Mechanics_Ramkumar_Chapter_03_Equilibrium of Coplanar Forces

., ,!Equilibrium: A body (or particle) is said to be in static equilibrium if it is as rest. For a body (or particle) to bein equilibrium, resultant of the force system acting on it should be completely zero i.e. sum of all forces shouldbe zero ( F = 0) and sum of all moments should be zero ( M = 0).Stable, Unstable and Neutral Equilibrium: When a system is disturbed from its position of equilibriumby the slightest force, it produces a small displacement from equilibrium position. If the system returns to theoriginal position as soon as the disturbing force is removed, the equilibrium is said to be a stable (Fig. 1). On the removal of the disturbing force, if the system moves away from the original position, it is calledto be in unstable equilibrium. (Fig. 2). On the removal of disturbing force, if the system neither returns nor moves away from its originalposition, it is called to be in neutral equilibrium (Fig. 3)Prof. Ramkumar'sFig. 1 Stable Equilibrium Fig. 2 Unstable Equilibrium Fig. 3 Neutral EquilibriumEquilibrium Conditions for Concurrent Force System:Set I: (i) Fx = 0, (ii) Fy = 0Where Fx and Fy are algebraic sum of x and y components of forces respectively.Set II: (i) MA = 0 (ii) MB = 0Where MA and MB are algebraic sum of moments of the forces about points A and B which do not lie onthe line passing through point of concurrence of force system.Equilibrium Conditions for Parallel Force System:Set I:(i) F = 0 (ii) MA = 0Where F = algebraic sum of the forces of the system parallel to the action lines of the forces. MA = algebraic sum of the moment of the forces of the system about any point A in the plane.Set II: (i) MA = 0 (ii) MB = 0Where MA and MB = algebraic sum of the moments of the forces of the system about A and B which maybe chosen anywhere in the plane provided the line joining A and B is not parallel to the forces of the system.Equilibrium Conditions for General Force System (Non-concurrent and Non-parallel):Set I: (i) Fx = 0 (ii) Fy = 0 (iii) MA = 0Where Fx = algebraic sum of x-components of the forces.Fy = algebraic sum of y-components of the forces.MA = algebraic sum of the moment of the forces of the system about any point A in the plane.Set II: (i) Fx = 0 (ii) MA = 0 (iii) MB = 0Where Fx = as stated above, MA and MB = algebraic sum of the moment of the forces of the system about any two points A andB in the plane, provided that the line joining A and B is not perpendicular to the x-axis.Set III: (i) MA = 0, (ii) MB = 0, (iii) MC = 0Where MA, MB and MC = algebraic sums of the moments of forces of the system about any three pointsA, B and C in the plane, provided that A, B and C do not lie on the same straight line. 28

Types of Supports1) Hinge Support: A hinge allows the free rotation of the body but it does not allow the body to move in xand y directions. It therefore offers reacting forces (reactions) in x and y directions as shown. i.e number ofreactions exerted by hinge support = 2.2) Roller Support: A roller support is free to roll on the surface on which it rests. Roller exerts only oneforce reaction in the direction normal to the surface on which roller rests. Various ways of showing rollersupport are3) Fixed Support (Built-in Support): A fixed support neither allows any linear motion nor allows anyrotation. It therefore offers two force reactions in x and y direction and also a moment reaction as shown. i.e.Number of reaction exerted by fixed support = 3Prof. Ramkumar'sFree Body Diagram (FBD): In solving problems concerning the equilibrium of a rigid body, first step in thesolution of problem consists in drawing a free-body diagram of the rigid body and then only equations ofequilibrium can be applied. To draw F.B.D. of rigid body, it is detached or isolated from supporting strings, ground, wall, hinge orany other body in contact and replaced them by reactions which these supports exert on the body. Suchdiagram of the isolated body with all forces acting on it is known as free body diagram.Rigid Body (Structure) Free Body Diagram (F.B.D.)1] Beam with one end hinge and other roller 1]2] Beam with one end fixed and other end free 2]3] Sphere resting on smooth planes 3] 29

