BA/BSc II (Mathematics) Paper: Mechanics Topic of Presentation: Statics Dr. Baljeet Singh Department of Mathematics Post Graduate Government College Sector 11, Chandigarh-160011

Terminology Statics: deals with the equilibrium of bodies, that is those that are either at rest or move with a constant velocity. Particle: A particle has a mass but a size that can be neglected. When a body is idealised as a particle, the principles of mechanics reduce to a simplified form, since the geometry of the body will not be concerned in the analysis of the problem.

Force A force is a vector quantity and must have magnitude, direction and point of action. F P Here, the point P is the point of action of force and and are directions. To notify that F is a vector, it is printed as bold. Its magnitude is denoted as |F| or simply F.

Two vectors are equal if they are equal in magnitude and act in the same direction. P Q p

Forces equal in magnitude can act in opposite directions

Types of Force Systems Collinear: all forces acting along the same straight line. Coplanar: all forces acting in the same plane. Concurrent: the lines of action of all the forces intersect at a common point Parallel: the lines of action of all the forces are parallel Non-concurrent: If not concurrent or parallel Non coplanar: the lines of actions of all the forces do not lie in the same plane.

Resultant of forces: The resultant of a system of forces on a particle is the single force which has the same effect as the system of forces.

Addition of Vectors Law of sines, Trapezoid rule for vector addition Triangle rule for vector addition B C Law of cosines, Law of sines,

Resolution of force A given force, F can be resolved into components. There are two major cases: (a) When one of the two components, P is known: The second component Q is obtained using the triangle rule. Join the tip of P to the tip of F. The magnitude and direction of Q are determined graphically or by trignometry. Q P F = P + Q

Forces in a plane 1. Resultant of 2 forces acting on a particle Two forces P and Q acting on a particle A, Figure 1 (a) can be replaced by a single force R which has the same effect on the particle, Figure 1 (c). This force R is called the resultant of the forces P and Q and this method for finding the resultant of the force is called the parallelogram law for the addition of two forces. (b) (c) Figure 1

Find the resultant force and the angle . Solution: Example 1: If the angle between F1 and F2 is 60o and F1 = 54N and F2 = 60N. Find the resultant force and the angle . Solution: R F1 F2 120o 60o R F1 F2 120o 60o

Example 2. The two forces P and Q acts on a bolt A as shown in Figure below (a) Determine the resultants. (b) Figure 2

Solution:

Example 3. Determine graphically, the magnitude and direction of the resultant of the two forces using (a) Paralleolegram law and (b) the triangle rule. 600 N 900 N 45o 30o

600N 900N 45o 30o

600N R 135o 30o 900 N B

Example 4. Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 30 kN and Q = 20 kN, determine the magnitude and direction of the resultant force exerted on the bracket. P 25o 50o Q

Resultant of a system of coplanar forces In order to determine the resultant of several coplanar forces, each force is first resolved into its x and y components. Then, the respective components are added using scalar algebra since they are collinear. The resultant force can be then formed by adding the resultants of x and y components using the parallelogram law.

Rectangular components of a force y Fy = Fy j F j x Fx = Fx i i

In many problems, it is desirable to resolve force F into two perpendicular components in the x and y directions. Fx and Fy are called rectangular vector components. In two-dimensions, the cartesian unit vectors i and j are used to designate the directions of x and y axes. Fx = Fx i and Fy = Fy j i.e. F = Fx i + Fy j Fx and Fy are scalar components of F

q While the scalars, Fx and Fy may be positive or negat ive, depending on the sense of Fx and Fy , their absolute values are respectively equal to the magnitudes of the component forces , Scalar components of F have magnitudes: Fx = F cos q and Fy = F sin F is the magnitude of force .

Example 5: Determine the resultant of the three forces below. y 600 N 800 N 350 N 45o 60o 25o x

Solution: 600 N y 800 N 350 N 45o 60o 25o x

Example 6. A hoist trolley is subjected to the three forces shown. Knowing that = 40o , determine (a) the magnitude of force, P for which the resultant of the three forces is vertical (b) the corresponding magnitude of the resultant. P 2000 N 1000 N

Solution: P 40o 40o 2000 N 1000 N

Moment of a Force The tendency of a force to produce rotation about some axis is called the moment of a force. The magnitude of this tendency is equal the magnitude of the force times the perpendicular distance between the axis and the line of action of the force (moment arm). M = F * d Unit: force x distance =F*L = lb-in, k-in, lb-ft, k-ft N-m, kN-m (SI unit) Sign conven.: clockwise (-), counterclockwise (+)

Varignon’s Theorem It states that the moment of a force about a point is equal to the sum of the moments of its components about the same point. Consider the Force, F with two components: F = F 1 + F 2 . Using the distr ibutive law o f vectors Mo = r x F + r x F = r x (F ) = r x F y Moment of F about 0 Mo = r x F r o x

Moment of couple Introduction: A couple is defined as two parallel forces which have the same magn itude, opposite directions and are separated by a perpendicular distance, d - F Since the resultant force of the two forces of the couple is zero, the only effect of a couple is to produce a rotation or tendency of a rotation in a specified direction.

