SlideShare une entreprise Scribd logo
1  sur  4
Télécharger pour lire hors ligne
EQUILIBRIUM OF NON-CONCURRENT COPLANAR FORCES

      The figure shows a general coplanar force system. It was shown in our previous discussion that such
      general coplanar force system may be reduced to a resultant force R and/or a resultant couple M acting in
      the same plane.




      For the body to be in equilibrium, both linear displacement due to R, and the rotation due to couple M
      should be zero.

Example:

      A boom AC hinged at A supports a 400 N load as shown. Find the force in cable BC, which is attached to
      the wall at B. Neglect self weight of boom. And also solve for the reactions at hinge A.




      Solution:

              From our previous example, we solved this problem knowing that the forces are concurrent at
              point C and we found out that the tension in the cable is T = 292.82 N.

              This time, we will consider the FBD of the boom AC to solve for the tension in the wire T, as well
              as the reaction of the wall at hinge A. Notice that the forces at the boom are non-current.

              FBD:
∑ MA = 0

                                      (Consider CCW rotation Positive)

                         (T sin 75 o )(2 m) − (400N sin 45 o )(2m) = 0

                         T = 292.82 N

                To solve for the reactions at the hinge A; we will take the summation of forces at the horizontal
                and vertical directions.

                         [∑ F   x   = 0] → +

                         R AH − (292.82N) cos 30 o = 0

                         RAH = 253.59 N




                         [∑ Fy         ]
                                    =0 ↑+


                         R AV + (292.82N) sin 30 o − 400 = 0

                         RAV = 253.59 N




Assignment:

   (1) An 8 kg slender rod AB is attached to two collars which may slide along guides without friction shown.
       Determine for equilibrium of rod AB (a) Angle θ (b) the reactions at A and B.

       Ans. 49.11o, RA = 45.31 N, RB = 90.62 N




   (2) A slender rod of length “L” and weight “W” is held in equilibrium position as shown. Determine the angle θ
       and tension in each cable.

       Ans. TAB = 0.596W, θ = 59.21o, TCD = 1.164W
(3) A 3m bar AB of negligible weight is to be kept in horizontal position as shown; Neglecting friction,
    determine distance ‘x’ if P = 300N. Also find the value of ‘P’ if x = 1.5m.




(4) A 200 N cylinder is supported on an 80 N member AB. Find ‘W’ to be attached to the rope neglecting
    friction mass and size of pulley. Take diameter of cylinder 750 mm and length of member AB to be 3m.
    What will be components of reaction at end A?

    Ans. W=117.65N, HA= 98.1N, VA= 221.18N




(5) A 150 N weight is held in equilibrium by a 5 kg uniform bar AB hinged at A as shown. Determine (a)
    components of reactions at A and (b) tension in cable DC. Take length of bar 1m.

    Ans. T = 604.6 N, HA = 302.3 N, VA = 324.5 N
(6) Rod AB is bent in the form of arc of a circle of 0.6m radius. Neglecting friction, obtain components of
        reactions at A and reaction at roller C.

        Ans. RC = 2W, HA = 0, VA = 1.732W




    (7) A 18 kg bar has small wheels at ends A and B. Knowing that length of bar AB is 1m, determine the
        reactions at A and B. Also find tension in cord. Neglect size and weight of rollers.

        Ans. RB = 152.9N, RA = 0, T = 233.6N




Reference:

Elements of Civil Engineering and Engineering Mechanics by SP Nitsure © 2007 Technical Publications Pune

Textbook in Applied Mechanics by MM Malhotra © 1994 New Age International (P) Ltd., Publishers

Contenu connexe

Tendances

Chapter 07 impulse and momentum
Chapter 07 impulse and momentumChapter 07 impulse and momentum
Chapter 07 impulse and momentum
Darwin Quinsaat
 
Friction And Wedges
Friction And WedgesFriction And Wedges
Friction And Wedges
vinaykumars
 
engineering statics: distributed forces-1
engineering statics: distributed forces-1engineering statics: distributed forces-1
engineering statics: distributed forces-1
musadoto
 

Tendances (20)

