(8) Replace the two forces and a coupleacting on the beam by an equivalentresultant force Ŕ and resultant moment MAat point A.Ans: R=1.644 +1.159 kN,M₁ = 2.22 kNm CCWA2 kN0.5 m1.5 m70°0.15 m0.15 m500 N·m-x1.2 kN

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

Answered: (8) Replace the two forces and a couple…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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(1) Determine the two components Fa and Fb of the 4 kN force along the oblique a- and b-axes. Also determine the two projections Pa and Pb of the 4 kN force onto both a- and b-axes. Ans: Fa 1.935 kN, Fb = 2.39 kN, Pa = 3.63 kN, Pb = 3.76 kN 15° 40° F= 4 kN x
(3) The gusset plate is subjected to two forces F₁ = 0.8 KN and F2 = 0.9 kN at point A as shown. If they are replaced with Fa and Fb and the resultant forces from the two pairs are identical, determine the magnitudes of Fa and Fb. Ans: Fa=1.199 kN, Fb = 0.433 kN F₂ 30°-40° x
(5) A force F = 50 N is exerted on the automobile parking-brake lever at the position x = 250 mm. Replace the force by an equivalent force-couple system at the pivot point O. Ans: F-17.10 +47.0ĵN, Mo 17.29 N·m CCW OA = OB F 20° 15° 100 mm 10°
Part A F₂= 260 N 5 13 12 A O R = 421 N, 0 = 112.3° OR = 251 N, 0 = 85.5° O R = 421 N, 0 = 67.7° OR = 251 N, 0 = 94.5° 30 F₁ = 300 N Determine the magnitude and direction of the resultant force. Give the direction as an angle measured counterclockwise from the positive x-axis.
A door is swung open by applying a 60 N force as shown, 2.0 m from the hinges. 2.0 m 100° 60 N Find the magnitude and direction of the torque through the door hinges.
A sheet of water of uniform thickness (h = 0.03 m) flows from the device shown in the figure below. The water enters vertically through the inlet pipe and exits horizontally with a speed that varies linearly from 0 to 11 m/s along the 0.2-m length of the slit. Determine the y component of anchoring force necessary to hold this device stationary. FAY = 0 m/s- i 0.2m 0.03m N 11m/s
45. Find the components for each vector by resolving the vector into its horizontal and vertical components. a) 3-

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(8) Replace the two forces and a couple acting on the beam by an equivalent resultant force Ŕ and resultant moment MA at point A. Ans: R=1.644 +1.159 kN, M₁ = 2.22 kNm CCW A 2 kN 0.5 m 1.5 m 70° 0.15 m 0.15 m 500 N·m -x 1.2 kN