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oldhw 24 - alexander(jra2623 oldhomework 24 Turner(92510...

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alexander (jra2623) – oldhomework 24 – Turner – (92510) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points An 8 kg watermelon is placed at one end of a 7 . 4 m, 203 N scaffolding supported by two cables. One supporting cable is at the opposite end of the scaffolding, and the other is 0 . 55 m from the watermelon. How much tension is in the cable at the end of the scaffolding? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 87 . 0555 N. Explanation: Let : m = 8 kg W = 203 N , = 7 . 4 m , x = 0 . 55 m , and g = 9 . 8 m / s 2 . Let the fulcrum be at the point of attach- ment of the cable closest to the watermelon. mg W T 1 The watermelon mg acts down (CCW) at a distance x from the fulcrum. The weight W acts down (CCW) at a distance 2 - x from the fulcrum, and the rightmost tension T 1 acts up (CCW) at a distance - x from the fulcrum. Thus summationdisplay τ = summationdisplay τ CW - summationdisplay τ CCW = 0 W parenleftbigg 2 - x parenrightbigg - m g x - T 1 ( - x ) = 0 W ( - 2 x ) - 2 m g x = 2 T 1 ( - x ) T 1 = W ( - 2 x ) - 2 m g x 2 ( - x ) = (203 N) [7 . 4 m - 2 (0 . 55 m)] 2 (7 . 4 m - 0 . 55 m) - 2 (8 kg) (9 . 8 m / s 2 ) (0 . 55 m) 2 (7 . 4 m - 0 . 55 m) = 87 . 0555 N . 002 (part 2 of 2) 10.0 points How much tension is in the cable closest to the watermelon? Correct answer: 194 . 345 N. Explanation: Let the fulcrum be at the point of attach- ment of the rightmost cable. mg W T 2 The watermelon m g acts down (CCW) at a distance from the fulcrum. The weight W acts down (CCW) at a distance 2 from the fulcrum, and the tension T 2 acts up (CCW) at a distance - x from the fulcrum. Thus T 2 ( - x ) - m g ℓ - W parenleftbigg 2 parenrightbigg = 0 . Multiplying by 2, 2 T 2 ( - x ) = 2 m g ℓ + W T 2 = 2 m g ℓ + W 2 ( - x ) = 2 (8 kg) (9 . 8 m / s 2 ) (7 . 4 m) 2 (7 . 4 m - 0 . 55 m) + (203 N) (7 . 4 m) 2 (7 . 4 m - 0 . 55 m) = 194 . 345 N .
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alexander (jra2623) – oldhomework 24 – Turner – (92510) 2 003 (part 1 of 3) 10.0 points A 1560 N uniform boom is supported by a cable as shown. The boom is pivoted at the
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