Chapter 2
Chapter 2Chapter 2
Chapter 2
 
Solution manual for mechanics of materials 10th edition hibbeler sample
Solution manual for mechanics of materials 10th edition hibbeler  sampleSolution manual for mechanics of materials 10th edition hibbeler  sample
Solution manual for mechanics of materials 10th edition hibbeler sample
 
Problems on support reaction.pdf
Problems on support reaction.pdfProblems on support reaction.pdf
Problems on support reaction.pdf
 
Chapter 03
Chapter 03Chapter 03
Chapter 03
 
Solutions manual for statics and mechanics of materials 5th edition by hibbel...
Solutions manual for statics and mechanics of materials 5th edition by hibbel...Solutions manual for statics and mechanics of materials 5th edition by hibbel...
Solutions manual for statics and mechanics of materials 5th edition by hibbel...
 
Dynamics of rigid bodies
Dynamics of rigid bodiesDynamics of rigid bodies
Dynamics of rigid bodies
 
OTA.pdf
OTA.pdfOTA.pdf
OTA.pdf
 
Hook law (stem school).
Hook law (stem school).Hook law (stem school).
Hook law (stem school).
 
Chapter 07 impulse and momentum
Chapter 07 impulse and momentumChapter 07 impulse and momentum
Chapter 07 impulse and momentum
 
Friction And Wedges
Friction And WedgesFriction And Wedges
Friction And Wedges
 
Lec5 total potential_energy_method
Lec5 total potential_energy_methodLec5 total potential_energy_method
Lec5 total potential_energy_method
 
Calculating truss forces
Calculating truss forcesCalculating truss forces
Calculating truss forces
 
Mechanics of Materials 9th Edition Hibbeler Solutions Manual
Mechanics of Materials 9th Edition Hibbeler Solutions ManualMechanics of Materials 9th Edition Hibbeler Solutions Manual
Mechanics of Materials 9th Edition Hibbeler Solutions Manual
 
Engineering Mechanics First Year
Engineering Mechanics First YearEngineering Mechanics First Year
Engineering Mechanics First Year
 
engineering statics: distributed forces-1
engineering statics: distributed forces-1engineering statics: distributed forces-1
engineering statics: distributed forces-1
 
Chapter 03 MECHANICS OF MATERIAL
Chapter 03 MECHANICS OF MATERIALChapter 03 MECHANICS OF MATERIAL
Chapter 03 MECHANICS OF MATERIAL
 
01 01 chapgere[1]
01 01 chapgere[1]01 01 chapgere[1]
01 01 chapgere[1]
 
Friction full
Friction fullFriction full
Friction full
 
5. radial and transverse compo. 2 by-ghumare s m
5. radial and transverse compo.  2 by-ghumare s m5. radial and transverse compo.  2 by-ghumare s m
5. radial and transverse compo. 2 by-ghumare s m
 
Engineering Mechanics Fundamentals
Engineering Mechanics FundamentalsEngineering Mechanics Fundamentals
Engineering Mechanics Fundamentals
 

En vedette

Concurrent Forces
Concurrent ForcesConcurrent Forces
Concurrent Forces
guestb54490
 
MECHANICS ENGINEERING - Equilibrium
MECHANICS ENGINEERING - EquilibriumMECHANICS ENGINEERING - Equilibrium
MECHANICS ENGINEERING - Equilibrium
Cik Aisyahfitrah
 
Fundamentals of statics
Fundamentals of statics Fundamentals of statics
Fundamentals of statics
sujay762
 
Hibbeler chapter5
Hibbeler chapter5Hibbeler chapter5
Hibbeler chapter5
laiba javed
 

En vedette (14)

Concurrent Forces
Concurrent ForcesConcurrent Forces
Concurrent Forces
 
Coplanar concurrent forces
Coplanar concurrent forcesCoplanar concurrent forces
Coplanar concurrent forces
 
1 force sys type
1 force sys type1 force sys type
1 force sys type
 
Trusses and its applications
Trusses and its applicationsTrusses and its applications
Trusses and its applications
 
Application of the trusses
Application of the trussesApplication of the trusses
Application of the trusses
 
equilibrium-of-rigid-body
equilibrium-of-rigid-bodyequilibrium-of-rigid-body
equilibrium-of-rigid-body
 
MECHANICS ENGINEERING - Equilibrium
MECHANICS ENGINEERING - EquilibriumMECHANICS ENGINEERING - Equilibrium
MECHANICS ENGINEERING - Equilibrium
 
Presentation (truss) by imran khan.
Presentation (truss) by imran khan.Presentation (truss) by imran khan.
Presentation (truss) by imran khan.
 
