net,net,0coscos0sinsinxCAyACBFFFFFφθθφ==−==+−¦¦To solve for BF, we first compute .φWith 220 NAF=, 170 NCF=and 47 ,θ=°we getcos(220 N)cos47.0cos0.88328.0170 NACFFφφ°===¡Substituting the value into the second force equation, we find sinsin(220 N)sin 47.0(170 N)sin 28.0241 N.
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.