k s 13500 13500 Nm 3 22 points 14 submissions 50 identical springs are placed

# K s 13500 13500 nm 3 22 points 14 submissions 50

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k s = 13500 13500 N/m 3. 2/2 points | 1/4 submissions 50 identical springs are placed side-by-side (in parallel), and connected to a large massive block. The stiffness of the 50 -spring combination is 19000 N/m. What is the stiffness of one of the individual springs? k s = 380 380 N/m
4/25/08 10:26 AM Ch4 HW1 S2008 Page 2 of 3 Solution or Explanation N side-by-side (parallel) springs, each stretched the same amount s , exert a total force N*k s *s , so effectively they act like a single spring of stiffness N*k s . One spring has 1/ N the stiffness of the combination. 4. 2/2 points | 1/4 submissions Suppose you are going to measure Young's modulus for three rods by measuring their stretch when they are suspended vertically and weights are hung from them, as shown. Rod 1 is 2.9 meters long and cylindrical with radius 8 mm (1 millimeter is 0.001 m). Rod 2 is 2.6 meters long by 9 mm wide by 6 mm deep. Rod 3 is 2.1 meters long by 4 mm wide by 4 mm deep. The definition of Young's modulus, Y = (F/A)/( L/L) , includes the quantity A , the cross-sectional area. (a) What is the cross-sectional area of rod 1? A = 2.011e-4 0.000201 m 2 (b)) What is the cross-sectional area of rod 2? A = 5.4e-5 5.4e-05 m 2 (c)) What is the cross-sectional area of rod 3?
4/25/08 10:26 AM Ch4 HW1 S2008 Page 3 of 3 A = 1.6e-5 1.6e-05 m 2

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