Fundamentals of statics
Fundamentals of statics Fundamentals of statics
Fundamentals of statics
 
Trusses
TrussesTrusses
Trusses
 
PPT OF TRUSSES
PPT OF TRUSSESPPT OF TRUSSES
PPT OF TRUSSES
 
Lec.2 statically determinate structures & statically indeterminate struct...
Lec.2 statically determinate structures & statically indeterminate struct...Lec.2 statically determinate structures & statically indeterminate struct...
Lec.2 statically determinate structures & statically indeterminate struct...
 
Hibbeler chapter5
Hibbeler chapter5Hibbeler chapter5
Hibbeler chapter5
 
Engineering Mechanics
Engineering MechanicsEngineering Mechanics
Engineering Mechanics
 

Similaire à Engmech 06 (equilibrium of non_concurrent force system)

Engineering mechanics-question-and-answers-for-gate-ias
Engineering mechanics-question-and-answers-for-gate-iasEngineering mechanics-question-and-answers-for-gate-ias
Engineering mechanics-question-and-answers-for-gate-ias
hitusp
 
2nd codition of equilibrium
2nd codition of equilibrium2nd codition of equilibrium
2nd codition of equilibrium
Nestor Enriquez
 
6161103 5.3 equations of equilibrium
6161103 5.3 equations of equilibrium6161103 5.3 equations of equilibrium
6161103 5.3 equations of equilibrium
etcenterrbru
 
III - resultant of non-concurrent forces
III -  resultant of non-concurrent forcesIII -  resultant of non-concurrent forces
III - resultant of non-concurrent forces
djpprkut
 
Solução mecânica vetorial para engenheiros.
Solução mecânica vetorial para engenheiros.Solução mecânica vetorial para engenheiros.
Solução mecânica vetorial para engenheiros.
wedson Oliveira
 

Similaire à Engmech 06 (equilibrium of non_concurrent force system) (20)

Equilibrium 3
Equilibrium 3Equilibrium 3
Equilibrium 3
 
Equilibrium
EquilibriumEquilibrium
Equilibrium
 
ME 245_ 2.pptx
ME 245_ 2.pptxME 245_ 2.pptx
ME 245_ 2.pptx
 
Stresses and strains (Part 1)
Stresses and strains (Part 1)Stresses and strains (Part 1)
Stresses and strains (Part 1)
 
Lecture-3-1.pptx
Lecture-3-1.pptxLecture-3-1.pptx
Lecture-3-1.pptx
 
Fundamental of Statics (Part 2)
Fundamental of Statics (Part 2)Fundamental of Statics (Part 2)
Fundamental of Statics (Part 2)
 
Mechanic Machine
Mechanic MachineMechanic Machine
Mechanic Machine
 
Ch12 ssm
Ch12 ssmCh12 ssm
Ch12 ssm
 
Ch14 ssm
Ch14 ssmCh14 ssm
Ch14 ssm
 
Engineering mechanics-question-and-answers-for-gate-ias
Engineering mechanics-question-and-answers-for-gate-iasEngineering mechanics-question-and-answers-for-gate-ias
Engineering mechanics-question-and-answers-for-gate-ias
 
6. forces, density and pressure examples
6.   forces, density and pressure examples6.   forces, density and pressure examples
6. forces, density and pressure examples
 
Gr
GrGr
Gr
 
2nd codition of equilibrium
2nd codition of equilibrium2nd codition of equilibrium
2nd codition of equilibrium
 
