A rigid system OA is subjected to two forces at point A as shown in the figure below. Given FAB= 0.912i +1.713j-2.286k (KN) and FAC = -1.078i +1.01j -1.346k (KN). 2m 1.5 m 1.6 m FAC B 0.8 m 1.6 m a) The magnitude of moment caused by force FAB about y-axis is b) The magnitude of moment caused by force FAC about OB is c) The resultant moment due to FAB and FAC about origin O is KN.m and its direction is KN.m and its direction is KN.m
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- Use the method of SECTIONS to find the true magnitude and direction in bar HG of the truss shown below in Fig 3.5 -500lb -100lb A 45° -200lb H -100lb 2 B -200lb -200lb MAGNITUDE AND DIRECTION OF SUPPORTS -500lb H -100lb 45° 2¹ C2D 2¹ Fig 3.5. -200lb -100lb LI 2 B 2₁ 2₁ D 2₁ Et = A rigid system DA is subjected to two forces at point A as shown in the figure below. Given FAB= -0.47i +0.47j - 1.884k (KN) and FAC -2.34i +1.872k (KN). D 4m Z A FAB 4m. 4m FAC B(4,5,0) a) The magnitude of moment caused by force FAB about y-axis is b) The magnitude of moment caused by force FAC about OB is c) The resultant moment due to FAB and FAC about origin O is KN.m and its direction is KN.m and its direction is KN.mFor the frame system given below, determine a) the forces at point B, b) internal normal force, shear force, and moment at point E of the two-member frame (also indicate negative/positive with respect to internal force sign convention). Show your work and box your results. 250 N/m ID -2 m 1.5 m E 300 N/m 4 m
- * O l 42%016:05 R The metal plate is held by two cables as shown in the following figure. If the force of each cable acting at A is FAB = 100 Ib and Fac = 200 Ib. Determine the magnitude of the resultant force of the two forces acting at A. -8 ft- 8 ft 6 ft B Y FAB FAC A(10;6;0) Select one: a. FR = 258.96 lb b. FR = 100 Ib c. FR = 200 Ib d. FR = 300 Ib Ouootion 7 II%A I. Untitled Section F1 F2 a C F3 A b a F4 D E B Consider the following values: - F5 a = 16 m; b = 11 m; c = 13 m; d = 8 m; F1 - 3 kN; F2 - 11 kN; F3 - 2 kN; F4 - 20 kN; F5 - 2 KN; a = 30°, 0 = 60° , B= 30° 11 What is the resultant moment of the five forces acting on the rod about point A? a) 88.8 kN.m b) 76.3 kN.m c) 19.6 kN.m d) 55.5 kN.m e) 59.1 kN.m f) 81.3 kN.m 2] What is the resultant moment of the five forces acting on the rod about point B? a) 88.8 kN.m b) 120.8 kN.m c) 31.6 kN.m d) 98.9 kN.m e) 41.1 kN.m f) 24.2 kN.m 31 What is the resultant moment of the five forces acting on the rod about point C? a) 125.7 kN.m b) 178.1 kN.m c) 254.6 kN.m d) 439.1 kN.m e) 144.1 kN.m ) 215.3 kN.m 4] What is the resultant moment of the five forces acting on the rod about point D? a) 77.8 kN.m b) 98.3 kN.m c) 100.6 kN.m d) 18.9 kN.m e) 97.1 kN.m H 60.1 kN.m 5] What is the moment of the force F2 about point E? c) 66.6 kN.m Activate W f) 151.3 kN.m a) 122.7 kN.m b) 108.1 kN.m d) 83.1 kN.m e)…Calculate the total moment of the system about point A, when P1 = 8 kN, P2 = 10 kN and 0 = 45°. %3D %3D 25 kN.m 1.5 m A –1.5 m- P2 →1.5 m- Select one: a. 23.21 kN.m (CCW) b. 48.21 kN.m (CW) c. 23.21 kN.m (CW) d. 48.21 kN.m (CCW)
- Three frictionless cylinders are supported in a cradle as r(A)=90 cm, r(B)=r(C)=60 cm, m(A)=200 kg, m(B)=m(C)=100 kga) Draw separate FBDs of each cylinder.b) Find all the forces acting on cylinder B. Be sure to unambiguously indicate the directions of each force.*c) For the case when there is no contact force between A and C, find the remaining reactions on cylinder A.A transmission tower (a truss) for supporting electric wires is shown. The load of wire at D is shown as force F1 at an angle of a and the load of wire at A is shown as force F2 at an angle of B. Using the values given in the table below: F1 (N) F: (N) a (deg) B (deg) w (m) h (m) Value 150 70 30 60 2 a) Find the reaction forces at supports Y and Z. b) Determine the forces supported by members DO, DM, SV, and UX and specify whether the member is in Tension or Compression. M Q B F1 F2 T W X hF 0.3 m mathalino.com 0.3 m Figure P-226 Given that F=1000Kn and P-1434. Determine the following: a. Components of forces F and P b. Moment at Point A, B, C, and D c. Moment at the Tail of Force F d. Moment at the Tail of Force P e. Moment at the Head of Force F f. Moment at the Head of Force P C. mathalino,cbm B.
- A folding tray mechanism is attached to a wall as shown. Find the internal forces and bending moment in the lower support arm at section s-s, located midway between points B and E, when a force of F = 250 N is applied at an angle of 0 = 32°. A B. 's E D k a- b – Values for dimensions on the figure are given in the following table. Note the figure may not be to scale. Variable Value a 17 cm b 26 cm 35 cm hi 25 cm h2 18 cm The internal axial load at section s-s is A = N. The internal shear load at section s-s is V = N. The internal bending moment at section s-s is M = N-m.Consider the given figure, Assumo L= 16 0 A, P 1200 ib and w 900 lb/t. Thon answer the following questions. Click here for the oquatoris providod w Ib/ft P lb |C L/2 ft L/2 ft Find the moment halfway between A and C. O -7560 lb.ft O-2.48x104 lb.ft O-1.84x10 lb.ft O -1.19x10 Ib.ft O -2.16x10 Ib.ft B.A folding tray mechanism is attached to a wall as shown. Find the internal forces and bending moment in the lower support arm at section s-s, located midway between points B and E, when a force of F = 270 N is applied at an angle of 0 = 28°. 2013 Michael Swanbom Ⓒ030 BY NC SA k s -Ś a b G с h₁ h₂ ка E S- B Values for dimensions on the figure are given in the following table. Note the figure may not be to scale. Variable Value 16 cm 19 cm 25 cm 23 cm 17 cm The internal axial load at section s-s is A = The internal shear load at section s-s is V = The internal bending moment at section s-s is M = N. N. C. N-m. 0