4] Bar AB of weight W hinged at A and supported by 4]string BD.Prof. Ramkumar's5] Rod Resting inside a sphere 5]6] Bar supported by smooth wall and a peg (nail) 6]7] Bar supported by smooth floor and Edge (Fulcrum) 7]Important points related to F.B.D.(1) Direction of reactions at hinge, roller and fixed supports are arbitrarily assumed in F.B.D.(2) Smooth plane exerts normal reaction towards the body.(3) Elements like string, wire, cable, thread and rope etc. are always subjected to tensile force (never compressive). Also Tension means away from body or joint and compression means towards the body or joint.(4) For solving problems of equilibrium, F.B.D. of rope, cable etc. is never required.(5) Pegs, nails and edge (fulcrum) exert normal reactions towards the body as shown in case [6] and[7]. of F.B.D.(6) In case of frictionless pulley, tension in the string on two sides of pulley will be same.(7) If plane is making angle ‘ ’ with horizontal, force or reaction normal to plane will make angle ‘ ’ with vertical and vice-versa.(8) Reactions between two bodies in contact are always equal and opposite and such reactions come into existence only when bodies are separated. 30

LAMI’S THEOREM: If three forces acting at apoint are in equilibrium, the ratio of any of theforces to the sine of the angle between theremaining two forces is the same.Prof. Ramkumar'sP = Q = R sin α sinβ sin γDerivation: Let the three forces P, Q and R are inequilibrium acting at point A be represented bythree adjacent side of the triangle ABC (as per lawof polygon). Let the angle between the forces Q and P be γLet the angle between forces P and R be β Let the angle between forces R and Q be αApplying Sine Rule in ∆ ABC P − α) = Q − β) = R − γ)sin(180 sin(180 sin(180 P = Q = R sin α sin β sin γLimitations: 1) It is not applicable for parallel and general force systems in equilibrium. 2) It is not applicable when more than three forces acting at a point are in equilibrium. 3) Lami’s Theorem does not give direction of unknown forces. 4) Angles α, β and γ should not be more than 180°Two Force Body in Equilibrium: A two force body will be in equilibrium only when the two forcesare equal in magnitude, opposite in direction and have the same line of action i.e. along the linejoining the two points. Two Force Body Two Force Body (Not in Equilibrium) (In Equilibrium)Three Force Body in Equilibrium: When a body is acted upon by three coplanar forces, it can be inequilibrium if either the lines of action of the three forces intersect at one point (concurrent) or theyare parallel. Let us consider a rigid body with three non-parallel forces F1, F2 and F3 acting at points A, B and C respectively (refer figure). Let the lines of action forces F1 and F2 intersect at point D. Transmit the force F1 and F2 to act through the point D. Replace now the forces F1 and F2 by their resultant R acting at the point D. The third force F3 and resultant R can keep the body in equilibrium if they have the same line of action or are collinear. So the third force must also pass through the point D. So, we can conclude that three non-parallel forces will keep the body in equilibrium only whenthey are concurrent i.e Three Force body cannot form general force system. 31

Beam: It is a bar carrying external forces that are vertical or inclined to its axis. It is generally considered ashorizontal and straight. Types of Beams1) Simply Supported Beam: A beam with itsends resting feely on supports that providesonly the vertical reactions and no restraint toends from rotating to any slope at thesupports.2) Cantilever: A beam which is fixed at oneend, the other end being free, is called ascantilever.3) Propped Cantilever: Additional support,called as prop is provided at the free end.4) Fixed Beam: A beam with both endsfixed.5) Continuous Beam: Beam having morethan two supports is called as continuousbeam, being continuous over intermediatesupports (more than one span also)6) Overhanging Beam: Beam having somepart of span projected beyond support iscalled as overhanging beam.Prof. Ramkumar's- ., ,! , !A 30 kg pipe is supported at ‘A’ by a system of !five cords. Determine the force in each cord for A roller of weight W = 1000 N rests on a smooth inclined plane. It is kept from rolling down the planeequilibrium. (May’09-5 marks) by string AC. Find the tension in the string and reaction at the point of contact D. (Dec’08−5marks)TAE = 169.91N, TAB = 339.83 N, T = 732 N and RD = 896.6 N 45°TBC = 562.31N, TBD = 490.5 N 32

