MECHANICAL DEPARTMENT

SUBJECT: Fundamental of Statics  SUBMITTED TO: Mr. A.J. PATEL

MECHANICAL D NAMEENROLL NO RAJ MISTRY RANA JAY RAVAL YASH SHAH SMIT THACKER SHREYASH THAKOR VISHAL

TOPICS  Forces  Moments  Equilibrium

Types of Forces  Concurrent coplanar  Non-Concurrent coplanar  Concurrent non-coplanar  Non-Concurrent non-coplanar  External forces  Internal forces

System of forces: Force System Characteristicexample Collinear forcesLine of action of all the forces act along the same line. Forces on a rope in a tug of war Coplanar concurrent forcesLine of action of all forces pass through a single point and forces lie in the same plane. Forces on a rod resting against a wall Coplanar non-concurrent.All forces do not meet at a point, forces but lie in a single plane Forces on a ladder resting against a wall when a person stands on a rung which is not at its centre of gravity Non-coplanar parallel forcesAll the forces are parallel to each other, but not in the same plane. The weight of benches in a class room

Non-coplanar concurrent forces All forces do not lie in the same plane, but their lines of action pass through a single point. A tripod carrying a camera Non-coplanar non- concurrent forces All forces do not lie in the same plane and their lines of action do not pass through a single point. Forces acting on a moving bus

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). Types of forces:

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 | Website for Students | VTU NOTES | QUESTION PAPERS 10

A vector is represented graphically by an arrow which defines the magnitude direction and sense  Representation of Forces

Vector Operations

Cartesian Vector NotationResultant of Coplanar forces F 1 = F 1x i + F 1y j F 2 = -F 2x i + F 2y j F 3 = -F 3x i - F 3y j F 1 = F x i + F y j Component vectors

Vector Resultant F R = F 1 + F 2 + F 3 = (F Rx )i + (F Ry )j F Rx =  F x F Ry =  F y Resultant - Scalar notation Resultant - Vector notation F R = F 1 + F 2 + F 3 = (F Rx )i + (F Ry )j

Rectangular Components of a Vector A = A x + A y + A z Unit vector u A = A A => A = A u A Cartesian vector representation A = A x i + A y j+ A z k Magnitude of a cartesian vector

Equilibrium of Forces 1 Two dimensional Forces  F = 0  F x i +  F y j = 0  F x = 0  F y = 0 => 2 Three dimensional Forces  F = 0  F x i +  F y j +  F z k = 0  F x = 0 SF y = 0 SF z = 0 =>

Moment of a Forces -Vector M 0 = r B x F M 0 = r C x F 1) Magnitude 0 = r x F Cartesian Vector Formulation M 0 = rF sin =F( r sin ) = Fd 2) Direction determined by right hand rule Transmissibility M 0 = (r y F z – r z F y )i – (r x F z – r z F x )j + (r x F y – r y F x )k

M Ro =  ( r x F) Resultant Moment of Forces -Vector Moment of a Forces -Vector M 0 = Fd Resultant Moment of Forces -Scalar M R0 =  Fd

Moment of a Couple M = F d 1 Scalar analysis M = r x F 2 Vector analysis - A Couple is a pair of equal and opposite parallel forces - Two Couples producing the same moment are equivalent

Equilibrium of a Rigid Body  F = 0  M = 0 Equilibrium in 2D

Principle of Transmissibility  According to this law the state of rest or motion of the rigid body is unaltered if a force acting on the body is replaced by another force of the same magnitude and direction but acting anywhere on the body along the line of action of the replaced force. Principle of Transmissibility - Conditions of equilibrium or motion are not affected by transmitting a force along its line of action. NOTE: F and F’ are equivalent forces.

Parallelogram Law  According to this law the state of rest or motion of the rigid body is unaltered if a force acting on the body is replaced by another force of the same magnitude and direction but acting anywhere on the body along the line of action of the replaced force.

Triangle Law  If two forces acting on a body are represented one after another by the sides of a triangle, their resultant is represented by the closing side of the triangle taken from first point to the last point.

Problems 1. Find the projection of a force on the line joining A = (-1, 2, 2) and B (2, -1, -3) Solution: The position vector = (2i – j -3k) – (-+2+2) = Magnitude of AB = Unit vector AB = Projection of on the line AB = unit vector along AB = 2   – 5  =

Free-body diagrams Free-body diagrams are pictures that show the size and direction of all forces acting on an object.

Steps to drawing a free body diagram 1. Pick one object to analyze 2. Draw a box to represent the object 3. Draw an arrow to represent each force acting on the object 4. Make sure the arrow shows the direction and relative size of the force

Problem 1 A book is at rest on a table top. Diagram the forces acting on the book.

Problem 1 In this diagram, there are normal and gravitational forces on the book.

Problem 1 The forces are balanced (they cancel each other out)

Problem 2 An egg is free-falling from a nest in a tree. Neglect air resistance. Draw a free-body diagram showing the forces involved.

Problem 2 Gravity is the only force acting on the egg as it falls.

Problem 2 The forces are unbalanced, so the egg will accelerate downward.

Problem 3 A flying squirrel is gliding (no wing flaps) from a tree to the ground at constant velocity. Consider air resistance. A free body diagram for this squirrel looks like…

Problem 3 Gravity pulls down on the squirrel while air resistance keeps the squirrel in the air for a while.

Problem 4 A rightward force is applied to a book at rest, in order to move it across a desk. Consider frictional forces. Neglect air resistance. Construct a free-body diagram for the book.

Note the applied force arrow pointing to the right. Notice how friction force points in the opposite direction. Finally, there are still gravity and normal forces involved.

Problem 5 A skydiver is falling with a constant velocity. Consider air resistance. Draw a free-body diagram for the skydiver.

Gravity pulls down on the skydiver, while air resistance pushes up as she falls.

Problem 6 A man drags a sled across loosely packed snow with a rightward acceleration. Draw a free-body diagram of the forces acting on the sled.

The rightward force arrow points to the right. Friction slows his progress and pulls in the opposite direction. Since there is not information that we are in a blizzard, normal forces still apply as does gravitational force since we are on planet Earth.

Problem 7 A football is moving upwards toward its peak after having been booted by the punter. Neglect air resistance. Draw a free-body diagram of the football in mid-air.

The force of gravity is the only force described. It is not a windy day (no air resistance).

Problem 8 A car runs out of gas and coasts to a stop on flat ground. Draw a free body diagram of the forces acting on the car.

Even though the car is coasting down the hill, there is still the dragging friction of the road (left pointing arrow) as well as gravity and normal forces.

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:F AC =0.058 kN(T),R=14.98 kN] C 7kN 5kN Fig(1) A

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 = 30 0 wrt Hz.] 15 Fig (2) θ 46

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 = ; T= 9.7kN.] 2m Fig(3). 10kN P θ 47

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: F AC =4.62kN, R A1 = 2.46kN, R A2 = 7.16kN( ) F BC =?, R B1 =?, R B2 =? ( ) ] B 70 A C 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: F AB =257.11N, F AC =162.50N, R A1 = N, R A2 = N, R B =306.42N, R C = N. ] 50 Fig(5). 40 B C A 49

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 = ,R A = R B = N] Fig(6). θ 30 B A 50

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.] Fig(7) P 2kN 51

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 = ] Loads are in kN W B C D A Fig(8) θ