Please use attached figure, Thank you Problem 2) As a result of loading, the rectangle with a = 25 mm and 6 =18 mm deforms into a paral-lelogram in which sides QA and BC shorten by 0.003 mm and rotate by 550µɛ rad counterclockwise.Sides AB and QC elongate by 0.004 mm and rotate by 1000 µɛ rad clockwise. Find the principalstrains and principal axes at point Q. Sketch a body showing the orientation and deformation of eachprincipal strain. IA- a

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Answered: Problem 2) As a result of loading, the…

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter8: Centroids And Distributed Loads

Section: Chapter Questions

Problem 8.73P

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STATICS OF RIGID BODIES RATE WILL BE GIVEN! NO LONG EXPLANATION NEEDED! SHOW THE COMPLETE SOLUTIONS. ANSWER IN 4 DECIMAL PLACES.
Compare the magnitudes of the equilibrant vectors measured from the experiment with those obtained from the graphical and component methods. Example: A: 200 g 60° above +x axis B: 300 g 45° above -x axis C: 400 g 30°below -x-axis A, A cos a = 1.96 N x cos 60° = 0.98 N B₂B cos b = 2.94 N x cos 135º = -2.08 N C, C cos g = 3.92 N x cos 210°= -3.39 N R₂-A, + B + C₂ = -4.49 N A, A sin a = 1.96 N x sin 60° = 1.70 N By B sin b = 2.94 N x sin 135º = 2.08 N C, C sin g = 3.92 N x sin 210° = -1.96 N Ry= Ay+ By + Cy = 1.82 N Questions: I: (a) A: 200 g along +x axis B: 100 g 45° above -x axis A₂ = A cos a = B₂ =B cos b = R₂-A₂+ B₁₂= A₂ = A sin a = B, = B sin b = R₂ = A + B₂ = R=(R₂. R₂):_ Quadrant R = √ (R₂²+R₂²) = Direction:q=tan [R, /R.] (c) A: 100 g along -y axis B: 200 g along -x axis A = A cos a = B = B cos b = R₂-A₂+ B₂ = A = A sin a = B, B sin b = R=A, +B₂ = R=(R₁, R₂): Quadrant R = √ (R₂²+R₂²) = Direction:q tan¹ [R, /R]
PROBLEM 2: The crank shown is free to rotate at B about the x-axis only. Given the following dimensions: x1 = 53 cm x2 = 25 cm Y₁ = 74 cm 2₁ = 29 cm A 100-N force is applied at A with the following spatial angles: From positive x-axis: a = 64° From positive y-axis: B = 119° It is known that the angle from positive z-axis is obtuse, 90° ≤ y ≤ 180⁰ Required: 1. Determine the moment about B due to the applied force at A. INCLUDE (-) NEGATIVE SIGN IF NECESSARY, FOR THIS ITEM ONLY. MA = k N-m use 2 decimal places 2. Determine the component of the force along line AB. 1+ 8= j+ PAB = Nuse 2 decimal places 3. Find the angle between the force applied and its component along line AB. use 3 decimal places
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.
PLEASE ANSWER NUMBER 3 MECHANICS OF DEFORMABLE BODIES: PLEASE GIVE DETAILED SOLUTIONS AND CORRECT ANSWERS. I WILL REPORT TO BARTLEBY THOSE TUTORS WHO WILL GIVE INCORRECT ANSWERS.(USE 4 DECIMAL PLACES FOR ALL YOUR COMPUTATIONS)
The 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. CG 1. How much is the torque done by object 1 around point P? A. 55,2 N*m B. 7,36 N*m C. 62,5 N*m D. Object 1 doesn't exert a torque. 2. How much is the torque done by the force of gravity of the beam around point P? A. 20,8 N*m N-m ך177 .B C. 156 N*m D. 313 N*m 3. If you needed to cancel the nomal forces of the two objects, where you should place object 2? The axis of rotation is point P. A. 3,30 m from point B. 9,18 m from point P C. 5,69 m from point P D. 3,62 m from point P
Answer for the 1st part of the problem. magnitude of force A = 74 N direction of force A = 25o north of west second force = P direction of force P = α south of west arrow_forward The resultant of two forces is horizontal , it means that the y component of resultant = 0 Ry = ASin(25o) - PSin(α) 0 = (74×0.42) - PSin(α) PSin(α) = 31.08 NFor, the smallest force P , the value of Sinα must be maximum which is 1 when the value of αis 90 degree Thus, P×Sin(90o) = 31.08 N P = 31.08 N The magnitude of smallest force P = 31.08 N the direction of force P = 90 degree south of west = along - y axis= arrow pointing vertically downw
Need help please Include all units and steps
Review Learning Goal: To use the vector cross product to calculate the moment produced by a force, or forces, about a specified point on a member. Part A - Moment due to a force specified by magnitude and endpoints The moment of a force F about the moment axis passing through O and perpendicular to the plane containing O and F can be expressed using the vector cross product, Mo =r x F. In a properly constructed Cartesian coordinate system, the vector cross product can be calculated using a matrix determinant: As shown, a member is fixed at the origin, point O, and has an applied force F, the tension in the rope, applied at the free end, point B. (Figure 1) The force has magnitude F = 180 N and is directed as shown. The dimensions are ¤1 = 0.350 m, x2 = 1.90 m, y1 = 2.30 m, and z1 = 1.20 m. What is the moment about the origin due to the applied force F? i j M =rx F =|rz k Express the individual components of the Cartesian vector to three significant figures, separated by commas. ry F F,…
Kindly solve the vector addition. Show your complete solution including the magnitude and direction. Also illustrate your work. You may use law of sin and cosines
Determine angle about P does the cone travel.

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