mA= 10 kg), and mgBGis the force of gravity on block B.θ= 30° is the angle of incline. ForAwe take the +xto be uphill and +yto be in the direction of the normal force; the positive direction is chosen downwardfor block B.SinceAis moving down the incline, the force of friction is uphill with magnitude fk= μkFN(where k= 0.20). Newton’s second law leads to sin0cos00kAANABBT fmaFmgmg T ma−+==−==for the two bodies (where a= 0 is a consequence of the velocity being constant). We
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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.