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Unformatted text preview: Pickering, Tracey – Homework 26 – Due: Apr 5 2006, noon – Inst: Drummond 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Given: g = 9 . 8 m / s 2 . A cylindrical stone column of diameter 2 R = 1 . 26 m and height H = 3 . 34 m is trans- ported in standing position by a dolly. a side view When the dolly accelerates or decelerates slowly enough, the column stands upright, but when the dolly’s acceleration magnitude exceed a critical value a c , the column top- ples over. (For a > + a c the column topples backward; for a <- a c the column toppes forward.) Calculate the magnitude of the critical ac- celeration a c of the dolly. Correct answer: 3 . 69701 m / s 2 . Explanation: In the non-inertial frame of the accelerating dolly, the column is subject to the horizontal inertial force ~ F in =- m~g. Together, the gravity and the inertial force combine into the apparent weight force ~ W app = m ( ~g- ~a ) in the direction θ = arctan a g from the vertical. From the torque point of view, this appar- ent weight force applies at the center of mass of the column. The column is stable in the vertical position when the line of this force goes through the column’s base CM W app but when this line misses the base, the column topples over CM W app Pickering, Tracey – Homework 26 – Due: Apr 5 2006, noon – Inst: Drummond 2 For the critical acceleration a c , the line goes through the edge of the base, hence the direc- tion of the apparent weight force must deviate from the vertical by the angle θ c = arctan R h cm = arctan R H/ 2 = arctan 2 R H = 20 . 6688 ◦ . Consequently, the critical acceleration of the dolly is a c = g × tan θ c = g × 2 R H = 3 . 69701 m / s 2 . 002 (part 1 of 1) 10 points Two objects, of masses 24 kg and 18 kg, are hung from the ends of a stick that is 70 cm long and has marks every 10 cm, as shown. 24 kg 18 kg A B C D E F G 10 20 30 40 50 60 If the mass of the stick is negligible, at which of the points indicated should a cord be attached if the stick is to remain horizontal when suspended from the cord? 1. A 2. E 3. D 4. F 5. C correct 6. G 7. B Explanation: Let : ‘ = 70 cm , m 1 = 24 kg , and m 2 = 18 kg . For static equilibrium, τ net = 0. Denote x the distance from the left end point of the stick to the point where the cord is attached. m 1 g x- m 2 g [ ‘- x ] = τ = 0 ( m 1- m 2 ) x = m 2 ‘ x = m 2 ‘ m 1- m 2 = (18 kg) (70 cm) 24 kg- 18 kg = 30 cm . Therefore the point should be point C . 003 (part 1 of 1) 10 points Two weights attached to a uniform beam of mass 23 kg are supported in a horizontal po- sition by a pin and cable as shown in the figure....
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This note was uploaded on 06/23/2008 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
- Fall '08