TIPLER CAP r25
TIPLER CAP r25TIPLER CAP r25
TIPLER CAP r25
 
6161103 5.3 equations of equilibrium
6161103 5.3 equations of equilibrium6161103 5.3 equations of equilibrium
6161103 5.3 equations of equilibrium
 
III - resultant of non-concurrent forces
III -  resultant of non-concurrent forcesIII -  resultant of non-concurrent forces
III - resultant of non-concurrent forces
 
Lecture--15, 16 (Deflection of beams).pptx
Lecture--15, 16 (Deflection of beams).pptxLecture--15, 16 (Deflection of beams).pptx
Lecture--15, 16 (Deflection of beams).pptx
 
Problems on simply supported beams (udl , uvl and couple)
Problems on simply supported beams (udl , uvl and couple)Problems on simply supported beams (udl , uvl and couple)
Problems on simply supported beams (udl , uvl and couple)
 
Solução mecânica vetorial para engenheiros.
Solução mecânica vetorial para engenheiros.Solução mecânica vetorial para engenheiros.
Solução mecânica vetorial para engenheiros.
 
Shear force & Bending Moment
Shear force & Bending MomentShear force & Bending Moment
Shear force & Bending Moment
 

Plus de physics101

Bm 10 mr and e-marketing
Bm 10 mr and e-marketingBm 10 mr and e-marketing
Bm 10 mr and e-marketing
physics101
 
Bm 9 marketing element for be and csr
Bm 9 marketing element for be and csrBm 9 marketing element for be and csr
Bm 9 marketing element for be and csr
physics101
 
Bm 8 brand equity & management
Bm 8 brand equity & managementBm 8 brand equity & management
Bm 8 brand equity & management
physics101
 
Bernoulli's equation
Bernoulli's equationBernoulli's equation
Bernoulli's equation
physics101
 
Fundamental law of gearing
Fundamental law of gearingFundamental law of gearing
Fundamental law of gearing
physics101
 
Centrifugal pumps in series and parallel
Centrifugal pumps in series and parallelCentrifugal pumps in series and parallel
Centrifugal pumps in series and parallel
physics101
 
Laminar and turbulent f low
Laminar and turbulent f lowLaminar and turbulent f low
Laminar and turbulent f low
physics101
 
Engine systems
Engine systemsEngine systems
Engine systems
physics101
 
Bomb calorimeter experiment
Bomb calorimeter experimentBomb calorimeter experiment
Bomb calorimeter experiment
physics101
 
Flash and fire point
Flash and fire pointFlash and fire point
Flash and fire point
physics101
 
Calibration of pressure gauges
Calibration of pressure gaugesCalibration of pressure gauges
Calibration of pressure gauges
physics101
 
Hazardous substances 01
Hazardous substances 01Hazardous substances 01
Hazardous substances 01
physics101
 
Hazard communication lect 1
Hazard communication lect 1Hazard communication lect 1
Hazard communication lect 1
physics101
 
Scientific methods
Scientific methodsScientific methods
Scientific methods
physics101
 

Plus de physics101 (20)

Bm 10 mr and e-marketing
Bm 10 mr and e-marketingBm 10 mr and e-marketing
Bm 10 mr and e-marketing
 
Bm 9 marketing element for be and csr
Bm 9 marketing element for be and csrBm 9 marketing element for be and csr
Bm 9 marketing element for be and csr
 
Bm 8 brand equity & management
Bm 8 brand equity & managementBm 8 brand equity & management
Bm 8 brand equity & management
 
Manuala hw
Manuala hwManuala hw
Manuala hw
 
F1
F1F1
F1
 
F6
F6F6
F6
 
Fire safety
Fire safetyFire safety
Fire safety
 
Bernoulli's equation
Bernoulli's equationBernoulli's equation
Bernoulli's equation
 
Gear trains
Gear trainsGear trains
Gear trains
 
Fundamental law of gearing
Fundamental law of gearingFundamental law of gearing
Fundamental law of gearing
 
Gears
GearsGears
Gears
 
Centrifugal pumps in series and parallel
Centrifugal pumps in series and parallelCentrifugal pumps in series and parallel
Centrifugal pumps in series and parallel
 
