P2_4_JK

# P2_4_JK - Julie Kokinos y F2 x F1 1 The...

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Unformatted text preview: Julie Kokinos +y F2 +x F1 1. The coordinate system is set up in the above diagram with the positive x ­axis pointing to the left and the positive y ­axis pointing upward. The x ­axis is defined as opposite of the “normal” way because the angle is to be defined from this axis. 2. Break F1 and F2 into the x and y components: (note: angles are in degrees and were determined by using the fact that angles within a right angle add up to 90 degrees) F1,x= 200sin(45) F1,y=  ­200cos(45) **negative because the projection of F1 on the y ­axis points downward F2,x= 150sin(30) F2,y= 150cos(30) 3. The x and y components of the resultant vector FR are: FR,x= F1,x + F2,x = 200sin(45) + 150sin(30) = 216.421 lb FR,y = F1,y + F2,y =  ­200cos(45) + 150cos(30) =  ­11.518 lb 4. The magnitude of the resultant vector FR is found by using the following equation: FR = squareroot(FR,x2 + FR,y2) Solve to find that FR = 216.728 lb 5. Find the direction of the resultant force vector by determining its angle from the positive x ­axis. θ= tan ­1(FR,y/FR,x) Solve to find that θ=  ­3.046 degrees ...
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