Rectangular components = cos( ) F F z j x x y k = - - PowerPoint PPT Presentation

rectangular components
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Rectangular components = cos( ) F F z j x x y k = - - PowerPoint PPT Presentation

Rectangular components = cos( ) F F z j x x y k = cos( ) F F F y r y y j = cos( ) F F F F z z z k z = + + 2 2 2 F F F F y x y z x r = + + F F i F


slide-1
SLIDE 1

Rectangular components

y x z

F r i Fxˆ j Fy ˆ k Fz ˆ k ˆ j ˆ i ˆ

z

θ

x

θ

y

θ

2 2 2 z y x

F F F F + + = ) cos(

x x

F F θ =

) cos(

y y

F F θ = ) cos(

z z

F F θ =

k F j F i F F

z y x

ˆ ˆ ˆ + + = r ) ˆ cos ˆ cos ˆ (cos k j i F F

z y x

θ θ θ + + = r ) ˆ ˆ ˆ ( k n j m i l F F + + = r

F

n F F ˆ = r

x

l θ cos =

y

m θ cos =

z

n θ cos = Direction cosine 1

2 2 2

= + + n m l

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SLIDE 2

Writing a vector in 3D (1)

1 Specification by two points on the line of action

x y z A (x1, y1, z1)

F r

B (x2, y2, z2)

A B AB

r r r r r r − = k z j y i x rA ˆ ˆ ˆ

1 1 1

+ + = r k z j y i x rB ˆ ˆ ˆ

2 2 2

+ + = r

| | ˆ

AB AB F

r r F n F F r r r = =

k z z j y y i x x rAB ˆ ) ( ˆ ) ( ˆ ) (

1 2 1 2 1 2

− + − + − = r

2 1 2 2 1 2 2 1 2 1 2 1 2 1 2

) ( ) ( ) ( ˆ ) ( ˆ ) ( ˆ ) ( z z y y x x k z z j y y i x x F F − + − + − − + − + − = r

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SLIDE 3

Writing a vector in 3D (2)

2 Specification by two angles which orient the line of action

y x z

F r

x

F r

y

F r

z

F r

k ˆ j ˆ i ˆ

θ φ

xy

F r

Fxy = F cos(φ) Fz = F sin(φ) Fx = Fxy cos(θ) = F cos(φ) cos(θ) Fy = Fxy sin(θ) = F cos(φ) sin(θ)

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SLIDE 4

Dot product

(scalar) P r Q r

α

Pcos(α) Projection of in the direction of P r Q r . Q ) cos(α PQ Q P = ⋅ r r

F r n r n F Fn ˆ ⋅ = r n n F Fn ˆ ˆ ⋅ = r r

If is a unit vector , dot product expresses the projection of vector in the unit vector direction Q r ) ˆ (n ) ˆ ˆ ˆ ( ) ˆ ˆ ˆ ( ˆ k j i k n j m i l F n F Fn γ β α + + ⋅ + + = ⋅ = r ) ( γ β α n m l F + + = 1 ˆ ˆ ˆ ˆ ˆ ˆ = ⋅ = ⋅ = ⋅ k k j j i i ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ = ⋅ = ⋅ = ⋅ = ⋅ = ⋅ = ⋅ j k k j i k k i i j j i

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SLIDE 5

Angle between two vectors

) cos(θ PQ Q P = ⋅ r r From relation of dot product The angle between vectors and is P r Q r PQ Q P r r ⋅ =

−1

cos θ The angle between vectors and is F r n ˆ F n F ˆ cos 1 ⋅ =

r θ ˆ = ⋅n F r n F ˆ ⊥ r

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SLIDE 6

Sample 1

A force F 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 F passes through a point A whose coordinates are 3m, 4m and 5m. Determine (a) the x, y and z scalar components of F, (b) the projection Fxy of F on the x-y plane, and (c) the projection FOB of F along the line OB.

slide-7
SLIDE 7

Sample 2

The cable BC carries a tension of 750 N. Write this tension as a force T acting on point B in terms of the unit vector i, j and k. The elbow at A forms a right angle.

slide-8
SLIDE 8

Sample 3

The tension in supporting cable BC is 3200 N. Write the force which this cable exerts on the boom OAB as a vector T. Determine the angles θx, θy and θz which the line of action of T forms with the positive x-, y- and z-axes.