University:
Michigan State University
Course:
ERG 160 | Success in Science, Technology, Engineering, and Mathematics
Academic year:

2021
Views:

253

Pages:

3
Author:

Tarime9wn

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 106. Free-Body Diagram: The free-body diagram indicates the forces on the camera and tripod slid along their lines of action to the plane ABCD. Note that the x-coordinate of the center of mass of the camera is: xCAM = − ( 2.4 in. − 1 in.) = −1.4 in. ΣM B = 0: or ( − 3 in.) k × C y j + ( −1.5 in.) k − (1.4 in.) i  × ( − 0.44 lb ) j + ( −1.5 in.) k − ( 2.8 in.) i  × ( − 0.53 lb ) j + ( −1.5 in.) k − ( 3.2 in.) i  × Ay j = 0 ( 3 in.) C yi − (1.5 in.)( 0.44 lb ) i + (1.4 in.)( 0.44 lb ) k − (1.5 in.)( 0.53 lb ) i + ( 2.8 in.)( 0.53 lb ) k + (1.5 in.) Ay i − ( 3.2 in.) Ayk = 0 Setting the coefficients of the unit vectors equal to zero: k: − Ay ( 3.2 in.) + ( 0.53 lb )( 2.8 in.) + ( 0.44 lb )(1.4 in.) = 0 Ay = 0.65625 lb or A y = 0.656 lb continued COSMOS: Complete Online Solutions Manual Organization System i: ( 3 in.) C y − (1.5 in.)( 0.44 lb ) − (1.5 in )( 0.53 lb ) + (1.5 in.) (0.65625 lb) = 0 C y = 0.156875 lb or C y = 0.1569 lb or B y = 0.1569 lb ΣFx = By − 0.53 lb − 0.44 lb + 0.156875 lb + 0.156875 lb = 0 By = 0.156875 lb (b) Free-Body Diagram: Condition for no tipping: By > 0 ΣM A = 0: ( 3.2 in.) i + ( −1.5 in.) k  × C y j + ( 3.2 in.) i + (1.5 in.) k  × B y j { } + (1.8 in.) i × ( − 0.44 lb ) j + 1.8 in. − (1.4 in.) cosθ  i − (1.4 in.) sinθ k × ( − 0.53 lb ) j = 0 or ( 3.2 in.) C yk + (1.5 in.) C yi + ( 3.2 in.) By k − (1.5 in.) Byi − (1.8 in.)( 0.44 lb ) k + 1.8 in. − (1.4 in.) cosθ  ( − 0.53 lb ) k + (1.4 in.) sinθ ( − 0.53 lb ) i = 0 Setting the coefficients of the unit vectors equal to zero: i: C y (1.5 in.) − By (1.5 in.) − ( 0.53 lb )(1.4 in.) sin θ = 0 C y = B y + ( 0.53 lb ) k: 1.4 sin θ 1.5 ( By + C y ) ( 3.2 in.) − ( 0.44 lb )(1.8 in.) − ( 0.53 lb ) 1.8 in. − (1.4 in ) cosθ  = 0 or  1.4   2By ( 3.2 in.) + ( 0.53 lb )   ( 3.2 in.) sin θ + (1.4 in.) cosθ  − ( 0.44 lb + 0.53 lb )(1.8 in.) = 0  1.5   continued COSMOS: Complete Online Solutions Manual Organization System Solving for By : By =   1.4    1 ( 0.44 lb + 0.53 lb )1.8 in. − 0.53 lb   ( 3.2 in.) sin θ + (1.4 in.) cosθ   2 ( 3.2 in.)   1.5    and By > 0 2.3531 > 2.1333sin θ + cosθ To solve for θ : 2.1333sin θ + cosθ = A cos (θ + α ) = A ( cosθ cos α − sin θ sin α ) where A= ( 2.1333)2 + (1)2 = 2.3561, and  − 2.1333   , which (noting that cos α > 0) gives 1   α = tan −1  α = − 64.885° The inequality for By becomes: 2.3531 > 2.3561cos(θ − 64.885°) or cos (θ − 64.885° ) < 0.99873 or θ − 64.885° < cos −1 ( 0.99873) or θ < 64.885° ± 2.8879° θ max = 62.00°