oldhw 24 - alexander (jra2623) – oldhomework 24 –...

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Unformatted text preview: 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 , ℓ = 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 ℓ 2- x from the fulcrum, and the rightmost tension T 1 acts up (CCW) at a distance ℓ- x from the fulcrum. Thus summationdisplay τ = summationdisplay τ CW- summationdisplay τ CCW = 0 W parenleftbigg ℓ 2- x parenrightbigg- mg x- T 1 ( ℓ- x ) = 0 W ( ℓ- 2 x )- 2 mg x = 2 T 1 ( ℓ- x ) T 1 = W ( ℓ- 2 x )- 2 mg x 2 ( ℓ- x ) = (203 N) [7 . 4 m- 2 (0 . 55 m)] 2 (7 . 4 m- . 55 m)- 2 (8 kg) (9 . 8 m / s 2 ) (0 . 55 m) 2 (7 . 4 m- . 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 mg acts down (CCW) at a distance ℓ from the fulcrum. The weight W acts down (CCW) at a distance ℓ 2 from the fulcrum, and the tension T 2 acts up (CCW) at a distance ℓ- x from the fulcrum. Thus T 2 ( ℓ- x )- mg ℓ-W parenleftbigg ℓ 2 parenrightbigg = 0 . Multiplying by 2, 2 T 2 ( ℓ- x ) = 2 mg ℓ + W ℓ T 2 = 2 mg ℓ + W ℓ 2 ( ℓ- x ) = 2 (8 kg) (9 . 8 m / s 2 ) (7 . 4 m) 2 (7 . 4 m- . 55 m) + (203 N) (7 . 4 m) 2 (7 . 4 m- . 55 m) = 194 . 345 N . alexander (jra2623) – oldhomework 24 – Turner – (92510) 2 003 (part 1 of 3) 10.0 points(part 1 of 3) 10....
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This note was uploaded on 10/20/2011 for the course PHY 302K taught by Professor Kaplunovsky during the Summer '08 term at University of Texas.

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oldhw 24 - alexander (jra2623) – oldhomework 24 –...

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