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Based on the figure below, find an equation in which you can determine x as a function of only z, y, a, and b
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- IA y 50 mm O y = 48.3 mm, I = 15.08(106) mm* O y = 48.3 mm, I = 17.51(106) mm¹ O y = 116.7 mm, I = 15.08(106) mm² O y = 48.3 mm, I = 13.49(106) mm¹ 120 mm -C 50 mm -x 20 mm 20 mm 125 mm 20 mm Determine the location y of the centroid C of the beam's cross-sectional area. Then compute the moment of inertia of the area about the XX axis.y a b b b -b --b-2-9. Resolve F, into componcnts along the u and v axcs and determine the magnitudes of these components. 2-10. Resolve F2 into components along the u and v axes and determine the magnitudes of these components. F = 250 N F = 150 N 30° 30 105°
- •2-17. Resolve F, into components along the u and v axes and determine the magnitudes of these components. F = 250 N F, = 150 N 30 30° 105°- 25. (y² cos x-3x²y – 2x) dx + (2y sin x-r³+ In y) dy-0, y(0) = eIf W=3.000 inches, Y=1.750 inches, and Z=0.875 inch, find X. TE- X www A Į A „Y. W 1 B ļ B ·Z www.t C
- The Force F, acts on the bracket within the octant shown. If the magnitude of the x and z components of F are Fx= -300N and Fz= 800N respectively and beta= 45 degrees. What is the Magnitude of the force F acting on the bracket in (kN) What is the y component of the force F acting on the bracket and what is the coordinate direction of angle y in degrees.x² + xy + y² = 1В F=105 N 4 m y 2 m A 2 m 4 m Compute the following quantities. Note: Drawing is not to scale. N
- Two forces act on the hook shown in Fig. 2-32a. Specify the magnitude of F2 and its coordinate direction angles of F, that the resultant force FR acts along the positive y axis and has a magnitude of 800 N. 120° y 60° 45 ARA. W .. * itar F, = 300 N shɔ'nu ig ^-*3a i+ ian vec I as LLe V uat E. r *2. 2_9.Resolve the 300 N into two components at the u and v axis 300N 20° 70° n.Question 2: Resolve F2 into components along the u and v axes and determine the magnitudes of these components. F = 250 N F= 150 N 30° 30 105"