Laminar and turbulent f low
Laminar and turbulent f lowLaminar and turbulent f low
Laminar and turbulent f low
 
Engine systems
Engine systemsEngine systems
Engine systems
 
Bomb calorimeter experiment
Bomb calorimeter experimentBomb calorimeter experiment
Bomb calorimeter experiment
 
Flash and fire point
Flash and fire pointFlash and fire point
Flash and fire point
 
Calibration of pressure gauges
Calibration of pressure gaugesCalibration of pressure gauges
Calibration of pressure gauges
 
Hazardous substances 01
Hazardous substances 01Hazardous substances 01
Hazardous substances 01
 
Hazard communication lect 1
Hazard communication lect 1Hazard communication lect 1
Hazard communication lect 1
 
Scientific methods
Scientific methodsScientific methods
Scientific methods
 

Engmech 06 (equilibrium of non_concurrent force system)

  • 1. EQUILIBRIUM OF NON-CONCURRENT COPLANAR FORCES The figure shows a general coplanar force system. It was shown in our previous discussion that such general coplanar force system may be reduced to a resultant force R and/or a resultant couple M acting in the same plane. For the body to be in equilibrium, both linear displacement due to R, and the rotation due to couple M should be zero. Example: A boom AC hinged at A supports a 400 N load as shown. Find the force in cable BC, which is attached to the wall at B. Neglect self weight of boom. And also solve for the reactions at hinge A. Solution: From our previous example, we solved this problem knowing that the forces are concurrent at point C and we found out that the tension in the cable is T = 292.82 N. This time, we will consider the FBD of the boom AC to solve for the tension in the wire T, as well as the reaction of the wall at hinge A. Notice that the forces at the boom are non-current. FBD:
  • 2. ∑ MA = 0 (Consider CCW rotation Positive) (T sin 75 o )(2 m) − (400N sin 45 o )(2m) = 0 T = 292.82 N To solve for the reactions at the hinge A; we will take the summation of forces at the horizontal and vertical directions. [∑ F x = 0] → + R AH − (292.82N) cos 30 o = 0 RAH = 253.59 N [∑ Fy ] =0 ↑+ R AV + (292.82N) sin 30 o − 400 = 0 RAV = 253.59 N Assignment: (1) An 8 kg slender rod AB is attached to two collars which may slide along guides without friction shown. Determine for equilibrium of rod AB (a) Angle θ (b) the reactions at A and B. Ans. 49.11o, RA = 45.31 N, RB = 90.62 N (2) A slender rod of length “L” and weight “W” is held in equilibrium position as shown. Determine the angle θ and tension in each cable. Ans. TAB = 0.596W, θ = 59.21o, TCD = 1.164W
  • 3. (3) A 3m bar AB of negligible weight is to be kept in horizontal position as shown; Neglecting friction, determine distance ‘x’ if P = 300N. Also find the value of ‘P’ if x = 1.5m. (4) A 200 N cylinder is supported on an 80 N member AB. Find ‘W’ to be attached to the rope neglecting friction mass and size of pulley. Take diameter of cylinder 750 mm and length of member AB to be 3m. What will be components of reaction at end A? Ans. W=117.65N, HA= 98.1N, VA= 221.18N (5) A 150 N weight is held in equilibrium by a 5 kg uniform bar AB hinged at A as shown. Determine (a) components of reactions at A and (b) tension in cable DC. Take length of bar 1m. Ans. T = 604.6 N, HA = 302.3 N, VA = 324.5 N
  • 4. (6) Rod AB is bent in the form of arc of a circle of 0.6m radius. Neglecting friction, obtain components of reactions at A and reaction at roller C. Ans. RC = 2W, HA = 0, VA = 1.732W (7) A 18 kg bar has small wheels at ends A and B. Knowing that length of bar AB is 1m, determine the reactions at A and B. Also find tension in cord. Neglect size and weight of rollers. Ans. RB = 152.9N, RA = 0, T = 233.6N Reference: Elements of Civil Engineering and Engineering Mechanics by SP Nitsure © 2007 Technical Publications Pune Textbook in Applied Mechanics by MM Malhotra © 1994 New Age International (P) Ltd., Publishers