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308i7ZRoTV24_1210016284_jwk572

308i7ZRoTV24_1210016284_jwk572 - homework 09 KIM JI Due...

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homework 09 – KIM, JI – Due: Oct 31 2007, 4:00 am 1 Question 1, chap 14, sect 1. part 1 of 1 10 points A string provides a horizontal force which acts on a rectangular block of weight 490 N at top right-hand corner as shown in the figure below. 1 m 0 . 7 m F 490 N If the block slides with constant speed, find the tension in the string required to start to tip the block over. Correct answer: 171 . 5 N (tolerance ± 1 %). Explanation: Let : F = 171 . 5 N , W = 490 N , h = 1 m , and w = 0 . 7 m . h w F W Locate the origin at the bottom left corner of the block. In equilibrium (at constant velocity; i.e. , no acceleration), we have summationdisplay vector F = 0, so summationdisplay F x = F - μ N = 0 (1) summationdisplay F y = N - W = 0 , (2) The block is trying to tip over its lower right-hand corner (the fulcrum). The string tension T pulls to the right (clockwise) at a distance h from that corner. The weight W acts down (counter-clockwise) at a distance w 2 from the corner. From the torques summationdisplay vector τ = 0 , so summationdisplay vector τ = 1 2 W w - F h = 0 . (3) F h = W w 2 F = W w 2 h = (490 N) (0 . 7 m) 2 (1 m) = 171 . 5 N . Question 2, chap 14, sect 1. part 1 of 1 10 points A student of mass 54 kg wants to walk be- yond the edge of a cliff on a heavy beam of mass 300 kg and length 5 . 5 m. The beam lies on the horizontal surface of the clifftop (with- out any attachment), with one end sticking out beyond the cliff’s edge: d The student wants 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: 2 . 33051 m (tolerance ± 1 %). Explanation:
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homework 09 – KIM, JI – Due: Oct 31 2007, 4:00 am 2 Let : m = 54 kg , M = 300 kg , and = 5 . 5 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. Let the cliff’s edge be the origin of our co- ordinate system. The beam extends from X 1 = d - L < 0 (on the cliff) to X 2 = d > 0 (off the cliff), 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 neglects 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 cliff, so X overall c . m . < 0 and d < d max = M m + M L 2 = 300 kg 54 kg + 300 kg 5 . 5 m 2 = 2 . 33051 m . Question 3, chap 14, sect 2. part 1 of 2 10 points A 22 . 1 kg person climbs up a uniform 118 N ladder. The upper and lower ends of the lad- der rest on frictionless surfaces. The bottom of the ladder is fastened to the wall by a horizontal rope that can support a maximum tension of 81 . 6 N . The angle between the hor- izontal and the ladder is 64 .
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