problem11_96

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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11.96: The geometry of the 3-4-5 right triangle simplifies some of the intermediate algebra. Denote the forces on the ends of the ladders by R L F F and (left and right). The contact forces at the ground will be vertical, since the floor is assumed to be frictionless. a) Taking torques about the right end, R L L F F F N. 391 so ), m 90 . 0 )( N 360 ( ) m 40 . 3 )( N 480 ( ) m 00 . 5 ( = + = may be found in a similar manner, or from N. 449 N 840 = - = L R F F b) The tension in the rope may be found by finding the torque on each ladder, using the point A as the origin. The lever arm of the rope is 1.50 m. For the left ladder, N 1 . 322 so ), m 60 . 1 )( N 480 ( ) m 20 . 3 ( ) m 50 . 1 ( = - = T F T L (322 N to three figures). As a check, using the torques on the right ladder, ) m 90 . 0 )( N 360 ( ) m 80 . 1 ( ) m 50 . 1 ( - = R F T gives the same result. c) The horizontal component of the force at A must be equal to the tension found in part (b). The
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Unformatted text preview: vertical force must be equal in magnitude to the difference between the weight of each ladder and the force on the bottom of each ladder, 480 N-391 N = 449 N-360 N = 89 N. The magnitude of the force at A is then N. 334 ) N 89 ( ) N 1 . 322 ( 2 2 = + d) The easiest way to do this is to see that the added load will be distributed at the floor in such a way that N. 961 ) N 800 )( 64 . ( and N, 679 ) N 800 )( 36 . ( = + = ′ = + = ′ R R L L F F F F Using these forces in the form for the tension found in part (b) gives N, 53 . 936 ) m 50 . 1 ( ) m 90 . )( N 360 ( ) m 80 . 1 ( ) m 50 . 1 ( ) m 60 . 1 )( N 480 ( ) m 20 . 3 ( =-′ =-′ = R L F F T which is 937 N to three figures....
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