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Solution, Prob. 1_3201 a. Choosing the positive direction of the x axis to the left, we have for mass A: F x A 0 mB g F cos f s Notice that the static friction was selected as positive. Its direction is actually determined by the competition between the other forces in the problem, such that- f s F cos mB g may be positive (friction pointing to the left) or negative (friction pointing to the right). b. We now demand that the friction is indeed in the static regime: N F cos mB g N mB mA mB mA gF g cos cos cos cos Where here we substituted N mA g F sin from the y axis equation. CHECK UNITS AND LIMITING CASES!! c. We now consider the case where F 0 . Newton's law states: F x A mAa mB g f k m a B g mA Where here we substituted f k mA g . Forces on m: Fx( m ) ma F f s mg sin (m) Fy 0 N m M mg cos Forces on M: Fx( M ) Ma f s f k Mg sin (M ) Fy 0 N surface N m M Mg cos And in addition it is known that f k k N surface . Solve for f s and demand the condition f s s N m M to get- m F Fmax 1 mg cos s k M