Explanation Given: The magnitude of component force, F 2 = 200 N The magnitude of component force, F 3 = 180 N The magnitude of resultant force along the positive x ′ axis, F R = 800 N The angle between the magnitudes of component force F 1 and x ′ axis, θ The angle between the magnitudes of component force F 2 and x ′ axis is 60 ° Show the component of forces F 1 , F 2 , and F 3 into x and y -axis components as in Figure (1). Use Figure (1) and find out cos θ and sin θ . cos θ = 12 13 sin θ = 5 13 Express the magnitude of resultant force acting along the positive x ′ axis. + → F R x = ∑ F x F R x = F 1 [ sin ( 60 ° + θ ) ] − F 3 cos θ (I) Express the magnitude of resultant force acting along the positive x ′ axis. + ↑ F R y = ∑ F y F R y = F 1 ( cos 60 ° + θ ) + F 2 + F 3 sin θ (II) Write the components forces of resultant force in x and y axis. ( F R ) x = F R sin 60 ° ( F R ) y = F R cos 60 ° Conclusion: Substitute F R sin 60 ° for ( F R ) x , 180 N for F 3 , and 12 13 for cos θ in Equation (I). F R sin 60 ° = F 1 [ sin ( 60 ° + θ ) ] − ( 180 ) 12 13 0.866 F R = F 1 [ sin ( 60 ° + θ ) ] − 166.153 (III) Substitute 800 N for F R in Equation (III). 0.866 ( 800 N ) = F 1 [ sin ( 60 ° + θ ) ] − 166

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Chapter 2.4, Problem 48P

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The magnitude and direction of the component force (F1) measured along the positive x′ axis.

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The magnitude and direction of the component force ( F 1 ) measured along the positive x ′ axis.

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