Qbb) Calculate support reactions in A and C, remember the sign of the force.Reaction force at point A:x-direction: XA=N, y-direction: YA=Reaction force at point C:x-direction: Xc=N, y-direction: Yc=N A wooden beam is loaded by a distributed load w = 8 N/mm as shown in Figure Q4(a).The cross-section of the beam is shown in Figure Q4(b). Assume that LAB = 2400 mm,LBC = 2600 mm, W=200 mm, Bo = 10 mm, and H= 240 mm, ho=30 mm.Units: use N, mm, MPa as consistent units for all calculations.Decimal points: any decimal points are fine and all the numerical questions have a 10%error tolerance.Steps: Fill the gaps in the "steps" to get some marks from the working-out process.Adaptive marking is still used.(а)(b)whotBoHLABLBCCross-sectionWFigure Q4: (a) A wooden beam is loaded by a distributed load, (b) cross-section of thebeam.

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b) Calculate support reactions in A and C, remember the sign of the force. Reaction force at point A: x-direction: XA= N, y-direction: YA= Reaction force at point C: x-direction: Xc= N, y-direction: Yc=

b) Calculate support reactions in A and C, remember the sign of the force. Reaction force at point A: x-direction: XA= N, y-direction: YA= Reaction force at point C: x-direction: Xc= N, y-direction: Yc=

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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H.W.pdf -> Q) A simplified model as shown in figure below of a planar robotic arm. You may assume that each segment has length 1.5 m, and the applied force is 250 N. 0, = 35" and O3 = 0 = 2r. Calculate the moment that the force, F. exerts about the point A. B 0:
What is the magnitude of the moment done by the force about point "O" in the following figure. Hint: Determine the moment of the force 'F' in Cartesian vector form then deduce the magnitude. 4 m F = [60i – 30j – 20k} N| 3 m -7m 4 m 6 m- 2 m Select one: a. 260 N.m b. 180 N.m c. 600 N.m d. 510 N.m
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Q) A simplified model as shown in figure below of a planar robotic arm. You may assume that each segment has length 1.5 m, and the applied force is 250 N. 01 = 35° and 02 = 03 = 20°. Calculate the moment that the force, F, exerts about the point A. F B D.
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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]
Problem 5 For the following figure, ● Find the reactions expressed as x and y force components Find the reactions expressed as a vector with magnitude and direction ● F = 650 N 100 45° F₂ = 750 N F = 900 N
The 80-lb door has its center of gravity at G. (Figure 1) Figure 1 of 1 18 in. 24 in. 24 in. 18 in.
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