ch06-p001 - F N mg F f F = µ s F N = µ s mg =(0 45(45...

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1 (a) The free-body diagram for the bureau is shown on the right. ± F is the applied force, ± f is the force of friction, ± F N is the normal force of the floor, and m±g is the force of gravity. Take the x axis to be horizontal and the y axis to be vertical. Assume the bureau does not move and write the Newton’s second law equations. The x component is F f =0andthe y component is F N mg =0 . The force of friction is then equal in magnitude to the applied force: f = F . The normal force is equal in magnitude to the force of gravity: F N = mg .A s F increases, f increases until f = µ s F N . Then the bureau starts to move. The minimum force that must be applied to start the bureau moving is . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Unformatted text preview: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ................................................. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .......... F N mg F f F = µ s F N = µ s mg = (0 . 45)(45 kg)(9 . 8 m / s 2 ) = 2 . × 10 2 N . (b) The equation for F is the same but the mass is now 45 kg − 17 kg = 28 kg. Thus F = µ s mg = (0 . 45)(28 kg)(9 . 8 m / s 2 ) = 1 . 2 × 10 2 N ....
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