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- 1. 5.3 Equations of EquilibriumFor equilibrium of a rigid body in 2D, ∑Fx = 0; ∑Fy = 0; ∑MO = 0∑Fx and ∑Fy represent the algebraic sums of thex and y components of all the forces acting onthe body∑MO represents the algebraic sum of the couplemoments and moments of the force componentsabout an axis perpendicular to x-y plane andpassing through arbitrary point O, which may lieon or off the body
- 2. 5.3 Equations of EquilibriumAlternative Sets of Equilibrium Equations For coplanar equilibrium problems, ∑Fx = 0; ∑Fy = 0; ∑MO = 0 can be used Two alternative sets of three independent equilibrium equations may also be used ∑Fa = 0; ∑MA = 0; ∑MB = 0 When applying these equations, it is required that a line passing through points A and B is not perpendicular to the a axis
- 3. 5.3 Equations of EquilibriumAlternative Sets of Equilibrium Equations Consider FBD of an arbitrarily shaped body All the forces on FBD may be replaced by an equivalent resultant force FR = ∑F acting at point A and a resultant moment MRA = ∑MA If ∑MA = 0 is satisfied, MRA = 0
- 4. 5.3 Equations of EquilibriumAlternative Sets of Equilibrium Equations If FR satisfies ∑Fa = 0, there is no component along the a axis and its line of axis is perpendicular to the a axis If ∑MB = 0 where B does not lies on the line of action of FR, FR = 0 Since ∑F = 0 and ∑MA = 0, the body is in equilibrium
- 5. 5.3 Equations of EquilibriumAlternative Sets of Equilibrium Equations A second set of alternative equations is ∑MA = 0; ∑MB = 0; ∑MC = 0 Points A, B and C do not lie on the same line Consider FBD, if If ∑MA = 0, MRA = 0 ∑MA = 0 is satisfied if line of action of FR passes through point B ∑MC = 0 where C does not lie on line AB FR = 0 and the body is in equilibrium
- 6. 5.3 Equations of EquilibriumProcedure for AnalysisFree-Body Diagram Establish the x, y, z coordinates axes in any suitable orientation Draw an outlined shape of the body Show all the forces and couple moments acting on the body Label all the loadings and specify their directions relative to the x, y axes
- 7. 5.3 Equations of EquilibriumProcedure for AnalysisFree-Body Diagram The sense of a force or couple moment having an unknown magnitude but known line of action can be assumed Indicate the dimensions of the body necessary for computing the moments of forces
- 8. 5.3 Equations of EquilibriumProcedure for AnalysisEquations of Equilibrium Apply the moment equation of equilibrium ∑MO = 0 about a point O that lies on the intersection of the lines of action of the two unknown forces The moments of these unknowns are zero about O and a direct solution the third unknown can be obtained
- 9. 5.3 Equations of EquilibriumProcedure for AnalysisEquations of Equilibrium When applying the force equilibrium ∑Fx = 0 and ∑Fy = 0, orient the x and y axes along the lines that will provide the simplest resolution of the forces into their x and y components If the solution yields a negative result scalar, the sense is opposite to that was assumed on the FBD
- 10. 5.3 Equations of EquilibriumExample 5.6Determine the horizontal and verticalcomponents of reaction for the beam loaded.Neglect the weight of the beam in thecalculations.
- 11. 5.3 Equations of EquilibriumSolutionFBD 600N force is represented by its x and y components 200N force acts on the beam at B and is independent of the force components Bx and By, which represent the effect of the pin on the beam
- 12. 5.3 Equations of EquilibriumSolutionEquations of Equilibrium + → ∑ M B = 0; 600 cos 45o N − Bx = 0 Bx = 424 N A direct solution of Ay can be obtained by applying ∑MB = 0 about point B Forces 200N, Bx and By all create zero moment about B
- 13. 5.3 Equations of EquilibriumSolution∑ M B = 0;100 N (2m) + (600 sin 45o N )(5m) − (600 cos 45o N )(0.2m) − Ay (7 m) = 0Ay = 319 N+ ↑ ∑ Fy = 0;319 N − 600 sin 45o N − 100 N − 200 N + B y = 0B y = 405 N
- 14. 5.3 Equations of EquilibriumSolutionChecking,∑ M A = 0;− (600 sin 45o N )(2m) − (600 cos 45o N )(0.2m) − (100 N )(5m)− (200 N )(7m) + B y (7m) = 0B y = 405 N
- 15. 5.3 Equations of EquilibriumExample 5.7The cord supports a force of 500N and wraps overthe frictionless pulley. Determine the tension in thecord at C and the horizontaland vertical components atpin A.
