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**Unformatted text preview: **m 00 . 4 ( m) (2.00 is = + B so the horizontal force at N. 232 m 31 . 4 ) m 00 . 2 ( N) 500 ( is = B Using the result of part (a), however, N 232 30.0 cos N) 268 ( = In fact, finding the horizontal force at B first simplifies the calculation of the tension slightly. c) N. 366 . 30 sin ) N 268 ( ) N 500 ( = -Equivalently, the result of part (b) could be used, taking torques about point , C to get the same result....

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