please box out or highlight all the answers A bent tube is attached to a wall with brackets as shown. . Aforce of F = 980 lb is applied to the end of the tube withdirection indicated by the dimensions in the figure.a.) Determine the force vector F in Cartesian components.→→b.) Resolve the force vector F into vector componentsparallel and perpendicular to the position vector rDA.Express each of these vectors in Cartesian components.2013 Michael Swanbomcc 10BY NC SAgxBAאZFKскаяbValues for dimensions on the figure are given in the tablebelow. Note the figure may not be to scale. Be sure to alignyour cartesian unit vectors with the coordinate axes shown inthe figure.VariableValuea8 in12 inс15 in36 inh23 ing28 ina. F =b. FDA == (+k) lbk) lbFIDA =2 +k) lb

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

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter9: Moments And Products Of Inertia Of Areas

Section: Chapter Questions

Problem 9.46P

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Figure (a) shows the cross section of a column that uses a structural shape known as W867 (wide-flange beam, nominally 8 in. deep, weighing 67 lb/ft). The American Institute of Steel Construction Structural Steel Handbook lists the following cross-sectional properties: A=19.7in.2,Ix=272in.4, and Iy=88.6in.4. Determine the dimensions of the rectangle in Fig. (b) that has the same Ix and Iy as a W867 section.
Find the internal force systems acting on sections 1 and 2.
A bent tube is attached to a wall with brackets as shown. . A force of F = 970 lb is applied to the end of the tube with direction indicated by the dimensions in the figure. a.) Determine the force vector F in Cartesian components. b.) Resolve the force vector F into vector components parallel and perpendicular to the position vector I DA- Express each of these vectors in Cartesian components. NOTE: The parallel component of the force is the same as the force along that direction. The perpendicular component of the force can be found from the relationship: Fparallel + Fperpendicular = Ftotal 2013 Michael Swanbom BY NC SA d A FK values for dimensions on the figure are given in the table below. Note the figure may not be to scale. Be sure to align your cartesian unit vectors with the coordinate axes shown in the figure. d 78 in h 57 in g 60 in Variable value 13 in 36 in 45 in F|DA = i. k) lb and b. FL DA =
Find the following for the vectors v = 10i+11j-2k and u=8i + 15j. a. v. u, |v|, |u| b. the cosine of the angle between v and u c. the scalar component of u in the direction of v d. the vector proj, u a. v. u= (Type an exact answer, using radicals as needed.) |v|= (Type an exact answer, using radicals as needed.) |u| = [ (Type an exact answer, using radicals as needed.) b. The cosine of the angle between v and u is (Type an exact answer, using radicals as needed.) c. The scalar component of u in the direction of v is (Type an exact answer, using radicals as needed.) d. proj, u = (Type your answer in terms of i, j, and k. Type an exact answer, using radicals as needed.)
5.) If the resultant force (single equivalent force) of the given system is known to pass through point B, what must the magnitude of the applied moment couple M be? Report a positive answer if M must be counterclockwise and a negative answer if M must be clockwise. [Answer: M = -8600 N m] 650 N 30° M -3 m 300 N -5 m- 500 N Ka -2 m- B
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 02 = 03 = 20°. Calculate the moment that the force, F, exerts about the point A. F 02 A
Express each result of Parts 1-5 in the form of a Cartesian vector. Part 4 requires the specification of 2 cartesian vectors: a force and a moment vector.
Problem Set No.5 - AutoRecovered - Compatibility Mode A Share P Comments AutoSave ff Review View Help O Search Design Layout References Mailings File Home Insert 4. Find the resultant of the three forces acting on the square plate. 70 lb 10 lb 0.75 ft 0.75 ft 60 lb 6 ft 6 ft + 160% 11:11 PM E English (United States) Recovered O G 4) ENG 4 Page 3 of 4 132 words 13/01/2021 O Type here to search
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Problem 6: Inputs: a = 600 mm b= 400 mm c = 200 mm F = 100 N The line of action of the force vector passes through points B and C. B N TB (a) Use the cross product to compute the moment about point A. For this part, you must use B in F your vector formulation. That is, compute MA = B x F. Express your answer as a vector and convert "mm" to "m". a b (b) Now use a different position vector, r, in the cross product. This time use rc to to compute the moment about point A. For this part, you must use rc in your vector formulation. Compute MA = rcx F. Express your answer as a vector and convert "mm" to "m". c) Do your answers agree? Yes or no. E By the way, how do you express the force as a vector when you are only given magnitude? Don't forget what we did on the last homework assignment: Position vector → Unit vector → Force vector. y
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