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As in Sec. 24.2, compute F using the trapezoidal rule and Simpson's
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Chapter 24 Solutions
EBK NUMERICAL METHODS FOR ENGINEERS
- 1 11:P1 STATIC MONTHL... Find the resultant force of the following? B. 2 m F = 600 N 3.5 m 0.5 m Fc= 450 N 1.5 m 1m 0.5 m 1.5 m y X IIarrow_forwardExercises 1. 4. 7. 10. j15 18 HE 5. 40 10 10 15 10 35 30 2. 5. 8. 11. 15 20 40 り 15 30 30 20 45 35 25 25 40 3. 6. 9. 12. 25 15 10 10 20 20 10 15arrow_forwardIn the crank-slider shown, OA = 5 cm, AB = 15 cm and the offset e angle a at which slider B reaches the extreme right position. Write your answer in degrees but do not write the unit. 7 cm. Find the value of 7777 x. Answer: DELLarrow_forward
- The figure below shows a steel structure made up of two pre-assembled segments that are hoisted by a mobile crane using a bespoke lifting frame. Each cable must withstand loads of FAB = FAC = FAD = FAE = 4000 kN. You must provide an accurate answer in scientific standard form with one decimal place only. In cartesian vector and in scalar form, compute the rAB, rAC, TAD & rAE length. 14 m 6m 6 m в 3 m 3 m Xarrow_forwardhe maghitude and direction of the resultant according to the chosen scale. = 246 N, 255° Test yourself 3: 1. Two forces of 60 N and 8O N respectively act simultaneously at a point. F = 60 N Sketch and determine their resultant by using the following methods: a) tail-to-head 60° b) tail-to-tail (122 N 25 2. Two forces, F, and F, act at a point as indicated. a) Draw a sketch diagram to show the resultant force when angle 6 = 120. The angle is now decreased from 120° to 20°. F = 80 N F (The magnitudes of the forces remain the same.) b) Draw a new sketch diagram to show the new resultant. Compare the sizes of the resultants and write your conclusion. Forces in equilibrium A stationary object might have no forces acting on it. Forces on the object can also be balanced so there is a zero resultant Inarrow_forward3 The equation of the curve, the slope of which is 4-2x, and which passes through the point (2,6) is_ -- a. y=2x - x² + 2 b. y=4x - x2 c. y=4x - x2 + 2 d. y=x - x2+1 e. y=x - x2 4 x2y" + 2xy` =1 is a DE in second order special case. The integrating factor is a. X b. x 2 C. x 2 d. x -1arrow_forward
- Adobe Acrobat Reader DC (32-bit) Window Help Quiz 1-163 - S3.pdf x 1 /2 138% Quiz 1- MENG163 Student Name Grade ID Number Section S3 Q.1) For the figure shown, compute the tensions in cables AC and BC if a = 60° and P = 500 N. Use sine and cosine law only. Draw whatever need. B 25 45 pe here to search IAarrow_forwardQuestion 2 Which of the following is the rectangular representation of the resultant? Not yet answered Marked out of 1.00 120 N P Flag question SO N 150 N 30° 35°/ 40° Select one: O a. -122. + 86.04j O b. -20.55i +250.22j O C. 54.38i -224j O d. 191.98i +-360jarrow_forward6 m! 1.0 m T3 2 m T2 3 m 2 m 2 m The magnitude of one of the forces is T1 = 100 kN. Find the rectangular representation of T1. * The magnitude of one of the forces is T¡ = 100kN. Find the rec tan gular representation of T1. O T = (15.62 i + 31.23 j – 93.70 k) N O T, = (15.62 i + 31.23 j + 93.70 k) N O T1 = (-15.62 i + 31.23 j + 93.70 k) N O T, = (-15.62 i – 31.23 j + 93.70 k) Narrow_forward
- S00 Ib In the fig. shown, compute the ff: (16-18) the resultant using cosine law (force polygon) 60 R = 35 (19-20) the angle of the R measured 500 lb cW from the x- axis.arrow_forwardThe following diagram corresponds to questions 1 to 3. A beam rests on two sharp edges as shown on the image. The beam has a length of 7,50 m and a mass of 4,25 kg. Object 1 has a mass of 1,50 kg; object 2 has a mass of 2,35 kg. Point P is 0,50 m form the center of the beam. d 2 Pl CG 1. How much is the torque done by object 1 around point P? A. 55,2 N*m B. 7,36 N*m С. 62,5 N*m D. Object 1 doesn't exert a torque.arrow_forwardIn the figure, let the mass of the block be 8.50 kg and the angle 0 = 30.0⁰ Step 1 m Create the free-body diagram. Use the checkboxes to select the correct vectors as instructed above. Move the vectors to the correct starting points and orient them in the correct direction as instructed. FN mg Frictionless T Draw free-body diagram here :+y +x Key to Force Labels FN = Normal force mg Gravitational force ƒ= Frictional force T = Tensionarrow_forward
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