104. We note that FN= mgin this situation, so fk= μkmg= (0.32) (220 N) = 70.4 N and fs,max= smg= (0.41) (220 N) = 90.2 N. (a) The person needs to push at least as hard as the static friction maximum if he hopes to start it moving. Denoting his force as P, this means a value of Pslightly larger than 90.2 N is sufficient. Rounding to two figures, we obtain
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