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Unformatted text preview: A. FBD on weight/man mg mg T T B. 8 kg g T A T C B T C A x 6 kg g 10 kg g T A C 1. 4N 2. 2N 4N 3. 2N 30 1N T T m 1 g m 2 g m 2 m 1 M = 3.3kg S T H F G.H = 2.1g F G.S = 3.3g = -F N.S m = 2.1kg F N T 30 30 mgcos mgsin F G 10-01-2009 171.101, General Physics for Physical Science Majors I In the pulley problems on 9-29, we checked the forces on Figure B using a curved coordinate system. In Figure A, we did not use the curved coordinate system, but looked at the forces individually for their magnitude and direction. Puck on Ice: Find accelerations 1. There is one force on the puck in the right direction with a magnitude of 4N, the puck weighs 0.2kg. a = F/m = 4N/.2kg = 20m/s 2 There are two horizontal forces on the puck, a 4N force directed to the right and a 2N force directed to the left. a = F/m = (4N -2N)/.2kg = 10m/s 2 The two horizontal forces on the puck are now a 2N force directed to the left and a 1N force directed to the right at an angle 30 below the positive horizontal.directed to the left and a 1N force directed to the right at an angle 30 below the positive horizontal....
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This note was uploaded on 10/01/2009 for the course DOGEE 171.101 taught by Professor Henry during the Fall '09 term at Johns Hopkins.
- Fall '09