Determine the horizontal distance x to which a Two identical rollers each of weight 500 N and5 m long inextensible string holding a weight of radius r are kept on a right angle frame ABC having negligible weight. Assuming smooth surface, find3 kN can be pulled before the string breaks. The the reactions induced at all contact surfaces.string can with stand a maximum force of 6 kN as (Dec’12-8 marks)Prof. Ramkumar'sshown in figure. Determine also the requiredforce F. (April’ 97 marks)x = 4.33 m and F = 5.196 kN R1 = 500N 30° ; R2 = 433N 60° R = 250N R3 = 433N 60° Three cylinders are piled up in a rectangularchannel as shown in figure. Determine thereactions between cylinders A, B and C withchannel surfaces. Assume all smooth surfaces.Given: WA = 150 N, rA = 4 cm ,WB = 400 N, rB = 6 cm, WC = 200 N, rC = 5 cm (May’12-10 marks)R1 = 102.08N(→); R2 = 224.55N;R3 = 552.09N(←); R4 = 750N;R5 = 750N (↑); R6 = 450N(→)Two spheres rests on a smooth surface as \" Determine the force P applied at 45 to theshown in figure. Find the reactions at points of horizontal just necessary to start a roller 100 cm in diameter over an obstruction 25 cm high, if thecontact 1, 2, 3 and 4. roller weighs 1000 N. Also find the magnitude and direction of P when it is minimum.Given: WA = 500N, WB = 200N,rA = 0.25 m, rB = 0.20 m (May’12-10 marks)R1 = 138.33N(←); R2 = 243.18N; 896.48 N and 866 N at 60° to horizontalR3 = 620.14N(↑); R4 = 159.73N 30°# Two rollers of weights 50 N and 100 N areconnected by a flexible string AB. The rollersrests on two mutually perpendicular DE and EFas shown in figure. Find the tension in the stringand the angle that it makes with the horizontalwhen the system is in equilibrium. 33

&'()*&+ For Roller A, Fx = 0 Consider F.B.D of two rollersT cos(60 − θ) = 50cos30 T = 50 cos30 ……… (1) cos(60 − θ)Prof. Ramkumar'sFor Roller B, Fx = 0 gives100cos 60 = T cos(30 + θ) T = 100 cos 60 ……… (2) cos(30 + θ)Equating (1) and (2) 50 × 0.866 = 100 × 0.5 cos(60 − θ) cos(30 + θ) 0.866 × cos(30 + θ) = cos(60 − θ)0.866 [cos 30 cos θ − sin 30 sin θ] = cos 60.cos θ + sin 60. sin θ0.866 [0.866 cos θ − 0.5 sin θ] = 0.5 cos θ + 0.866 sin θon solving we get, 1.299 sinθ = 0.25 cos θ tan θ = 0.25 θ = 10.9° 1.299From (1), T = 50 cos 30 = 66.15 N cos(60 − 10.9)$ Three identical right circular cylindersA, B and C each of weight W are arrangedon smooth inclined surfaces as shown.Determine the least value of angle thatwill prevent the arrangement fromcollapsing. (June’03 6 marks) &'()*&+ Note: (i) At the point of collapse, reaction between cylinders A and B will be zero.(ii) centres of three cylinders form an equilateral triangle. Fy = 0 (Cylinder − C) Consider F.B.D. of cylinder-CR1 sin 60 + R1 sin 60 = W 2R1 × 0.866 = W R1 = 0.577 W Fx = 0 (Cylinder − A) 0.577W cos 60 = R2 sin θ R2 sin θ = 0.2885 W ……… (1) Fy = 0 (Cylinder − A) Consider F.B.D. of cylinder-AR2 cos θ = 0.577 W sin60 + W ……… (2) ∴ R2 cos θ = 1.5 W(1)/(2) gives R 2 sin θ = 0.2885 W R2 cos θ 1.5 W tan θ = 0.1923 θ = 10.9° 34

% Two cylinders, having weights WA =2000 N WB = 1000 N are resting on asmooth inclined planes having inclinations60 and 45 with the horizontal respectivelyas shown in figure. They are connected by aweightless bar AB with hinge connections.The bar AB makes 15 angle with thehorizontal. Find the magnitude of the force Prequired to hold the system in equilibrium.Prof. Ramkumar's &'()*&+ F.B.D. of cylinder A Note: Bar AB is a Two-force members.F.B.D. of cylinder B Let F = Tensile force in bar AB Fx = 0 Fcos15 + R1 cos30 = 0 R1 = −1.115F Fy = 0 R1 sin30 = 2 + F sin15 −1.115F × 0.5 = 2 + F sin15 F = −2.45 = 2.45 kN (C) Fy = 0 gives R2 cos 45 = 1+ P cos 60 + 2.45 sin15 1+ Pcos 60 + 2.45 sin15 R2 = cos 45 Fx = 0 gives 2.45cos15 = P sin 60 + R2 sin 45 2.45cos15 = P sin60 + 1+ P cos 60 + 2.45 sin15 cot 45 P = 0.536 kN ! ! If the cords suspend the two buckets in In a ship unloading operation, a 15.6 kNequilibrium position shown in figure. Determine theweight of bucket-B. Bucket-A has a weight of 60 automobile is supported by a cable. A rope is tiedN. to a cable at A and pulled in order to centre the (Dec’05-5marks) automobile over intended position. What is the tension in the cable. (May’09−8 marks−VJTI)WB = 88.8 N TAB = 15.931 kN 35

