See Image Consider a simple model of an airplane which models the main bodyattached to the two wings as a cantilevered beam. The mass of the main body is4 times the mass of each of the wings. The equations of motion for the transversemotion of the airplane are given asm₁₁ =k(x2x1)--m2x2 = − k(x2 − x1) - k(x2 − x3)m33 =k(x2-x3)where the transverse bending stiffness k = 3EI/1³, m1= 3EI/1³, m1 = m3 = m and m2 = 4m. Theparameter values are as follows: mass m = 1000 kg, Young's modulus E = 6.9 × 109N/m², length of each wing = 2 m, and moment of inertial I = 5.2 × 10-4 m². Findand describe the mode shapes Confirm your description by plotting the position ofthe main body and wings in response to appropriate initial conditions for 1 second.(HINT: you may find the MATLAB command initial useful.)

modeExplanation:Step 1:Step 2:Step 3:Step 4:

Answered: Consider a simple model of an airplane…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Dynamic Lecture: The 2 kg BC ring can only move left and right on the frictionless rigid arm. Ring BC is connected to springs with spring constants k = 300 N / m and k ′ = 200 N / m at points AB and CD. The unstretched length of both springs is 600 mm. Since it is known that the springs are not tensioned and starts from rest, find the velocity of the ring BC at the moment when the external force F is applied 200 mm.
The super-container vessel has a total mass displacement of 23000 long tons. This vessel towed by a tugboat with α=20⁰ angle with horizontal. If a constant tension is applied as F=320 kN by the tugboat, please calculate the time to bring the vessel to a speed of 2 knot from the rest. At these low speeds, please neglect the hull resistance that created by the motion. Note: (1 long tons = 1016.05 kg, 1 knot=1.151 mi/hr= 0.5144 m/s)
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Find the minimum force P for impending motion of the refrigerator shown, and determine whether it will slip or tip, given:W = 180 lbs, L1 = 4.5 ft, L2 = 3.7 ft, L3 = 2.3 ft, L4 = 1.5 ft, μS = 0.5
Rod OA rotates counterclockwise at a constant angular rate θ˙ = 4 rad/s. The double collar B is pin-connected together such that one collar slides over the rotating rod and the other collar slides over the circular rod described by the equation r=(1.6cosθ)m. Both collars have a mass of 0.7 kg . Motion is in the horizontal plane. (Figure 1)
The spring constants are k1 = 140N / m, k2 = 240N / m and the unstretched lengths of the springs are 0.3 m. If the 6 kg ring is released from rest from point A, calculate its velocity when it reaches point B. According to the given datum line, the total potential energy (Ve) at point A is A = 1116.86 J and the total elastic potential energy (Ve) at point B is B = 370.8 J. Neglect the dimensions of the bracelet. (L1 = 0.90 m, L2 = 1.80 m, h1 = 1.20 m and h2 = 2.40 m)
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The double pulley shown in the figure is formed by two wheels that are coupled to each other. The complete pulley (formed by the two wheels) has a mass of 15 kg and a turning radius of 110mm. Block A has a mass of 40 kg. If a force of 2 kN is applied to the tied rope of the inner pulley wheel, determine the speed of block A after 3 seconds. At the beginning, the whole system was at rest. Disregard the mass of the string and consider that the moment of inertia (kg.m²) of the complete pulley is given by IP = mko²where m is the mass of the pulley and Ko is the radiusspinning
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