hw30 - homework 30 Turner(58185 1 This print-out should...

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This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A student oF mass 53 kg wants to walk beyond the edge oF a cli± on a heavy beam oF mass 260 kg and length 9 . 8 m. The beam is not attached to the cli± in any way, it simply lays on the horizontal surFace oF the cli±top, with one end sticking out beyond the cli±’s edge: d The students want to position the beam so it sticks out as Far as possible beyond the edge, but he also wants to make sure he can walk to the beam’s end without Falling down. How Far From the edge oF the ledge can the beam extend? Correct answer: 4 . 07029 m. Explanation: Let : m = 53 kg , M = 260 kg , and L = 9 . 8 m . An unsecured body supported From below remains stable iF and only iF its center oF grav- ity projects down inside the area supporting the body. IF the body in turn supports some- thing else above it, then the overall center oF gravity (oF the body plus everything on top oF it) must project down inside the supporting area. ²or the problem in question, the stability oF the beam – with the student standing on it – requires that the center oF gravity oF the beam+student system must lay above the cli± itselF. IF this overall center oF gravity were to move beyond the cli±’s edge, the beam would tilt down and Fall o± the cli± – and the student would Fall down too. Let the cli±’s edge be the origin oF our co- ordinate system. The beam extends From X 1 = d - L < 0 (on the cli±) to X 2 = d > 0 (o± the cli±), so assuming the beam is uni- Form, its center oF mass is located at X beam c . m . = X 1 + X 2 2 = d - L 2 . When the student walks all the way to the beam’s end, his own center oF mass is located at X student c . m . d, where the approximation is neglecting the size oF the student compared to the beam’s length. The overall center oF mass oF the beam + student system is thereFore located at X overall c . m . = m m + M × X student c . m . + M m + M × X beam c . m . = d - M m + M × L 2 . The stability condition is that this overall center oF mass should stay on the cli±, thus X overall c . m . < 0 and hence d < d max = M m + M × L 2 = 4 . 07029 m . 002 10.0 points A shelF bracket is mounted on a vertical wall by a single screw. 2 s s 1 s homework 30 – Turner – (58185) 1
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hinojosa (jlh3938) – homework 30 – Turner – (58185) 2 Neglecting the weight of the bracket, and if s = 6 . 7 cm, s 1 = 5 . 7 cm and s 2 = 3 . 3 cm, Fnd the horizontal force component exerted on the bracket by the screw when a 38 . 7 N vertical force is applied as shown. Imagine that the bracket is slightly loose.
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This note was uploaded on 05/02/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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hw30 - homework 30 Turner(58185 1 This print-out should...

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