Explanation Given: The magnitude of force, F 1 = 60 lb . The resultant force in Cartesian vector form, F R = { − 100 k } lb The angle between F 1 and F ′ is 50 ° The angle between F ′ and x axis is 30 ° Here, the sum of components of vector force in x and y direction is F ′ . Show the magnitude of components force F 1 and F 2 into x , y , and z direction as in Figure (1). Express the resultant force F R as a Cartesian vector using Figure (1). F R = { − 100 k } lb Here, the unit vector is k . Express the force F 1 as a Cartesian vector using Figure (1). F 1 = F x ( − i ) + F y ( + j ) + F z ( − k ) = { − F ′ 1 cos 30 ° i + F ′ 1 sin 30 ° j − F 1 sin 50 ° k } = − F 1 cos 50 cos 30 ° ( i ) + F 1 cos 50 ° sin 30 ° j − F 1 sin 50 ° k = F 1 { − cos 50 cos 30 ° ( i ) + cos 50 ° sin 30 ° j − sin 50 ° k } lb (I) Express the force F 2 as a Cartesian vector using Figure (1). F 2 = { F 2 x ( i ) + F 2 y ( j ) + F 2 z ( k ) } lb Express the resultant force as a Cartesian vector. F R = F 1 + F 2 (II) Express the magnitude of F 2 . F 2 = ( F 2 ) x 2 + ( F 2 ) y 2 + ( F 2 ) z 2 (III) Write the coordinate direction angles. cos α 2 = ( F 2 ) x F 2 (IV) cos β 2 = ( F 2 ) y F 2 (V) cos γ 2 = ( F 2 ) z F 2 (VI) Conclusion: Substitute 60 lb for F 1 in Equation (I). F 1 = 60 { − cos 50 cos 30 ° ( i ) + cos 50 ° sin 30 ° j − sin 50 ° k } lb = 60 { − ( 0.6427 ) ( 0.866 ) ( i ) + ( 0.6427 ) ( 0.5 ) j − 0.766 k } = 60 { ( − 0.556 ) ( i ) + ( 0.321 ) j − 0.766 k } = − 33.36 i + 19.26 j − 45.96 k (VII) Substitute { − 100 k } lb for F R , − 33

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Chapter 2.6, Problem 78P

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The magnitude of F2 and its coordinate direction angles (α2,β2,γ2).

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The magnitude of F 2 and its coordinate direction angles ( α 2 , β 2 , γ 2 ) .

Solution Summary: The author explains how to determine the magnitude of F_2 and its coordinate direction angles.

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The magnitude of F2 and its coordinate direction angles (α2,β2,γ2).

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