Explanation Given data: The force exerted by Adrienne F A on K along 30 ° below positive x -direction is 100 N . The force exerted by Jim along negative y -direction is 150 N . Formula used: The vector resolution of any vector force along the x -direction is: F x = F cos θ Here, F x is the force along the x -direction, F is the force vector acting on the body and θ is the angle between force vector and x -axis. The vector resolution of any vector force along y -direction is: F y = F sin θ Here, F y is the force along the y -direction, F is the force vector acting on the body and θ is the angle between force vector and x -axis. The magnitude of the force vector is: F = ( F x ) 2 + ( F y ) 2 Here, F is the resultant force, F x is the force along the x -direction and F y is the force along the y -direction. The direction of the resultant vector force is: tan ϕ = F y F x Here, ϕ is the angle between the resultant force and positive x -direction, F x is the force along the x -direction and F y is the force along the y -direction. Explanation: The force applied by the Luis to put a knot in equilibrium along the x -direction is equal and opposite to the total force exerted by Adrienne and Jim along the x -direction. Since Jim applied the force along the negative y- direction, so, the component of the force exerted by Jim along the x- direction is zero. Write the expression of vector resolution of force along the x -direction. F L on K x = − F A on K cos θ Here, F L on K x is the force applied by Luis on knot along the x -direction, F A on K is the force exerted by Adrienne and θ is the angle of F A on K with x -axis. Substitute 100 N for F A on K and − 30 ° for θ . F L on K x = − 10 0 N cos ( − 30 ° ) = − 86.6 N Thus, Luis has to apply 86.6 N force along negative x -direction. Again, the force exerted by Jim and Adrienne on the knot in the y -direction is equal and opposite to the force exerted by Luis along y -direction to put the knot in equilibrium. Write the expression of vector resolution of force along the y -direction. F L on K y = − ( F − y + F A on K sin θ ) Here, F L on K y is the force applied by Luis on knot along y -direction and F − y is the y -component of force exerted by Jim along negative y -direction and F A on K sin θ is the y -component of force exerted by Adrienne along the negative y -direction

College Physics: Explore And Apply, Volume 2 (2nd Edition)

2nd Edition

ISBN: 9780134862910

Author: Eugenia Etkina, Gorazd Planinsic, Alan Van Heuvelen, Gorzad Planinsic

Publisher: PEARSON

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Chapter 8, Problem 6P

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The magnitude and direction of force applied by Luis to put the knot at equilibrium if Adrienne exerts a 100 N force FA on K along 30° below the positive x-axis and Jim exerts a 150 N force in the negative y-direction.

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Chapter 8, Problem 5P

Chapter 8, Problem 7P

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Chapter 8 Solutions

College Physics: Explore And Apply, Volume 2 (2nd Edition)

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The magnitude and direction of force applied by Luis to put the knot at equilibrium if Adrienne exerts a 100 N force F A on K along 30 ° below the positive x -axis and Jim exerts a 150 N force in the negative y -direction.

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The magnitude and direction of force applied by Luis to put the knot at equilibrium if Adrienne exerts a 100 N force FA on K along 30° below the positive x-axis and Jim exerts a 150 N force in the negative y-direction.

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Chapter 8 Solutions