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y Q1. If 0 = 60° and F = 20 kN, determine the magnitude of the resultant force and its direction measured clockwise from the positive x axis. 60=0 F S √2 50 kN -x Fx = 20 cos (60) = 10 kN Fy = 20sin(60) = 17.3KN 2 FR = √(Fr) x² + (Fr) y² FR = √(10KN)² +(17.3kN)² = 19.98 KN 29KN = 40 kN

y Q1. If 0 = 60° and F = 20 kN, determine the magnitude of the resultant force and its direction measured clockwise from the positive x axis. 60=0 F S √2 50 kN -x Fx = 20 cos (60) = 10 kN Fy = 20sin(60) = 17.3KN 2 FR = √(Fr) x² + (Fr) y² FR = √(10KN)² +(17.3kN)² = 19.98 KN 29KN = 40 kN

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y Q1. If 0 = 60° and F = 20 kN, determine the magnitude of the resultant force and its direction measured clockwise from the positive x axis. 60=0 F S √2 50 kN -x Fx = 20 cos (60) = 10 kN Fy = 20sin(60) = 17.3KN 2 FR = √(Fr) x² + (Fr) y² FR = √(10KN)² +(17.3kN)² = 19.98 KN 29KN = 40 kN
Transcribed Image Text:y Q1. If 0 = 60° and F = 20 kN, determine the magnitude of the resultant force and its direction measured clockwise from the positive x axis. 60=0 F S √2 50 kN -x Fx = 20 cos (60) = 10 kN Fy = 20sin(60) = 17.3KN 2 FR = √(Fr) x² + (Fr) y² FR = √(10KN)² +(17.3kN)² = 19.98 KN 29KN = 40 kN
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