problem11_62

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

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11.62: A and B are straightforward, the tensions being the weights suspended; N. 588 . 0 ) s m 80 . 9 )( kg 0360 . 0 kg 0240 . 0 ( N, 353 . 0 ) s m 80 . 9 )( kg 0360 . 0 ( 2 2 A = + = = = B T Τ To find , and D C T T a trick making use of the right angle where the strings join is available; use a coordinate system with axes parallel to the strings. Then, N, 353 . 0 53.1 cos N, 470 . 0 36.9 cos = ° = = ° = B D B C T T T T To find , E T take torques about the point where string F is attached; m, N 0.833 m) 500
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Unformatted text preview: )( s m 80 9 kg)( 120 ( ) m 200 . ( 53.1 sin ) m 800 . ( 36.9 sin ) m 000 . 1 ( 2 ⋅ = + ° + ° = . . . T T T C D E so F E T T N. 833 . = may be found similarly, or from the fact that F E T T + must be the total weight of the ornament. N. 931 . which from N, 76 . 1 ) s m 80 . 9 )( kg 180 . ( 2 = = F T...
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  • 0.200 M, 0.180 kg, 0.500 M, 0.120 kg, 0.0240 kg

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