- 16. 5.3 Equations of EquilibriumSolutionFBD of the cord and pulley Principle of action: equal but opposite reaction observed in the FBD Cord exerts an unknown load distribution p along part of the pulley’s surface Pulley exerts an equal but opposite effect on the cord
- 17. 5.3 Equations of EquilibriumSolutionFBD of the cord and pulley Easier to combine the FBD of the pulley and contracting portion of the cord so that the distributed load becomes internal to the system and is eliminated from the analysis
- 18. 5.3 Equations of EquilibriumSolutionEquations of Equilibrium ∑ M A = 0; 500 N (0.2m) + T (0.2m) = 0 T = 500 NTension remains constant as cordpasses over the pulley (true forany angle at which the cord isdirected and for any radius ofthe pulley
- 19. 5.3 Equations of EquilibriumSolution + → ∑ Fx = 0; − Ax + 500 sin 30o N = 0 Ax = 250 N + ↑ ∑ Fy = 0; Ay − 500 N − 500 cos 30o N = 0 Ay = 933N
- 20. 5.3 Equations of EquilibriumExample 5.8The link is pin-connected at a and rest asmooth support at B. Compute the horizontaland vertical components of reactions at pin A
- 21. 5.3 Equations of EquilibriumSolutionFBD Reaction NB is perpendicular to the link at B Horizontal and vertical components of reaction are represented at A
- 22. 5.3 Equations of EquilibriumSolutionEquations of Equilibrium ∑ M A = 0;− 90 N .m − 60 N (1m) + N B (0.75m) = 0N B = 200 N+ → ∑ Fx = 0;Ax − 200 sin 30o N = 0Ax = 100 N
- 23. 5.3 Equations of EquilibriumSolution+ ↑ ∑ Fy = 0;Ay − 60 N − 200 cos 30o N = 0Ay = 233N
- 24. 5.3 Equations of EquilibriumExample 5.9The box wrench is used to tighten the bolt atA. If the wrench does not turn when the loadis applied to the handle, determine thetorque or moment applied to the bolt and theforce of the wrench on the bolt.
- 25. 5.3 Equations of EquilibriumSolutionFBD Bolt acts as a “fixed support” it exerts force components Ax and Ay and a torque MA on the wrench at A
- 26. 5.3 Equations of EquilibriumSolutionEquations of Equilibrium∑ M A = 0; 12 M A − 52 N (0.3m) − (30 sin 60o N )(0.7 m) = 0 13 M A = 32.6 N .m+ → ∑ Fx = 0; 5Ax − 52 N + 30 cos 60o N = 0 13 Ax = 5.00 N
- 27. 5.3 Equations of EquilibriumSolution+ ↑ ∑ Fy = 0; 12 Ay − 52 N − 30 sin 60o N = 0 13 Ay = 74.0 NCCW ↑ ∑ M y = 0; 12 M A − 52 N (0.3m ) − (30 sin 60° N )(0.7 m ) = 0 13 M A = 32.6 Nm
- 28. 5.3 Equations of EquilibriumSolution Point A was chosen for summing the moments as the lines of action of the unknown forces Ax and Ay pass through this point and these forces are not included in the moment summation MA must be included Couple moment MA is a free vector and represents the twisting resistance of the bolt on the wrench
- 29. 5.3 Equations of EquilibriumSolution By Newton’s third law, the wrench exerts an equal but opposite moment or torque on the bolt For resultant force on the wrench, FA = (5.00)2 + (74.0)2 = 74.1N For directional sense, 74.0 N −1 θ = tan = 86.1o 5.00 N FA acts in the opposite direction on the bolt
- 30. 5.3 Equations of EquilibriumSolution Checking, ∑MC = 0; 12N(0.4m) + 32.6N.m − 74.0N(0.7m) = 0 52 13 19.2N.m + 32.6N.m − 51.8N.m = 0
- 31. 5.3 Equations of EquilibriumExample 5.10Placement of concrete from thetruck is accomplished using thechute. Determine the force that thehydraulic cylinder and the truckframe exert on the chute to hold it inposition. The chute and the wetconcrete contained along its lengthhave a uniform weight of 560N/m.
