition hibbeler solutions manual.1.Engineering Mechanics Statics 14th Edition Hibbeler Solutions ManualFull clear download( no error farmatting) at:60° and 450 N, determine the magnitude of theresultant force and its direction, measured counterclockwise yfrom the positive x axis. FSOLUTIONThe parallelogram law of addition and the triangular rule are shown in Figs. A and b,respectively.Applying the law of consines to Fig.

B,700245022(700)(450) cos 45°x15700 N497.01 N 497 N Ans.This yieldssin700sin 45°497.0195.19°Thus, the direction of angle of Fpositive axis, ismeasured counterclockwise from the60° 95.19° 60° 155° Ans.Ans:FR = 497 Nf = 15522.2–2.If the magnitude of the resultant force is to be 500 N, ydirected along the positive y axis, determine the magnitude Fof force F and its direction u.SOLUTIONThe parallelogram law of addition and the triangular rule are shown in Figs. A and b,respectively.Applying the law of cosines to Fig. B,F = 25002+ 7002- 2(500)(700) cos 105°ux15700 N= 959.78 N = 960 NApplying the law of sines to Fig. B, and using this result, yieldsAns.sin (90° + u)700=sin 105°959.78u = 45.2° Ans.Ans:F = 960 Nu = 45.223.2–3.Determine the magnitude of the resultant force FR = F1 + F2 yand its direction,measured counterclockwise from the positivex axis. F1 250 lb30SOLUTIONFR = 2(250)2+ (375)2- 2(250)(375) cos 75° = 393.2 = 393 lbxAns. 45393.2 250sin 75°=sin uu = 37.89°f = 360° - 45° + 37.89° = 353° Ans.F2 375 lbAns:FR = 393 lbf = 35324.2–4.The vertical force F acts downward at on the two-memberedframe. Determine the magnitudes of the two components ofF directed along the axes of.

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Set 500 N.BSOLUTION AParallelogram Law: The parallelogram law of addition is shown in Fig. A.Trigonometry: Using the law of sines (Fig.

B), we haveF500 Csin 60° sin 75°448 N Ans.sin 45°500sin 75°366 N Ans.Ans:FAB = 448 NFAC = 366 N.25.2–5.Solve Prob. 2-4 with F = 350 lb.B45SOLUTIONParallelogram Law: The parallelogram law of addition is shown in Fig. A.Trigonometry: Using the law of sines (Fig. B), we haveAF 30FAB 350 Csin 60°=sin 75°FAB = 314 lb Ans.FAC 350sin 45°=sin 75°FAC = 256 lb Ans.Ans:FAB = 314 lbFAC = 256 lb26.2–6.Determine the magnitude of the resultant forceFR = F1 + F2 and its direction, measured clockwise fromthe positive u axis.v3075F1 4 kN30uF2 6 kNSolutionParallelogram Law. The parallelogram law of addition is shown in Fig. Applying Law of cosines by referring to Fig.

B,FR = 242+ 62- 2(4)(6) cos 105 = 8.026 kN = 8.03 kN Ans.Using this result to apply Law of sines, Fig. B,sin u6=sin 105; u = 46.228.026Thus, the direction f of FR measured clockwise from the positive u axis isf = 46.22 - 45 = 1.22 Ans.Ans:f = 1.2227.2–7.Resolve the force F1 into components acting along the uand v axes and determine the magnitudes of the components.v3075F1 4 kN30uF2 6 kNSolutionParallelogram Law. The parallelogram law of addition is shown in Fig. Applying the sines law by referring to Fig. B.(F1)v 4sin 45=sin 105; (F1)v = 2.928 kN = 2.93 kN Ans.(F1)u 4sin 30=sin 105; (F1)u = 2.071 kN = 2.07 kN Ans.Ans:(F1)v = 2.93 kN(F1)u = 2.07 kN28.2–8.Resolve the force F2 into components acting along the uand v axes and determine the magnitudes of the components.v3075F1 4 kN30uF2 6 kNSolutionParallelogram Law. The parallelogram law of addition is shown in Fig.

Rc Hibbeler Solutions 13th Free

Applying the sines law of referring to Fig. B,(F2)u 6sin 75=sin 75; (F2)u = 6.00 kN Ans.(F2)v 6sin 30=sin 75; (F2)v = 3.106 kN = 3.11 kN Ans.Ans:(F2)u = 6.00 kN(F2)v = 3.11 kN29.2–9.If the resultant force acting on the support is to be 1200 lb, Fdirected horizontally to the right, determine the force F in Arope A and the corresponding angle u.uB60900 lbSolutionParallelogram Law. The parallelogram law of addition is shown in Fig. Applying the law of cosines by referring to Fig. B,F = 29002+ 12002- 2(900)(1200) cos 30 = 615.94 lb = 616 lb Ans.Using this result to apply the sines law, Fig.

Engineering Mechanics Statics 14th Edition Solution Manual Pdf

B,sin u sin 30900=615.94; u = 46.94 = 46.9 Ans.Ans:F = 616 lbu = 46.930.2–10.Determine the magnitude of the resultant force and itsdirection, measured counterclockwise from the positive x axis.y800 lb40SolutionParallelogram Law. The parallelogram law of addition is shown in Fig. Applying the law of cosines by referring to Fig. B,FR = 28002+ 5002- 2(800)(500) cos 95 = 979.66 lb = 980 lb Ans.Using this result to apply the sines law, Fig. B,35x500 lbsin u sin 95500=979.66; u = 30.56Thus, the direction f of FR measured counterclockwise from the positive x axis isf = 50 - 30.56 = 19.44 = 19.4 Ans.Ans:FR = 980 lbf = 19.431.2–11.The plate is subjected to the two forces at A and B asshown. If u = 60°, determine the magnitude of the resultantof these two forces and its direction measured clockwisefrom the horizontal.FA 8 kNuASOLUTIONParallelogram Law: The parallelogram law of addition is shown in Fig. A.Trigonometry: Using law of cosines (Fig.

Hibbeler Statics 13th Edition Solutions Pdf

B), we haveFR = 282+ 62- 2(8)(6) cos 100°= 10.80 kN = 10.8 kNThe angle u can be determined using law of sines (Fig. B).Ans.40BFB 6 kNsin u6=sin 100°10.80sin u = 0.5470u = 33.16°Thus, the direction f of FR measured from the x axis isf = 33.16° - 30° = 3.16° Ans.Ans:FR = 10.8 kNf = 3.1632.2–12.Determine the angle of u for connecting member A to theplate so that the resultant force of FA and FB is directedhorizontally to the right. Also, what is the magnitude of theresultant force?FA 8 kNuASOLUTIONParallelogram Law: The parallelogram law of addition is shown in Fig. A.Trigonometry: Using law of sines (Fig.b), we havesin (90° - u)6=sin 50°8 40sin (90° - u) = 0.5745u = 54.93° = 54.9° Ans.BFB 6 kNFrom the triangle, f = 180° - (90° - 54.93°) - 50° = 94.93°.