CV2101 Chapt 2 - Force Vectors

# CV2101 Chapt 2 - Force Vectors - Chapter 2 h Force Vectors...

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Chapter 2 orce Vectors Force Vectors 2-1

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1 Scalars & Vectors 2.1 Scalars & Vectors Mechanics deals with two kinds of quantities: ± Scalars - quantities with magnitudes only. eg, mass (kg), length (m), speed (m/s) ± Vectors - magnitudes, directions and sense eg, force (N), displacement (m), moment (N ڄ m) 2-2 Chapter 2 Force Vectors
2.1 Scalars & Vectors ont. Cont. The magnitude of the vector is the length of the arrow, the direction is defined by the angle between a reference axis and the arrow’s line of action, and the sense is indicated by the arrowhead. 2-3 Chapter 2 Force Vectors

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ypes of Vectors Types of Vectors 1. Coplanar vectors lie in the same plane. 2. Collinear vectors act along the same line of action. 3. Concurrent vectors have lines of action that pass rough the same point. through the same point. 2-4 Chapter 2 Force Vectors
2 Vector Operations 2.2 Vector Operations ± Vector addition: R = A + B = B + A Vectors obey parallelogram law. The resultant R can also be obtained by the triangle rule. 2-5 Chapter 2 Force Vectors

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2 Vector Operations 2.2 Vector Operations Cont. ± Vector subtraction: R’ = A - B = A + (-B) ± Resolution of a vector A vector may be resolved into two components by drawing a //gram. For a given vector, there exist an infinite number of possible sets of components. 2-6 Chapter 2 Force Vectors
artesian vector form (2- ) Cartesian vector form (2 D) A force F may be resolved into two components along the x and y axes: F = F x i + F y j where F = F cos θ ; = F sin θ i is a unit vector in the x direction j is a unit vector in the y direction We can also represent F with a agnitude: ( nd magnitude: F = (F 2 + F 2 ) and direction tan θ = F / F 2-7 F’ = F’ i –F’ j Chapter 2 Force Vectors

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esultant of several coplanar forces Resultant of several coplanar forces Where three or more forces passing through the same point, it is convenient to determine their resultant by adding the respective components of all forces to obtain the magnitude and direction of e resultant the resultant. F R = F 1 + F 2 + F 3 = (F x -F x + F x ) i + (F y + F y y ) j 1x 2x 3x 1y 2y 3y = (F Rx ) i + (F Ry ) j Magnitude and direction of F R F = (F x 2 + F y 2 ); R Rx Ry θ = tan -1 (F Ry / F Rx ) 2-8 Chapter 2 Force Vectors
ine law and Cosine law Sine law and Cosine law 2-9 Chapter 2 Force Vectors

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xample 2 1 Example 2.1 Two forces act on the hook. Determine the magnitude and irection of the resultant force direction of the resultant force. Solution: Method 1: Addition by trigonometry Method 2: Algebraic solution using x, y components 2-10 Chapter 2 Force Vectors
xample 2 1 Example 2.1 Cont. Solution 1: Addition by trigonometry Construct the force triangle. The angle between F 1 and F 2 = 90°- 10°- 15° = 65° =>supplementary angle =180°-65° =115° aw of cosines to determine F Law of cosines to determine F R : aw of sines to obtain the direction of F Law of sines to obtain the direction of F R : o o 150N 212.6N = θ = 39.8 sin θ sin115 Thus the direction 2-11 φ =39.8 +15 =54.8 oo o Chapter 2 Force Vectors

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xample 2 1 Example 2.1 Cont.
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## This note was uploaded on 09/06/2009 for the course CEE CV2101 taught by Professor Soh during the Spring '09 term at Nanyang Technological University.

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CV2101 Chapt 2 - Force Vectors - Chapter 2 h Force Vectors...

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