A smooth circular cylinder of weight W and A circular roller of weight 1000 N and radiusradius R rests in a V shape groove whose sides 20 cm hangs by a tie rod AB = 40 cm and restsare inclined at angles α and β to the horizontal as against a smooth vertical wall at C as shown inshown. Find the reactions RA and RB at the points figure. Determine the tension in the rod andof contact. Given: α = 20°, β = 60° reaction at point C. Hint: Tie rod means rod subjected to tensile forces. (Dec’12-4 marks)Prof. Ramkumar'sRA = 0.879 W 70°, RB = 0.347W 30° T = 1154.7 N and RC = 577.35 N ( → ) A ball of weight W rests upon a smooth horizontal plane and hasattached to its centre two strings AB and AC which pass over africtionless pulley at B and C and carry load P and Q as shown in thefigure. If the string AB is horizontal, find the angle α that the string ACmakes with the horizontal when the ball is in position of equilibrium.Also determine the pressure between the ball and the plane in termsof P, Q and W. Hint: Consider F.B.D. of ball. (May’07−5 marks) α = cos−1 P , R = W − Q2 − P2 Q Two identical rollers each of weight Q = 500 N are supported byan inclined plane and a vertical wall as shown in figure. Assumingsmooth surfaces, find the reactions induced at the points of supportA, B and C.HC = 577.36 N (→), RA = 433 N 60°RB = 721.7 N 60°\" Two smooth spheres A and B of weight 200 N and 100 Nrespectively are resting against two smooth vertical walls and smoothhorizontal floor as shown in figure. The radius of sphere A is100 mm and the radius of sphere B is 50 mm. Find the reactions fromthe vertical walls and horizontal floor. Also find the reaction exertedby each sphere on the other. (Dec.’00 6marks)RC = 89.440N (→), RD = 300N (↑)RE = 89.44 N (←), R = 134.164N# Two cylinders each of diameter 100 mm and each weighing 200 Nare placed as shown in figure. Assuming that all the contact surfacesare smooth, find the reactions at A, B and C. (Dec’09-12 marks) R A = 186 N(→), RB = 406.17 N 80° RC = 115.47 N (←) 36

$ Two spheres A and B of weight 1000 N and % Determine the reactions at point of contact 1,750 N respectively are kept as shown in figure. 2 and 3. Assume smooth surfaces. WA = 1 kg, rA = 1 cm, WB = 4 kg, rB = 4 cm.Determine the reactions at all contact points 1,2, 3 Hint: Take inclination of line joining centres as unknown.and 4. Radius of A = 400 mm and Radius of (Dec’11-10 marks, May’08−10 marks−old & Dec’04-8 marks)Prof. Ramkumar'sB = 300 mm.(May’11-10 marks)R1 = 496.65 N (→), R2 = 1463.26 N (↑), R1 = 19.75 N 65°, R2 = 11.62N,R3 = 869.14 N, R4 = 573.5 N 30° R3 = 32.24N 75° A roller of weight 500 N has a radius of120 mm and is pulled over a step at height 60 mmby a horizontal force P. Find magnitude of P to juststart the roller over the step.Hint: RB will pass through A. (Dec’06−5 marks) 288.67 N- ., ,! ! , ! Find reactions for the beam shown in figure.VA = 42.17 kN (↑), VE = 57.83 kN (↑) Two smooth circular cylinder of weight W = 500 Neach and radius r = 150 mm are connected at theircentres by a string AB of length l = 400 mm and restupon a horizontal plane supporting above them a thirdcylinder of weight 1000 N and radius r = 150 mm asshown in figure. Find the force in the string andpressure produced at the points of contact D and E. RD = RE = 1000 N(↑) and T = 447.2 N A heavy rod AB of length 3 m lies on horizontal ground. To lift the end B off the ground needs a verticalforce of 200 N. To lift A end off the ground needs a force of 160 N. Find the weight of the rod and the positionof centre of mass. (May’08−5 marks) W = 360N, x = 1.667 m from A 37

