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problem11_10

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

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11.10: a) Since the wall is frictionless, the only vertical forces are the weights of the man and the ladder, and the normal force. For the vertical forces to balance, N, 900 N 740 N 160 m 1 2 = + = + = w w n and the maximum frictional forces is N 360 ) N 900 )( 40 . 0 ( 2 s = = n μ (see Figure 11.7(b)). b) Note that the ladder makes contact with the wall at a height of 4.0 m above the ground. Balancing torques about the point of contact with the ground, m, N 684 ) N 740 ))( 5 3 )( m 0 . 1 ( ) N 160 )( m 5 . 1 ( ) m 0 . 4 ( 1
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Unformatted text preview: = + = n so N, . 171 1 = n keeping extra figures. This horizontal force about must be balanced by the frictional force, which must then be 170 N to two figures. c) Setting the frictional force, and hence 1 n , equal to the maximum of 360 N and solving for the distance x along the ladder, ), N 740 )( 5 3 ( ) N 160 )( m 50 . 1 ( ) N 360 )( m . 4 ( x + = so x = 2.70 m, or 2.7 m to two figures....
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