-FF200 NO A) (0.8) FINM -(0.8) Fj NmB) -(0.8) Fj Nm + (0.6) F k Nm1000 mm600 mm800 mm200 N-C) -(0.8) Fj Nm-P800 mmO D) -(0.8) F k NmWhat is the couple moment of forces F and -F ?O E) -(0.6) Fi Nm

To find moment we need to do cross product.. As the alignments of the axis such that cross gives…

F 200 N O A) (0.8) FiNm -(0.8) Fj Nm B O B) -(0.8) Fj Nm + (0.6) F k Nm 1000 mm 600 mm 800 mm 200 N- O ) (0.8) Fj Nm 800 mm O D) -(0.8) Fk Nm What is the couple moment of forces F and -F ? O E) -(0.6) F i Nm

F 200 N O A) (0.8) FiNm -(0.8) Fj Nm B O B) -(0.8) Fj Nm + (0.6) F k Nm 1000 mm 600 mm 800 mm 200 N- O ) (0.8) Fj Nm 800 mm O D) -(0.8) Fk Nm What is the couple moment of forces F and -F ? O E) -(0.6) F i Nm

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

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter1: Introduction To Statics

Section: Chapter Questions

Problem 1.12P: A differential equation encountered in the vibration of beams is d4ydx4=2D where x = distance...

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A differential equation encountered in the vibration of beams is d4ydx4=2D where x = distance measured along the beam; [x]=[L] y = displacement of the beam; [y]=[L] = circular frequency of vibration; []=[T1] = mass of the beam per unit length; []=[ML1] D = bending rigidity of beam; [D]=[FL2] Show that the equation is dimensionally homogeneous. Note that [F]=[MLT2] see Eq. (1.21:).
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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]