PROBLEM 2.77 A horizontal circular plate is suspended as shown from three wires which are attached to a support at D and form 30° angles with the vertical. Knowing that the x component of the force exerted by wire AD on the plate is 220.6 N, determine (a) the tension in wire AD, (b) the angles θx, θy, and θz that the force exerted at A forms with the coordinate axes.

SOLUTION

(a) sin 30 sin 50 220.6 NxF F= ° ° = (Given)

220.6 N 575.95 Nsin30 sin50

= =° °

F

576 N=F

(b) 220.6cos 0.3830575.95

θ = = =xx

FF

67.5xθ = °

cos30 498.79 NyF F= ° =

498.79cos 0.86605575.95

yy

FF

θ = = =

30.0yθ = °

sin 30 cos50zF F= − ° °

( )575.95 N sin 30 cos50= − ° °

185.107 N= −

185.107cos 0.32139575.95

zz

FF

θ −= = = −

108.7zθ = °

81

PROBLEM 2.78 A horizontal circular plate is suspended as shown from three wires which are attached to a support at D and form 30° angles with the vertical. Knowing that the z component of the force exerted by wire BD on the plate is –64.28 N, determine (a) the tension in wire BD, (b) the angles θx, θy, and θz that the force exerted at B forms with the coordinate axes.

SOLUTION

(a) sin 30 sin 40 64.28 NzF F= − ° ° = − (Given)

64.28 N 200.0 N

sin30 sin40= =

° °F 200 NF =

(b) sin 30 cos 40xF F= − ° °

( )200.0 N sin 30 cos 40= − ° °

76.604 N= −

76.604cos 0.38302200.0

xx

FF

θ −= = = − 112.5xθ = °

cos30 173.2 NyF F= ° =

173.2cos 0.866200

yy

FF

θ = = = 30.0yθ = °

64.28 NzF = −

64.28cos 0.3214200

zz

FF

θ −= = = − 108.7zθ = °

82

PROBLEM 2.79 A horizontal circular plate is suspended as shown from three wires which are attached to a support at D and form 30° angles with the vertical. Knowing that the tension in wire CD is 120 lb, determine (a) the components of the force exerted by this wire on the plate, (b) the angles θx, θy, and θz that the force forms with the coordinate axes.

SOLUTION

(a) ( )120 lb sin 30 cos60 30 lbxF = − ° ° = −

30.0 lbxF = −

( )120 lb cos30 103.92 lbyF = ° =

103.9 lb= +yF

( )120 lb sin 30 sin 60 51.96 lbzF = ° ° =

52.0 lbzF = +

(b) 30.0cos 0.25120

xx

FF

θ −= = = −

104.5xθ = °

103.92cos 0.866120

yy

FF

θ = = =

30.0yθ = °

51.96cos 0.433120

zz

FF

θ = = =

64.3zθ = °

83

PROBLEM 2.80 A horizontal circular plate is suspended as shown from three wires which are attached to a support at D and form 30° angles with the vertical. Knowing that the x component of the forces exerted by wire CD on the plate is –40 lb, determine (a) the tension in wire CD, (b) the angles θx, θy, and θz that the force exerted at C forms with the coordinate axes.

SOLUTION

(a) sin 30 cos60 40 lbxF F= − ° ° = − (Given)

