Advanced Mechanics Analysis Guide
"Learn about resolving forces in 3D space, expressing forces as vectors, and solving complex mechanical problems with detailed examples and visuals."
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RECTANGULAR COMPONENTS Many problems in mechanics require analysis in three dimensions, and for such problems it is often necessary to resolve a force into its three mutually perpendicular components. + + = z F F i F j F k x y Z Fz k = + + ( cos ) ( cos ) ( cos ) F F i F j F k x y z F z = = = cos ; cos ; cos F F F F F F x x y y z z y y 2 = + + 2 2 F F F F Fx i Fy j x y z = cos = l x x x y cos Direction cosines of m F F y = cos n cos F j z y y + + = 2 2 2 = + + Note : 1 l m n ( i l ) F F m j n k x x cos F i x
= + n + ( i l ) F F m j n k = ( ) F F F n Where is unit vector in the direction of force + + = F z nF i l m j n k Fz k F z y y Fx i Fy j x x
Specification of a force vector (a) Specification by two points on the line of action of the force. (b) Specification by two angles which orient the line of action of the force. a) Two points:
b) Two angles: = + + = F F i F j F k x y z = + + (cos cos cos sin sin ) F i j k
Problem-1 The turnbuckle is tightened until the tension in the cable AB equals 24 kN. Express the tension T acting on point A as a vector z 18 m A T=24 kN 30 m D y O 6m C B 5m x
z 18 m A=A(0, 18, 30) B=B(6, 13, 0) A T=24 kN 30 m = = + + ; T T n n i l m j n k AB AB D y O = + + = 2 2 2 Distance between tw points o (6 - 0) 13 ( 18 ) 0 ( 30 ) 31 m 6m C 6 0 13 18 0 30 B 5m x = = = = = = . 0 . 0 194 ; . 0 161 ; 968 l m n 31 i l = 31 31 + + = . 0 194 . 0 161 . 0 968 n m = j n k i j k AB = = 24 . 0 ( 194 . 0 161 . 0 968 ) . 4 64 . 3 87 23 22 . T T n i j k i j k AB = . 4 64 . 3 87 23 22 . kN T i j k
Problem-2 Consider a force as shown in the Figure. The magnitude of this force is 10 kN. Express it as a vector. z 10 kN 0 30 y 0 45 x
z = = 0 10 cos 30 . 8 66 kN F xy z F = = 0 10 sin 30 0 . 5 kN F z 10 kN = = = 0 0 cos 45 . 8 66 cos 45 . 6 12 kN F F x xy F 0 30 y y = = = 0 0 sin 45 . 8 66 sin 45 . 6 12 kN F F y xy 0 45 x F F x xy = + + = F F i F j F k x y z = + + . 6 12 . 6 12 . 5 00 kN F i j k
(Orthogonal) Projection F n C B ProjectionA = of on line in the or BC n - direction F A n F unit vecto = in the r direction of line n ABC n = + + where, n Here , , direction the are cosines of unit vecto r i l m j n k l m n n ( , , ) B x y z 2 2 2 ( = , , ) A x y z ( ) ( ) ( ) x x y y z z 1 1 1 = = = ; ; 2 1 2 1 2 1 l m n AB AB AB + + 2 2 2 ( ) ( ) ( ) AB x x y y z z 2 1 2 1 2 1
Problem-1 For the shown force: a. Determine the magnitudes of the projection of the force F = 100 N onto the u and v axes. b. Determine the magnitudes of the components of force F along the u and v axes. v F = 100 N 15 45 u O
Components of the force along u and v axes Projections of the force onto u and v axes v v 100 N 15 100 N 15 45 u O 45 u O ( ) 15 F ( ) sin = F F 100 u v comp comp 45 = = ( ( ) F ) = = 100 cos 45 70 7 . N F sin sin 120 F u proj ( ( ) F ) ( ( )proj v F ) 29 9 . N = = 100 cos 15 96 6 . N u u comp proj v proj = 81 6 . N v comp
Note Rectangular components of a force along the two chosen perpendicular axes, and projection of the force onto the same axes are the same. y 100 N 40 x O ( ( ) F ) ( ) = = = 100 cos 40 76 6 . N F F x x comp proj ( ) F = = = 100 sin 40 64 3 . N y y comp proj 2 2 = + = + = 2 2 : 76 6 . 64 3 . 100 N Check F F F x x
Problem-2 force A applied is F at the origin O of the axes shown as x-y-z in the Figure. Determine the vector form of projection . . ( i OA find along ) the line . F e F OA OA z = + + 106 141 176 N F i j k y A 3 m O 4 m x
= = + + 2 2 2 ( ) ( ) ( ) OA L x x y y z z 2 1 2 1 2 1 = + + = 2 2 2 6 ( ) 0 4 ( ) 0 3 ( ) 0 . 7 81 m OA 6 ( ) 0 4 ( ) 0 = = = = . 0 768 ; . 0 512 ; l m z . 7 81 . 7 81 = + + 106 141 176 N F i j k 3 ( ) 0 = = . 0 384 n . 7 81 y A Therefore, = unit vecto + along r the line OA is : F 3 m OA + n i l m j n + k OA O = + . 0 768 . 0 512 . 0 384 n i j k 4 m OA x
= = + + + + = . 106 ( 141 176 ).( . 0 768 . 0 512 . 0 384 ) 221 18 . N F F n i j k i j k OA OA = = = + + ( . ) 221 18 . . 0 ( 768 . 0 512 . 0 384 ) F F n n + F n i j k OA OA OA OA OA OA = + 169 . 87 113 24 . 84 . 93 N F i j k OA z = + + 106 141 176 N F i j k y A F 3 m OA O 4 m x
Problem-3 A force with a magnitude of 100 N is applied at the origin O of the axes x-y-z as shown. The line of action of force passes through a point A. Determine the projection Fxy of 100N force on the x-y plane. z y 100 N A 3 m O 4 m x
z y ) 0 4 ( + ) 0 + ) 0 2 2 2 6 ( 0 ( + . 7 211 = = = cos . 0 923 = (6,4,3) 100 N F A xy . 7 810 ) 0 4 ( + ) 0 ) 0 2 2 2 6 ( 3 ( 3 m xy (6,4,0) = = = cos 100 . 0 ( 923 ) 92 3 . N F F O B xy xy F (0,0,0) xy 4 m x
Alternative Solution ) 0 ) 0 4 ( + + ) 0 6 ( 3 ( + i j k ) k j = + + = ( i l 100 F F m n z ) 0 4 ( + ) 0 ) 0 2 2 2 6 ( 3 ( i j + + 6 4 3 k k 8 . j = = + + 100 76 51 2 . 38 4 . N F i y . 7 810 = (6,4,3) 100 N F A k ) 0 ) 0 4 ( + 0 ( + ) 0 6 ( i j k j = + + = n i l m n OB 3 m ) 0 4 ( + ) 0 0 ( + ) 0 2 2 2 6 ( xy (6,4,0) k i j + + 6 4 0 O i j = = + B . 0 832 . 0 554 n F (0,0,0) OB . 7 211 xy 4 m x = = . 0 + . 0 + = 76 8 . 832 51 2 . 554 38 4 . 0 92 3 . N F F n xy OB = 92 3 . N F xy
