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Homework 29

# Homework 29 - homework 29 BAUTISTA ALDO Due 4:00 am Version...

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homework 29 – BAUTISTA, ALDO – Due: Apr 12 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 2. 10 points. A sphere of weight W is held against a wall by a string being pulled at an angle θ as shown. θ R O F The torque equation about point O , the center of the sphere, leads to : ( f is the magnitude of the frictional force, and weight W = Mg ) 1. F sin θ cos θ = f 2. F = f correct 3. W = f 4. F cos 2 θ = f 5. F + W = f 6. F sin θ = f Explanation: For static Equilibrium, X i F i = 0 , X i ~ τ i = 0 . From X i ~ τ i = 0, we get τ c = τ cc about O , the center of the sphere. That is, F R = f R, where τ c = F R and τ cc = f R . Or F = f. Question 2 Part 2 of 2. 10 points. The vertical component of the force equa- tion leads to : 1. F cos θ + W = f 2. F sin θ = W 3. F sin θ = f + W 4. F sin θ = f 5. F sin θ + f = W correct Explanation: From X i ( F i ) y = 0, we get F sin θ + f = W. Question 3 Part 1 of 1. 10 points. In the figure the radius of the rod is 0 . 221 cm, the length of aluminum part is 1 . 3 m, the copper part is 2 . 79 m. For alu- minum, Young’s modulus is 70000000000 Pa, for copper, 110000000000 Pa. Aluminum Copper r L a L b Determine the elongation of the rod if it is under a tension of 7790 N Correct answer: 2 . 23056 cm (tolerance ± 1 %). Explanation: The cross-sectional area of the wire A = π (0 . 221 cm) 2 = 1 . 53439 × 10 - 5 m 2 m 2 , and the tension = 7790 N N throughout both

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homework 29 – BAUTISTA, ALDO – Due: Apr 12 2006, 4:00 am 2 pieces. Let us compute the elongation of each part separately: For aluminum: L al = L 0 F Y A = (1 . 3 m)(7790 N) (70000000000 Pa)(1 . 53439 × 10 - 5 m 2 ) = 0 . 00942862 m . Similarly, for the copper part: L cu = L 0 F Y A = (2 . 79 m)(7790 N) (110000000000 Pa)(1 . 53439 × 10 - 5 m 2 ) = 0 . 012877 m . So, the total elongation = 0 . 0223056 m = 2 . 23056 cm . Question 4 Part 1 of 1. 10 points. A stainless-steel orthodontic wire is applied to a tooth that is out of line by 40 . The wire has an unstretched length of 4 . 5 cm and a diameter of 0 . 22 mm. 40 40 If the wire is stretched 0 . 11 mm, find the magnitude of the force on the tooth. (Disre- gard the width of the tooth). Young’s modu- lus for stainless steel is 180000000000 Pa. Correct answer: 21 . 5023 N (tolerance ± 1 %). Explanation: Given : L 0 = 4 . 5 cm = 0 . 045 m , θ = 40 , r = 0 . 11 cm = 0 . 00011 m , L = 0 . 11 mm = 0 . 00011 m , and Y = 180000000000 Pa . 40 40 F F We need the tension required to stretch the wire by 0 . 11 mm: Y = F L 0 A L F = Y A L L 0 = Y ( π r 2 ) L L 0 The resultant force exerted on the tooth has components X F x = F cos θ - F cos θ = 0 and X F y = - F sin θ - F sin θ so F net = - 2 F sin θ = - 2 π Y r 2 ∆L sin θ L 0 = - 2 π (180000000000 Pa) (0 . 00011 m) 2 · (0 . 00011 m) sin 40 0 . 045 m = - 21 . 5023 N . which has a magnitude of 21 . 5023 N , and is directed down the page. Question 5 Part 1 of 1. 10 points. A small piece of wood is dropped from a height 4 . 1 m m above water. The wood has density density 790 kg / m 3 while the water’s density is 1000 kg / m 3 . How deep would the wooden piece sink into the water before floating back to the surface if there were no viscosity or any other kind of water resistance to its motion?
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Homework 29 - homework 29 BAUTISTA ALDO Due 4:00 am Version...

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