40 lb 160 lbsin30 cos60

= =° °

F

160.0 lbF =

(b) 40cos 0.25160

xx

FF

θ −= = = −

104.5xθ = °

( )160 lb cos30 103.92 lbyF = ° =

103.92cos 0.866

160y

yFF

θ = = =

30.0yθ = °

( )160 lb sin 30 sin 60 69.282 lbzF = ° ° =

69.282cos 0.433160

zz

FF

θ = = =

64.3zθ = °

84

PROBLEM 2.81 Determine the magnitude and direction of the force

( ) ( ) ( )800 lb 260 lb 320 lb .= + −F i j k

SOLUTION

( ) ( ) ( )2 2 22 2 2 800 lb 260 lb 320 lbx y zF F F F= + + = + + − 900 lbF =

800cos 0.8889900

xx

FF

θ = = = 27.3xθ = °

260cos 0.2889900

yy

FF

θ = = = 73.2yθ = °

320cos 0.3555

900z

zFF

θ −= = = − 110.8zθ = °

85

PROBLEM 2.82 Determine the magnitude and direction of the force

( ) ( ) ( )400 N 1200 N 300 N .= − +F i j k

SOLUTION

( ) ( ) ( )2 2 22 2 2 400 N 1200 N 300 Nx y zF F F F= + + = + − + 1300 NF =

400cos 0.307691300

xx

FF

θ = = = 72.1xθ = °

1200cos 0.92307

1300y

yFF

θ −= = = − 157.4yθ = °

300cos 0.23076

1300z

zFF

θ = = = 76.7zθ = °

86

PROBLEM 2.83 A force acts at the origin of a coordinate system in a direction defined by the angles θx = 64.5° and θz = 55.9°. Knowing that the y component of the force is –200 N, determine (a) the angle θy, (b) the other components and the magnitude of the force.

SOLUTION

(a) We have

( ) ( ) ( ) ( ) ( ) ( )2 2 22 2 2cos cos cos 1 cos 1 cos cosx y z y y zθ θ θ θ θ θ+ + = ⇒ = − −

Since 0yF < we must have cos 0yθ <

Thus, taking the negative square root, from above, we have:

( ) ( )2 2cos 1 cos64.5 cos55.9 0.70735yθ = − − ° − ° = − 135.0yθ = °

(b) Then:

200 N 282.73 N

cos 0.70735y

y

FF

θ−

= = =−

and ( )cos 282.73 N cos64.5x xF F θ= = ° 121.7 NxF =

( )cos 282.73 N cos55.9z zF F θ= = ° 158.5 NyF =

283 NF =

87

PROBLEM 2.84 A force acts at the origin of a coordinate system in a direction defined by the angles θx = 75.4° and θy = 132.6°. Knowing that the z component of the force is –60 N, determine (a) the angle θz, (b) the other components and the magnitude of the force.

SOLUTION

(a) We have

( ) ( ) ( ) ( ) ( ) ( )2 2 22 2 2cos cos cos 1 cos 1 cos cosx y z y y zθ θ θ θ θ θ+ + = ⇒ = − −

Since 0zF < we must have cos 0zθ <

Thus, taking the negative square root, from above, we have:

( ) ( )2 2cos 1 cos75.4 cos132.6 0.69159zθ = − − ° − ° = − 133.8zθ = °

(b) Then:

60 N 86.757 N

cos 0.69159z

z

FFθ

−= = =

− 86.8 NF =

and ( )cos 86.8 N cos75.4x xF F θ= = ° 21.9 NxF =

( )cos 86.8 N cos132.6y yF F θ= = ° 58.8 NyF = −

88

PROBLEM 2.85 A force F of magnitude 400 N acts at the origin of a coordinate system. Knowing that θx = 28.5°, Fy = –80 N, and Fz > 0, determine (a) the components Fx and Fz, (b) the angles θy and θz.

SOLUTION

(a) Have

( )cos 400 N cos 28.5x xF F θ= = ° 351.5 NxF =

Then:

2 2 2 2x y zF F F F= + +

So: ( ) ( ) ( )2 2 2 2400 N 352.5 N 80 N zF= + − +

Hence:

( ) ( ) ( )2 2 2400 N 351.5 N 80 NzF = + − − − 173.3 NzF =

(b)

80cos 0.20

400y

yFF

θ −= = = − 101.5yθ = °

173.3cos 0.43325400

zz

FF

θ = = = 64.3zθ = °

89

PROBLEM 2.86 A force F of magnitude 600 lb acts at the origin of a coordinate system. Knowing that Fx = 200 lb, θz = 136.8°, Fy < 0, determine (a) the components Fy and Fz, (b) the angles θx and θy.

SOLUTION

(a) ( )cos 600 lb cos136.8z zF F θ= = °

437.4 lb= − 437 lbzF = −

Then:

2 2 2 2x y zF F F F= + +

So: ( ) ( ) ( ) ( )22 2 2600 lb 200 lb 437.4 lbyF= + + −

Hence: ( ) ( ) ( )2 2 2600 lb 200 lb 437.4 lbyF = − − − −

358.7 lb= − 359 lbyF = −

(b)

200cos 0.333600

xx

FF

θ = = = 70.5xθ = °

358.7cos 0.59783600

yy

FF

θ −= = = − 126.7yθ = °

90