Force Vector niMAP Portal力向量德艺门.ppt

Force Vector (cont’) 2.8 Force Vector Directed along a Line Force F acting along the chain can be presented as a Cartesian vector by - Establish x, y, z axes - Form a position vector r along length of chain Unit vector, u = r/r that defines the direction of both the chain and the force We get F = Fu Example 2.13 The man pulls on the cord with a force of 350N. Represent this force acting on the support A, as a Cartesian vector and determine its direction. Solution End points of the cord are A (0m, 0m, 7.5m) and B (3m, -2m, 1.5m) r = (3m – 0m)i + (-2m – 0m)j + (1.5m – 7.5m)k = {3i – 2j – 6k}m  Magnitude = length of cord AB Unit vector, u = r /r = 3/7i - 2/7j - 6/7k Solution Force F has a magnitude of 350N, direction specified by u. F = Fu = 350N(3/7i - 2/7j - 6/7k) = {150i - 100j - 300k} N α = cos-1(3/7) = 64.6° β = cos-1(-2/7) = 107° γ = cos-1(-6/7) = 149° 2.9 Dot Product Dot product of vectors A and B is written as A·B (Read A dot B) Define the magnitudes of A and B and the angle between their tails A·B = AB cosθ where 0°≤ θ ≤180° Referred to as scalar product of vectors as result is a scalar 2.9 Dot Product Laws of Operation 1. Commutative law A·B = B·A 2. Multiplication by a scalar a(A·B) = (aA)·B = A·(aB) = (A·B)a 3. Distribution law A·(B + D) = (A·B) + (A·D) 2.9 Dot Product Cartesian Vector Formulation - Dot product of Cartesian unit vectors i·i = (1)(1)cos0° = 1 i·j = (1)(1)cos90° = 0 - Similarly i·i = 1 j·j = 1 k·k = 1 i·j = 0 i·k = 1 j·k = 1 2.9 Dot Product Cartesian Vector Formulation Dot product of 2 vectors A and B A·B = AxBx + AyBy + AzBz Applications The angle formed between two vectors or intersecting lines. θ = cos-1 [(A·B)/(AB)] 0°≤ θ ≤180° The components of a vector parallel and perpendicular to a line. Aa = A cos θ = A·u Example 2.17 The frame is subjected to a horizontal force F = {300j} N. Determine the components of this force parallel and perpendicular to the member AB. Solution Since Thus Soluti