A right angled lamina ABC with sidesAB = 12 cm and BC = 5 cm is suspended fromA as shown in the figure. Find the angle made bythe side AB with the horizontal through A. (May’07−5 marks) 78.23°Prof. Ramkumar's A uniform beam AB of weight W = 100 N rests on tworoller supports C and D as shown. If a force of 250 N isapplied to the end B, find the range of the values of forceF for which the beam will remain in equilibrium&'()*&+ Case (i) Fmax case: Put VD = 0 beam will be lifted off at D. MC = 0 100 × 20 + 250 × 70 = F × 30 F = 650 N (max)Case (ii) Fmin case: beam will be lifted off at C. Put VC = 0 MD = 0 250 × 20 = 100 × 30 + F × 80 F = 25 N (min) An isosceles triangle is to be cut from oneedge of a square plate of side 1 m, such thatthe remaining part of the plate remains inequilibrium in any position, when suspendedfrom the apex of the triangle. Find the area ofthe triangle to be removed.&'()*&+ Based on Two-forced body in equilibrium concept, apex (E) should be C.G of the plate. Hence y = h (1×1) × 0.5 − 1 ×1× h × h y= 2 3 =h (1×1) − 1 ×1× h 2 0.5 − h2 6 1− 0.5h =h 0.5 − 0.167 h2 = h − 0.5h2 0.333h2 − h + 0.5 = 0 On solving, we get h = 2.369 m and 0.634 m∴ Area of triangle = ½ × 1 × 0.634 = 0.317 m2 (Adopting h = 0.634 m as h = 2.369 m is not possible). 38

! !Find reactions for the beam shown in figure. Find reactions for the beam shown in figure.Prof. Ramkumar'sVA = 57.5 kN (↑), VB = 42.5 kN (↑) VA = 41.5kN (↑), VB = 56 kN (↑) Calculate the reactions of the beam loaded as Determine the uniform reaction from theshown in figure. surface on cylindrical log as shown. The self weight of the log is assumed to be 1300 N/m. There are two point loads of 8000 N each acting at I m from each end. Hint: Reaction is in the form of u.d.l. (May’05-4 marks)VA = 6.73 kN ( ↑ ), VB = 11.27 kN ( ↑ ) R = 4500 N/m ( ↑ )- ., ,! ! ! Find the reactions at the supports of the beamloaded as shown in figure. (Dec’09-8 marks) HB = 0, VB = 50 kN(↑), VC = 60 kN (↑) Find the support reactions for the beam loaded and supported as shown in figure. (Dec’12-8 marks) RF = 67.63 kN 60° HB = 104.53 kN (→) VB = 152.14kN (↑) 39

Determine the intensity of distributed load ‘w’ at the end Cof the beam ABC for which the reaction at C is zero. Alsocalculate the reaction at B (May’08−10 marks) w = 3 kN / m, RB = 21.6 kN(↑) Find the reactions at A. (May’12-10 marks)HA = 28.19kN(→), VA = 41.26 kN (↑)MA = 339.17 kNm ( ) A light bar AD is suspended from a cable BE and supportsa 200 N block at C. The extremities A and D of the bar are incontact with frictionless vertical walls. Determine the tension incable BE and the reactions at A and D (May’00 8 marks)RA = 75 N( → ), RD = 75 N( ← ), T = 200 NProf. Ramkumar's A uniform beam AB hinged at A, is kept horizontal bysupporting and settling a 50 kN weight with the help of stringtied at B and passing over a smooth peg at C, as shown infigure. The beam weighs 25 kN. Find the reactions at A and C.HA = 25 kN(→), VA = 25 kN( ↑ ), HC = 25 kN( ← ), VC = 50 kN( ↑ )\" A man raises a 12 kg joist of length 4m by pulling the rope.Find the tension in the rope and the reaction at A. (Dec’10-8 marks)T = 98.48 N, RA=177.44 N = 58.6°# The lever AB is hinged at C and attached to a cable at A. Ifthe lever is subjected at B to a 400 N horizontal force,determine (i) The tension in the cable, (ii) The reaction at C. (June’02 6 marks)(i) T = 300 N (ii) RC = 360.55 N 46.1° 40

