BHAGWAN MAHAVIR COLLEGE OF ENGG.& TECH.

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BHAGWAN MAHAVIR COLLEGE OF ENGG.& TECH. EQUILIBRIUM OF CONCURRENT FORCES Prepared by: Garg Pooja (130060102015) Kakadiya Pratik (130060119081)

EQUILIBRIUM OF CONCURRENT COPLANAR FORCE SYSTEMS CONTENTS: 1] BASIC CONCEPTS: (a) Definition & Conditions for Equilibrium (b) Space Diagram & Free Body Diagram (FBD) (c) A few guidelines for drawing FBD 2] NUMERICAL PROBLEMS: (a) Solved Examples (b) Exercise

EQUILIBRIUM OF CONCURRENT COPLANAR FORCE SYSTEMS Definition:- If a system of forces acting on a body, keeps the body in a state of rest or in a state of uniform motion along a straightline, then the system of forces is said to be in equilibrium.    ALTERNATIVELY, if the resultant of the force system is zero, then, the force system is said to be in equilibrium.  

Conditions for Equilibrium of Concurrent Coplanar Force System A coplanar concurrent force system will be in equilibrium if it satisfies the following two conditions: i)  Fx = 0; and ii)  Fy = 0   i.e. Algebraic sum of components of all the forces of the system, along two mutually perpendicular directions, is ZERO. Y X

Graphical conditions for Equilibrium Triangle Law: If three forces are in equilibrium, then, they form a closed triangle when represented in a Tip to Tail arrangement, as shown in Fig 1.(a). F3 F2 F2 F1 Fig 1.(a) F3 F1 Polygonal Law: If more than three forces are in equilibrium, then, they form a closed polygon when represented in a Tip to Tail arrangement, as shown in Fig 1.(b). F4 F3 F3 F2 F5 Fig 1.(b) F2 F1 F4 F1 F5

LAMI’S THEOREM If a system of Three forces is in equilibrium, then, each force of the system is proportional to sine of the angle between the other two forces (and constant of proportionality is the same for all the forces). Thus, with reference to Fig(2), we have, Note: While using Lami’s theorem, all the three forces should be either directed away or all directed towards the point of concurrence. F3 α F2 F1 Fig (2)

SPACE DIAGRAMS & FREE BODY DIAGRAMS  Space Diagram(SPD):The sketch showing the physical conditions of the problem, like, the nature of supports provided; size, shape and location of various bodies; forces applied on the bodies, etc., is known as space diagram. Eg. Fig 3 (a) is a space diagram Cable P = 2kN 3 m Sphere θ 30 R=1m. wall Fig 3 (a) SPD Weight of sphere = 0.5 kN

Eg. Fig 3 (b) is a Free Body Diagram. Free Body Diagram(FBD) : It is an isolated diagram of the body being analyzed (called free body), in which, the body is shown freed from all its supports and contacting bodies/surfaces. Instead of the supports and contacting bodies/surfaces, the reactive forces exerted by them on the free body is shown, along with all other applied forces. Eg. Fig 3 (b) is a Free Body Diagram. P=2kN 30 T Rw W=0.5kN θ Note: Free Body Diagrams should be NEAT, LEGIBLE & SUFFICIENTLY BIG. Only the details required for the analysis of the problem are to be shown. Fig 3(b) T = Tension in the cable Rw = Reaction of the wall

A Few Guidelines for Drawing FBD 1)      Tensile Force: It is a force trying to pull or extend the body. It is represented by a vector directed away from the body. 2)      Compressive Force: It is force trying to push or contract the body. It is represented by a vector directed towards the body. 3)      Reactions at smooth surfaces: The reactions of smooth surfaces, like walls, floors, Inclined planes, etc. will be normal to the surface and pointing towards the body. 4)      Forces in Link rods/connecting rods: These forces will be acting along the axis of the rod, either towards or away from the body. (They are either compressive or tensile in nature).

5) Forces in Cables (Strings or Chords): These can only be tensile forces. Thus, these forces will be along the cable and directed away from the body. 6) Tension in cables on either side of a smooth pulley will be equal in magnitude. (Eg. As shown in Fig) 40N P =40N

EXERCISE PROBLEMS 1] A 10kN roller rests on a smooth horizontal floor and is held by the bar AC as shown in Fig(1). Determine the magnitude and nature of the force in the bar AC and reaction from the floor under the action of the forces applied on the roller. [Ans:FAC=0.058 kN(T),R=14.98 kN] 7kN C 45 5kN A 30 Fig(1)

2] A 1kN roller resting on a smooth incline as shown in Fig (2) is held by a cable. If the tension in the cable is limited to 0.518kN, determine the maximum inclination to which the plane can be raised. [Ans: θ = 300 wrt Hz.] 15 θ Fig (2)

3] A 10 kN weight is suspended from a rope as shown in Fig(3) 3] A 10 kN weight is suspended from a rope as shown in Fig(3). Determine the magnitude and direction of the least force P required to pull the rope, so that, the weight is shifted horizontally by 0.5m. Also, determine, tension in the rope in its new position. [Ans: P= 2.43 kN, θ = 14.480 ; T= 9.7kN.] 2m P θ 10kN Fig(3).

4] Three spheres A, B, C of diameters, 500mm, 500mm, 800mm and weighing 4kN, 4kN, 8kN, respectively, are placed in a trench as shown in Fig(4). Find the reactions at all contact points. [Ans: FAC=4.62kN, RA1= 2.46kN, RA2= 7.16kN( ) FBC=?, RB1=?, RB2=? ( ) ] C A B 70 70 650 mm Fig(4).

5] Three cylinders A, B, C of diameters, 200mm, 200mm, 100mm and weighing 400N, 400N, 200N, respectively, are placed in a trench as shown in Fig(5). Find the reactions at all contact points. [Ans: FAB=257.11N, FAC=162.50N, RA1= 459.62N, RA2= 460.06N, RB=306.42N, RC= 182.72N. ] C B A 50 40 Fig(5).

6] Two rollers A and B of same diameter and weight 1000N, 600N, respectively, interconnected by a light weight rod are placed on smooth planes as shown in Fig(6). Determine the inclination θ of the rod and the reaction of the planes. [Ans : θ = 23.410,RA = RB = 923.7 N] B θ A 30 30 Fig(6).

7] Determine the value of P and the nature of the forces in the bars for equilibrium of the system shown in Fig(7). [Ans: P = 3.04 kN, Forces in bars are Compressive.] 60 45 45 75 P 2kN Fig(7).

8] A cable fixed as shown in Fig(8), supports three loads 8] A cable fixed as shown in Fig(8), supports three loads. Determine the value of the load W and the inclination of the segment BC. [Ans: W=25kN, θ = 54.780] A D 30 B 60 θ C 20 22.5 Loads are in kN W Fig(8)