CHAPTER 2: STATICS OF PARTICLES STATICS AND DYNAMICS PDT 101/3 CHAPTER 2: STATICS OF PARTICLES PREPARED BY: TAN SOO JIN sjtan@unimap.edu.my

CHAPTER OUTLINE Scalars and Vectors Vector Operations Vector Addition of Forces Addition of a System of Coplanar Forces

APPLICATION OF VECTOR ADDITION There are three concurrent forces acting on the hook due to the chains. We need to decide if the hook will fail (bend or break). To do this, we need to know the resultant or total force acting on the hook as a result of the three chains. FR

2.1 Scalars and Vectors Scalar – any positive or negative physical quantity that can be completely specified by its magnitude. e.g. Mass, time, and length

2.1 Scalars and Vectors Vector – A quantity that has magnitude and direction e.g. Position, force and moment – Represent by a letter with an arrow over it, – Magnitude is designated as – In this subject, vector is presented as A and its magnitude (positive quantity) as A

Scalar Multiplication 2.2 Vector Operations Vector A and its negative counterpart Scalar Multiplication and Division

2.2 Vector Operations Vector Addition Parallelogram Law Triangle construction Resultant vector R can be found by triangle construction

2.2 Vector Operations Vector Subtraction or A - B R’ = A – B = A + (-B)

2.3 Vector Addition of Forces Finding a Resultant Force Parallelogram law is carried out to find the resultant force Resultant, FR = (F1+ F2)

a/sin A = b/sin B = c/sin CC Law of Cosines Law of Sinus a2 = b2 + c2 – 2bc(cosA) b2 = a2 + c2 – 2ac(cosB) c2 = a2 + b2 – 2ab(cosC) a/sin A = b/sin B = c/sin CC

2.3 Vector Addition of Forces Procedure for Analysis Parallelogram Law Make a sketch using the parallelogram law 2 components forces add to form the resultant force Resultant force is shown by the diagonal of the parallelogram The components is shown by the sides of the parallelogram

2.3 Vector Addition of Forces Procedure for Analysis Trigonometry Redraw half portion of the parallelogram Magnitude of the resultant force can be determined by the law of cosines Direction if the resultant force can be determined by the law of sines Magnitude of the two components can be determined by the law of sines

EXAMPLE 1 The screw eye is subjected to two forces, F1 and F2. Determine the magnitude and direction of the resultant force.

SOLUTION Parallelogram Law Unknown: magnitude of FR and angle θ

How to solve the question?

Trigonometry Law of Cosines Law of Sines

Trigonometry Direction Φ of FR measured from the horizontal

EXERCISE 1 If the tension in the cable is 400 N, determine the magnitude and direction of the resultant force acting on the pulley. This angle is the same angle of line AB on the tailboard block. (Ans: 400N, 60o)

2.4 Addition of a System of Coplanar Forces Scalar Notation - x and y axes are designated positive and negative - Components of forces expressed as algebraic scalars

2.4 Addition of a System of Coplanar Forces Cartesian Vector Notation Cartesian unit vectors i and j are used to designate the x and y directions Unit vectors i and j have dimensionless magnitude of unity ( = 1 ) Magnitude is always a positive quantity, represented by scalars Fx and Fy

Coplanar Force Resultants To determine resultant of several coplanar forces: - Resolve force into x and y components - Addition of the respective components using scalar algebra - Resultant force is found using the parallelogram law - Cartesian vector notation:

2.4 Addition of a System of Coplanar Forces Coplanar Force Resultants - Vector resultant is therefore - If scalar notation are used

2.4 Addition of a System of Coplanar Forces Coplanar Force Resultants -In all cases we have * Take note of sign conventions - Magnitude of FR can be found by Pythagorean Theorem

EXAMPLE 2 Resolve the 500 N force exerted on the hook in Figure into horizontal and vertical components.

SOLUTION 2

EXERCISE 2 Determine the x and y components of the forces shown in Figure below. Fx = -200 kN Fy = 210 kN Fx = -143.40 kN Fy = 204.79 kN

EXAMPLE 3 Determine x and y components of F1 and F2 acting on the boom. Express each force as a Cartesian vector.

SOLUTION Scalar Notation Hence, from the slope triangle, we have

By similar triangles we have Scalar Notation: Cartesian Vector Notation:

EXERCISE 4 Determine the x and y components of the forces of the two forces F1 and F2 acting on the eye-bolt shown in Figure below. (Ans: F1x=173.21N, F1y=100N F2x=-76.6N, F2y= 64.28N)

EXAMPLE 4 The link is subjected to two forces F1 and F2. Determine the magnitude and orientation of the resultant force.

SOLUTION I Scalar Notation:

Resultant Force From vector addition, direction angle θ is

SOLUTION II Cartesian Vector Notation F1 = { 600cos30°i + 600sin30°j } N F2 = { -400sin45°i + 400cos45°j } N Thus, FR = F1 + F2 = (600cos30ºN - 400sin45ºN)i + (600sin30ºN + 400cos45ºN)j = {236.8i + 582.8j}N The magnitude and direction of FR are determined in the same manner as before.

EXERCISE 3 Find the x and y components of the forces P and Q shown in Figure below. Calculate magnitude and direction of the resultant force. (Ans: FR= 40.26kN, θ= 75.62o)

EXERCISE 4 If Φ=30° and the resultant force acting on the gusset plate is directed along the positive x-axis, determine the magnitudes of F2 and the resultant force. (Ans: F2=12.93 kN, FR=13.19 kN)