$ A uniform steel plate weighing 500 N has been % A pulley of 1 m radius, supporting a load ofcut in the shape shown in figure. 500 N is mounted at B on a horizontal beam as(i) Locate the centroid w.r.t. A shown in figure. If the beam weighs 200 N and(ii) Find support reactions. (Feb.’ 03 07 marks) pulley weighs 50 N, find the hinge force at C.Prof. Ramkumar's (i) x = 3.582 cm and y = 3.917 cm RC = 471.7 N 32°(ii) VA = 201.5 N (↑), VB = 298.5 N (↑),HB = 0 Find support reactions at A, B and C for the A cylinder weighing 1000 N and 1.5 m rigid link DEF supported by the cylinders at D anddiameter is supported by a beam AB of length 6 m F. The link is loaded by a single force of 20 kN asand weight 400 N as shown. Neglecting friction at shown in the figure. Neglect friction and self weightthe surface of contact of the cylinder, determine of link and cylinders. Take diameters of cylinders(i) wall reaction at D. as 200 mm and DE = EF = 300 mm.(ii) Tension in the cable BC and(iii) Hinged reaction at support A. (Feb.’02 8 marks) (June’02 9 marks)(i) RD = 1000 N ( → ), (ii) T = 588.3 N, RA = 12.144 k N( →) , RB = 15.176 kN (↑ ),(iii) RA = 1209.75 N 66.08° RC = 3.032kN 45° Find the reactions of the beam shown infigure.VA = 35.83 N( ↑) , VC = 11.67 N (↑ ),HA = 21.65 N( →),MA = 63.31 Nm ( ) 41

The beam AB and CF are arranged as shownin figure. Determine the reactions at A, C and Ddue to the forces acting on the beam as shown inthe figure. (Dec’06−10 marks)Prof. Ramkumar'sHA = 28.28 kN(←), VA = 9.43 kN(↓)HC = 10 kN(→), VC = 34.12 kN(↑),VD = 40.91 kN(↑) Find reactions at A and B for a bentbeam ABC loaded as shown in figure. (May’06−10 marks) &'()*&+ Fx = 0 gives,HA = 12 + 6 = 18 kN (→) Fy = 0 gives,VA + VB = 10 + 8 = 18 kN .........(1) MA = 0 gives,(10 × 2) + (8 × 6) = 10 + (VB × 7) + (6 ×1) + (12 ×1.5) VB = 4.86 kN (↑)From (1), VA = 13.14 kN (↑) For the system shown in figure, find thereactions at B and C. Also find the tension in thecable. Neglect friction. (Dec’07−10 marks−old) &'()*&+ MB = 0 gives,(T cos30 × 60 sin60) + T sin30 × (40 + 60cos 60) = (T × 40) + (200 × 40) T = 200 N Fx = 0 gives,HB = T cos30 = 173.2 N (←) Fy = 0 gives,VB + T + T sin30 = 200 VB = −100 = 100N (↓) 42

Fx = 0 gives,Prof. Ramkumar's F.B.D. of Pulley-CHB = 200cos30 = 173.2 N (→) Fy = 0 gives, VC = 200 + 200 sin30 = 300N(↑) \" A crane pivoted at the end B issupported by a guide at A. Determinethe reaction produced at A and B by avertical load W = 5 kN applied at C.(May’06−5 marks) &'()*&+ Note: support B will act as a hinge. Fy = 0 gives VB = 5 kN VB = 5 kN (↑) MB = 0 gives 5 × 2.4 = T ×1.8 T = 6.67 kN (←) Fx = 0 gives HB = T HB = 6.67 kN (→) RB = (5)2 + (6.67)2 = 8.33 kN α = tan−1 5 = 36.87° 6.67 F.B.D. of Crane # A uniform rod AB of length and weight Wremains in equilibrium in a vertical plane lodgedbetween a smooth vertical wall and a peg C.Ignoring friction, find the inclination between thewall and the rod.&'()*&+ Consider FBD of Rod ABNote: Based on 3 force body in equilibriumconcept, 3-forces meet at point D,sin θ = a AC = a AC sin θ Fy = 0 givesR sinθ = W R = W sin θ MA = 0 givesR × a θ = W × 2 sin θ sinPut R = W sin θ 43