Some remarks on couple A couple causes rotation about an axis perpendicular to its plane The moment of a couple is independent of the choice of the axis of moment (moment center) A couple cannot be replaced with a single equivalent resultant force A couple may be transferred to any location in its plane and still have the same effect

Example 7. Determine the moment of the couple shown in Figure and the perpendicular distance between the two forces 760 N A 760 N 200 mm B 350 100 mm

Solution: 760 N 350 B A 200 mm 100 mm FA = -760 cos(350) - 760 sin(350) =-622 i – 435.9 j N rBA = -0.1 i + 0.2 j m MB= i j k -0.1 0.2 0 -622 -435.9 0 = 168 k Nm |MB|= Mx2 + My2 + Mz2 = 168 Nm d=M/F=168/760=0.22 m

Resolution of a force into a force and couple acting at another point Any force F acting on a rigid body may be moved to any given point A (with a parallel line of action), provided that a couple M is added. The moment M of the couple is equal to F times the perpendicular distance between the original line of action and the new location A.

Resultant of two parallel forces The magnitude of the resultant R of the parallel forces A and B equals the algebraic summation of A and B, where R = A + B. Location of the resultant R is obtained by the principle of moments.

Equivalent couples Two couples are equivalent if they produce the same moment. Since the moment produced by a couple is always perpendicular to the plane containing the couple forces, it is necessary that the forces of equal couples lie either in the same plane or in planes that are parallel to one another.

Resolution of a Force into a Force at B and a couple A Force, F acting at point A on a rigid body can be resolved to the same force acting on another point B and in the same direction as the original force plus a couple M equal to r x F i.e. moment of F about B i.e Force in (a) equal to that in (b) equal to that in © M = 25 kN m 5 kN F = 5 kN 5 kN 5 kN B A B 5 kN 5 m (a) (c) (b)

Example A force and couple act as shown on a square plate of side a = 625 mm. Knowing that P = 300 N, Q = 200 N and  = 50o, replace the given force and couple by a single force applied at a point located (a) on line AB (b) on line AC. In each case, determine the distance for A to the point of application of the force.

Coplanar forces in equilibrium

Three forces in equilibrium

Lami’s theorem

Non-concurrent forces

Equilibrium under parallel forces

Condition for equilibrium of a particle A particle is said to be in equilibrium if the particle: • is at rest (if originally at rest); or • has a constant velocity (if originally in motion). Normally, however, the term ‘equilibrium’ of more precisely ‘static equilibrium’ is used to describe an object at rest. To maintain equilibrium, it is necessary to satisfy Newton’s first law of motion. Newton’s first law of motion requires that the resultant force acting on a particle to be equal to zero. This condition may be expressed as :

Newton’s first law of motion requires that the resultant force acting on a particle to be equal to zero. This condition may be expressed as F = 0 where  F is the vector sum of all the forces acting on a particle.

Coplanar Force Systems For equilibrium condition, the equation can be written as : F = 0 Fx i + Fy j = 0 For both of these vector equations above to be valid, then both the x and y components must be equal to zero. Hence, Fx = 0 = F1x + F2x + ….. Fy = 0 = F1y + F2y + ….. The scalar equations of equilibrium shown above require that the algebraic sum of the x and y components of all the forces acting on the particle must be equal to zero.

Working rule for numerical exercises Coplanar force equilibrium problems for a particle can be analyzed using the following procedure: 1. Free-Body Diagram 2. Establish the x and y axes is any suitable orientation 3. Label all the known and unknown force magnitudes and directions on the diagram. 4. The sense of a force having an unknown magnitude can be initially assumed. 5. Equations Of Equilibrium Apply the equations of equilibrium Fx = 0 and Fy = 0 . Components are positive if they are directed along a positive axis and negative if they are directed along the negative axis. 6. If the solution gives a negative result, this indicates that the sense of the force is reverse of that shown on the free-body diagram.

Conditions for Rigid body equilibrium Consider a rigid body which is at rest or moving with x y z reference at constant velocity rigid body

Free body diagram of ith particle of the body External force : gravitational, electrical, magnetic or contact force j Internal force i Force equilibrium equation for particle i (Newton first law)

Force equilibrium equation for the whole body (Newton’s 3rd law ) Moment of the forces action on the ith particle about pt. O

Moment equilibrium equation for the body Equations of equilibrium for a rigid body are

Equations of Equilibrium for 2D rigid body (1) Conditions of equilibrium Couple moment y x Here: algebraic sum of x components of all force on the body. algebraic sum of y components of all force on the body. algebraic sum of couple moments and moments of all the force components about an axis ⊥ xy plane and passing 0.

Example 600N 200N 2m 3m 100N y x Bx A B By Ay 3 unknown Ax, Bx, By Equations of equilibrium 3 equations for 3 unknowns

General conditions for equilibrium in a plane

Principle of Transmissibility The principle of transmissibility states that the condition of equilibrium or of motion of a rigid body will remain unchanged if a force F action at a given point of the rigid body is replace by a force F’ of the same magnitude and the same direction, but acting at a different point, provided that the two forces have the same line of action.