Newton's Three laws of Motion:- First law : A particle originally at rest, or moving in a straight line with constant velocity, tends to remain in this state provided the particle is not subjected to an unbalanced force.

Second law: A particle acted upon by an unbalanced force “F” experiences an acceleration “a” that has the same direction as the force and a magnitude that is directly proportional to the force.

Third Law: The mutual forces of action and reaction between two particles are equal, opposite, and collinear.

Newton’s law of gravitational attraction :

Transmissibility This maintains the equivalent of the system as the equilibrium remains unchanged.

Scalars and Vectors Scalar – A quantity characterized by a positive or negative number – Indicated by letters in italic such as A Eg: Mass, volume and length

Vector – A quantity that has both magnitude and direction Eg: Position, force and moment – Represent by a letter with an arrow over it such as or A – Magnitude is designated as or simply A

2.1 Scalars and Vectors Vector – Represented graphically as an arrow – Length of arrow = Magnitude of Vector – Angle between the reference axis and arrow’s line of action = Direction of Vector – Arrowhead = Sense of Vector

Magnitude of Vector = 4 units Example Magnitude of Vector = 4 units Direction of Vector = 20° measured counterclockwise from the horizontal axis Sense of Vector = Upward and to the right The point O is called tail of the vector and the point P is called the tip or head

Vector Operations Multiplication and Division of a Vector by a Scalar - Product of vector A and scalar a = aA - Magnitude = - If a is positive, sense of aA is the same as sense of A - If a is negative sense of aA, it is opposite to the sense of A

Multiplication and Division of a Vector by a Scalar - Negative of a vector is found by multiplying the vector by ( -1 ) - Law of multiplication applies Eg: A/a = ( 1/a ) A, a≠0

Vector Addition - Addition of two vectors A and B gives a resultant vector R by the parallelogram law - Result R can be found by triangle construction - Communicative Eg: R = A + B = B + A

Vector Addition

Vector Addition - Special case: Vectors A and B are collinear (both have the same line of action)

Vector Subtraction - Special case of addition Eg: R’ = A – B = A + ( - B ) - Rules of Vector Addition Applies

Procedure for Analysis Parallelogram Law - Make a sketch using the parallelogram law - Two 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

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

Procedure for Analysis Trigonometry - Magnitude of the two components can be determined by the law of sines

Example 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

Solution Trigonometry Law of Cosines

Solution Trigonometry Law of Sines

Solution Trigonometry Direction Φ of FR measured from the horizontal

Q = 60N P = 40N Sample Problem The two forces act on bolt at A Determine their resultant

Solution Q = 60N P = 40N

Solution Q = 60N P = 40N

Law of Cosines c A B a b C Solution a2 = b2 + c2 – 2bc(cosA) P = 40N Q = 60N c Law of Cosines A B a2 = b2 + c2 – 2bc(cosA) b2 = a2 + b2 – 2ac(cosB) c2 = a2 + b2 – 2ab(cosC) a b C

Law of Sinus Q = 60N P = 40N Q = 60N P = 40N Solution Law of Sinus Q = 60N P = 40N a/sin A = b/sin B = c/sin C Q = 60N P = 40N

Sample Problem SP 2.2 A barge is pulled by two tugboats. If the resultant of the forces exerted by the tugboats is a 25 kN force directed along the axis of the barge, determine the tension in each rope given α = 45, (b) the value of α for which the tension in rope 2 is minimum.

Solution 25kN Tension for Graphical Solution The parallelogram law is used; the diagonal (resultant) is known to be equal 25kN and to be directed to the right. This sides are drawn paralled to the ropes.If the drawing is done to scale, we measure T1 = 18.5 kN T2= 13.0 kN

25kN 18.30kN 25kN 12.94kN

25kN 12.5kN 25kN 25kN 21.65kN 25kN

Rectangular Components of a Force x- y components Perpendicular to each other Fx = F cos θ Fy = F sin θ y Fy F j θ O x i Fx

Example 1 A force of 800N is exerted on a bold A as shown in the diagram. Determine the horizontal and vertical components of the force. Fx = -F cosα= -(800 N ) cos 35º = -655N Fy = +F sinα= +(800 N ) sin 35º = +459N

Example 2 (a) (2.9) (b) Fx = + (300N) cos α Fy = - (300N) sin α A man pulls with a force of 300N on a rope attached to a building as shown in the picture. What are the horizontal and vertical components of the force exerted by the rope at point A? Fx = + (300N) cos α Fy = - (300N) sin α Observing that AB = 10m cos α = 8m = 8m = 4 AB 10m 5 sin α = 6m = 6m = 3 We thus obtain Fx = + (300N) = +240 N 4 5 Fy = - (300N) = -180 N 3 5 We write F = + (240N) i - (180 N) j tan θ= Fx Fy (2.9) F =

A force F = (3.1 kN)i + (6.7 kN)j is applied to bolt A. Example　3 6.7 A force F = (3.1 kN)i + (6.7 kN)j is applied to bolt A. Determine the magnitude of the force and the angle θit forms with the horizontal.

First we draw the diagram showing the two Solution First we draw the diagram showing the two rectangular components of the force and the angle θ. From Eq.(2.9), we write 6.7 θ= 65.17˚ Using the formula given before, Fx = F cos θ Fy = F sin θ

Pyj P Syj Ryi R S Pxi Sxi O O Qxi Rxi Q Qyj

ADDITION OF FORCE BY SUMMING X AND Y COMPONENTS How we calculate the force ?

2.8 ADDITION OF FORCE BY SUMMING X AND Y COMPONENTS

SAMPLE PROBLEM 2.3 Four forces act on A as shown. F2 = 80 N F1 = 150 N F4 = 100 N F3 = 110 N Four forces act on A as shown. Determine the resultant of the forces on the bolt.

Solution ( F2 cos 20˚ ) j ( F1 sin 30˚ ) j F2 = 80 N F1 = 150 N - F3 j - ( F2 sin 20˚ ) i ( F2 cos 20˚ ) j - ( F4 sin 15˚ ) j ( F4 cos 15˚ ) i ( F1 sin 30˚ ) j ( F1 cos 30˚ ) i F2 = 80 N F1 = 150 N F4 = 100 N F3 = 110 N

Solution ( F2 cos 20˚ ) j ( F1 sin 30˚ ) j F2 = 80 N F1 = 150 N - F3 j - ( F2 sin 20˚ ) i ( F2 cos 20˚ ) j - ( F4 sin 15˚ ) j ( F4 cos 15˚ ) i ( F1 sin 30˚ ) j ( F1 cos 30˚ ) i F2 = 80 N F1 = 150 N F4 = 100 N F3 = 110 N

- F3 j - ( F2 sin 20˚ ) i ( F2 cos 20˚ ) j - ( F4 sin 15˚ ) j ( F4 cos 15˚ ) i ( F1 sin 30˚ ) j ( F1 cos 30˚ ) i Ry = (14.3 N) j Rx = (199.1 N) i

PROBLEMS

2.21 Determine the x and y component of each of the forces shown.

Solution

Solution

Problems 2.23 Determine the x and y components of each of the forces shown. 1.2 m 1.4 m 2.3 m 1.5 m 2.0 m 1800 N 950N 900N

Solution 2.59m 2.69m 2.5 m

Solution 2.59m 2.69m 2.5 m

THE END