aW = W sin θ sin3 θ = 2a θ = sin−1 3 2asin2 θ 2Alternate solutionProf. Ramkumar'sIn ∆ AGD, AD = AG sin θ = sin θ 2In ∆ ACD, sin θ = AC = a / sinθ sin3 θ = 2a θ = sin−1 3 2a AD ( / 2)sinθ ! ! Find the reactions at the supports of the beam Find the reaction at the supports of the beamapplying conditions of equilibrium. AB loaded as shown in the figure below. (June’10-8marks) (May’11-8 marks)HA = 5.657 kN (←), VA = 8.44 kN (↑), HB = 2.12 kN(→), VB = 8.62 kN(↑),VB = 7.22 kN (↑) VA = 7.5 kN (↑) Calculate support reactions for following beam. Calculate support reactions for the following (May’08) beam: HA = 0.67 kN (←), VA = 31.16 kN (↑), HA = 0, VA = 104 kN( ↑ ), VB = 136 kN( ↑ )RE = 13.66 kN 60° Find analytically the support reaction at B and Forces act on the plate ABCD as shown in load P for the beam shown in figure if reaction atfigure. The distance AB is 4 m. Given that the support A is zero.plate is in equilibrium find, (i) force F (ii) angle αand (iii) the distance AD (May’08−5 marks) (Dec’11-10 marks & Dec’08−10 marks) VB = 72 kN↑, P = 26 kN ↓ (i) F = 13 N (ii) α = 67.38° (iii) AD = 8.667m 44

\" Find the support reactions for the beam loaded # A weightless bar is placed in a horizontaland supported as shown. position on the smooth inclines as shown in figure. Find x at which 200 N force should be placed from point B to keep the bar horizontal.Prof. Ramkumar'sHA = 10 kN (←), VA = 127.32 kN(↑) x = 1.608 mMA = 694.6 kNm ( ) % A man raises a 10 kg joist of length 4 m$ Determine the tension T in the cable and the pulling on a rope. Find the tension in the rope andreaction at pin support A for the beam loaded as the reaction at A. Hint: A will acts as hinge.shown in figure. Neglect the weight of the beamand the size of the pulley. Pulley D is frictionless.T = 11.186 kN, RA = 5.385 kN 21.8° T = 82.07 N, RA = 147.87N, α = 58.57° Determine the tension in cable BC for the bar A bar AB of weight 1 kN is hinged to a verticalloaded and supported as shown in figure. Bar AB wall at A and supported at the end B by a cableis of negligible weight. Hint: Apply only MA = 0 BD. Find the tension in the cable and the magnitude and direction of reaction at hinge. Hint: Consider length of bar as ‘ ’. T = 4.7 kN T = 866 N, RA = 500 N 30° A fixed crane has a mass of 1000 kg and isused to lift a 2400 kg crate. It is held in place by apin at A and a rocker at B. The C.G. of crane islocated at G. Determine the components ofreactions at A and B.Hint: Pin means hinge and rocker means roller.Consider FBD of crane.VA = 33.35 kN ( ↑ ), HA = 107.26 kN ( ← ), HB =107.26kN( → ) 45

Find the beam reactions at A and B shown in A smooth cylinder of weight 100 N radiusfigure. (May’01 8 marks) 10 cm resting on a horizontal surface supports a bar AB of length 30 cm which is hinged at A. The weight of bar is 50 N. The cylinder is kept from rolling away by a string AO of length 20 cm. Assuming all surfaces to be smooth, find the tension in the string.Prof. Ramkumar'sTension in cable = 75 kN, T = 21.65 NHA = 45 kN( ← ), VA = 120 kN ( ↑ ) A hollow right circular cylinder of diameter9 cm and height h =12 cm is open at both endsand rests on a smooth horizontal plane. A uniformbar of length l = 24 cm and weight 100 N is placedwithin the cylinder and rests as shown. Find thereactions at the points A and D.Hint: Consider FBD of bar AB. A will act as hingeand RD will be ⊥ to AB. RD = 48N θ, RA = 80.9 N 61.66 \" Find the reactions of the beam shown in figure.Hint: Load at internal hinge is taken on any one side.HA = 0 , VA = 10 kN (↑ ), VB = 60kN(↑), VC = 10 kN (↑) # Find the reactions for the beam shown in $ In the system shown in figure find the reactionfigure. at D caused by the 1200 N load. Neglect the weight of the members. Hint: Roller reaction at B will act ⊥ to CDHA = 10 k N( →) , VA = 13 kN (↓ ),VB = 53 kN(↑), VD = 30 kN (↑) RD = 350 N (←) 46