- 32. 5.3 Equations of EquilibriumSolution Idealized model of the chute Assume chute is pin connected to the frame at A and the hydraulic cylinder BC acts as a short link
- 33. 5.3 Equations of EquilibriumSolutionFBD Since chute has a length of 4m, total supported weight is (560N/m)(4m) = 2240N, which is assumed to act at its midpoint, G The hydraulic cylinder exerts a horizontal force FBC on the chuteEquations of Equilibrium A direct solution of FBC is obtained by the summation about the pin at A
- 34. 5.3 Equations of EquilibriumSolution∑ M A = 0;− FBC (0.5m) + 2240 cos 30o N (2m)+ 2240 sin 30o N (0.0625m) = 0FBC = 7900 N+ → ∑ Fx = 0;− Ax + 7900 N = 0Ax = 7900 N
- 35. 5.3 Equations of EquilibriumSolution+ ↑ ∑ Fy = 0;Ay − 2240 N = 0Ay = 2240 NChecking,∑ M B = 0;− 7900 N (0.5m) + 2240 N (1cos 30o m) +2240 cos 30o N (1m) + 2240 sin 30o N (0.0625m) = 0
- 36. 5.3 Equations of EquilibriumExample 5.11The uniform smooth rod is subjected to a force andcouple moment. If the rod is supported at A by asmooth wall and at B and C either at the top orbottom by rollers, determinethe reactions at these supports.Neglect the weight of the rod.
- 37. 5.3 Equations of EquilibriumSolutionFBD All the support reactions act normal to the surface of contact since the contracting surfaces are smooth Reactions at B and C are acting in the positive y’ direction Assume only the rollers located on the bottom of the rod are used for support
- 38. 5.3 Equations of EquilibriumSolutionEquations of Equilibrium∑ M A = 0;− B y (2m) + 4000 N .m − C y (6m) + (300 cos 30o N )(8m) = 0+ → ∑ Fx = 0;C y sin 30o + B y sin 30o − Ax = 0+ ↑ ∑ Fy = 0;− 300 N + C y cos 30o + B y cos 30o = 0
- 39. 5.3 Equations of EquilibriumSolution Note that the line of action of the force component passes through point A and this force is not included in the moment equation B y = −1000.0 N = −1kN C y = 1346.4 N = 1.35kN Since By’ is negative scalar, the sense of By’ is opposite to shown in the FBD
- 40. 5.3 Equations of EquilibriumSolution Top roller at B serves as the support rather than the bottom one 1346.4 sin 30o N − 1000.0 sin 30o N − Ax = 0 Ax = 173N
- 41. 5.3 Equations of EquilibriumExample 5.12The uniform truck ramp has a weight of1600N ( ≈ 160kg ) and is pinned to the bodyof the truck at each end and held in positionby two side cables.Determine the tensionin the cables.
- 42. 5.3 Equations of EquilibriumSolution Idealized model of the ramp Center of gravity located at the midpoint since the ramp is approximately uniformFBD of the Ramp
- 43. 5.3 Equations of EquilibriumSolutionEquations of Equilibrium∑ M A = 0;− T cos 20o (2 sin 30o m) + T sin 20o (2 cos 30o )+ 1600 N (1.5 cos 30o ) = 0T = 5985 NBy the principle of transmissibility, locate T at C d 2m =sin 10 sin 20o od = 1.0154m
- 44. 5.3 Equations of EquilibriumSolutionSince there are two cables supporting theramp, T’ = T/2 = 2992.5N

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