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Unformatted text preview: 11)a) F smax = F g sin 45 o F smax = mg sin 45 o F smax = (22 kg)(9.8 N/kg) sin 45 F smax = 152.5 N The maximum force applied upward, parallel to the ramp is 1.5 x 10 2 N. b) F n = (F smax ) / (u s ) F n = (152.5 N) / (0.78) F n = 195.5 N F a = F n F g cos45 o F n = 195.5 N (22 kg)(9.8 m/s 2 ) cos45 o F a = 195.5 N 152.5 N F a = 43 N The smallest force that can be applied to the top of the box is 43 N....
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This note was uploaded on 12/13/2009 for the course ENGG eg 404 taught by Professor Weisenhaupt during the Spring '09 term at University of the Bío-Bío.
- Spring '09