# phyhw15 - valdez (vv689) Homework 15 florin (58140) 1 This...

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Unformatted text preview: valdez (vv689) Homework 15 florin (58140) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A student of mass 69 kg wants to walk beyond the edge of a cliff on a heavy beam of mass 260 kg and length 8 . 2 m. The beam is not attached to the cliff in any way, it simply lies on the horizontal surface of the clifftop, with one end sticking out beyond the cliffs edge: d How far from the edge of the ledge can the beam extend if it sticks out as far as possible beyond the edge, but the student can walk to the beams end without falling down? Correct answer: 3 . 24012 m. Explanation: Let : m = 69 kg , M = 260 kg , and L = 8 . 2 m . The stability of the beam (with the student standing on it) requires that the center of gravity of the beam + student system must lie above the cliff itself. If this overall center of gravity were to move beyond the cliffs edge, the beam would tilt down and fall off the cliff. Let the cliffs edge be the origin of our co- ordinate system. The beam extends from X 1 = d- L &amp;lt; 0 (on the cliff) to X 2 = d &amp;gt; (off the cliff), so assuming the beam is uni- form, its center of mass is located at X beam CM = X 1 + X 2 2 = d- L 2 . When the student walks all the way to the beams end, his own center of mass is located at X student CM d, where the approximation neglects the size of the student compared to the beams length. The overall center of mass of the beam + student system is therefore located at X overall CM = m m + M X student CM + M m + M X beam CM = md m + M + M m + M parenleftbigg d- L 2 parenrightbigg = d- M m + M L 2 &amp;lt; for overall stability, so d &amp;lt; d max = M m + M L 2 = 260 kg 69 kg + 260 kg 8 . 2 m 2 = 3 . 24012 m . 002 (part 1 of 2) 10.0 points Consider the rectangular block of mass 39 kg height 0 . 72 m, length 0 . 71 m. A force F is applied horizontally at the upper edge. l m h F What minimum force is required to start to tip the block? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 188 . 446 N. Explanation: Let : = 0 . 71 m , h = 0 . 72 m , m = 39 kg , and g = 9 . 8 m / s 2 . valdez (vv689) Homework 15 florin (58140) 2 summationdisplay vector F = 0 and summationdisplay vector = 0 . Let the right lower corner of the block be denoted as A ; then summationdisplay A = hF- 2 mg = 0 F = mg 2 h = (39 kg) (9 . 8 m / s 2 ) . 71 m 2 (0 . 72 m) = 188 . 446 N . 003 (part 2 of 2) 10.0 points What minimum coefficient of static friction is required for the block to tip with the applica- tion of a force of this magnitude? Correct answer: 0 . 493056. Explanation: summationdisplay F y = N A- mg = 0 N A = mg and summationdisplay F x = F- f = F- N A = 0 = F N A = F mg = 188 . 446 N (39 kg) (9 . 8 m / s 2 ) = . 493056 ....
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## This note was uploaded on 09/26/2010 for the course PHY 58230 taught by Professor Svhets during the Spring '10 term at University of Texas at Austin.

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phyhw15 - valdez (vv689) Homework 15 florin (58140) 1 This...

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