3. With angle alpha = 30 degrees, find the reaction forces at A and C (35 points) %3D S00 mm 300 N 200 min 300 N 200 mni B. at
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A: for solution refer below images.
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Q: C(a) For the figure below take M number of you and find the reactions ? * 3 m 3 m 3 m E C 12 M= 32…
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Q: The steel box wrench (E=29,000ksi, v=0.32) is loaded with a 40-lb horizontal force and a 25-lb…
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Q: Q1: For structure shown in Figure, Find the reactions of the hinge force at A, B, and C. 2 m 200 N/m…
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- A homogenous ladder 10m long and weighing 35 kg leans against a vertical wall. The angle between it and the floor is 60 degrees. The coefficient of friction of the ladder against the ground as well as the wall is 0.25.1. How far up the ladder measured from the ground can a 70-kg man climb before it starts to slide? a. 5.506 mb. 6.671 mc. 4.493 md. 3.806 m2. Compute the reaction at the wall when it starts to slide. a. 17.59 kgb. 25.27 kgc. 29.66 kgd. 28.67 kg3. Compute the reaction between the ground and the ladder when it starts to slide. a. 88.82 kgb. 126.72 kgc. 112.7 kgd. 101.88 kg12 - Three dimensional F1 = 48i + 73j - 14k F2 = 73i - 11j + 131k for vectors F1 + F2 =? do the collection.A) 666i - 593j - 43kB) 121i + 62j + 117kC) 174i - 174j + 328kD) 154i - 49j + 160kE) 106i - 17j + 112kTrue or false Statics mechanics allows us to represent distributed loads as a concentrated force located at the centroid of the distributed load.
- review engineering statics Take d1 = 600 mm, d2 = 800 mm, θθ = 45° and F = 500 N (a) Show that member DB is a two-force member? Draw a qualitatively correct extended free-body diagram of DB. (b) Draw an extended FBD of member ABC. (c) Write the three equilibrium equations for member ABC (namely Fnet, x = 0, Fnet,y = 0, and Mnet/P,z = 0). Make clear the choice of your point P for the moment equation. (d) Find the forces acting on member ABC at A and B (give your answer in component form or as a direction and magnitude.) Show your work. These are my answers FBD=2F, RAX=353.5N RAy= 353.5N but am not sure if a correct any helpreview engineering statics Take d1 = 600 mm, d2 = 800 mm, θθ = 45° and F = 500 N (a) Show that member DB is a two-force member? Draw a qualitatively correct extended free-body diagram of DB. (b) Draw an extended FBD of member ABC. (c) Write the three equilibrium equations for member ABC (namely Fnet, x = 0, Fnet,y = 0, and Mnet/P,z = 0). Make clear the choice of your point P for the moment equation. (d) Find the forces acting on member ABC at A and B (give your answer in component form or as a direction and magnitude.) Show your work.What would be the weight of one 3-kg mass on planet where the acceleration due to gravity is 10 m/s2? Select the correct response 3 kN 29.43 N 30N 13 KN An open tank contains 5.7 m of water covered with 2.8 m of kerosene. Find the pressure at the interface of the two liquids. Specific weight of kerosene is 8.0 kN/m^3. Select the correct response 22.400 55.917 44.000 33.240
- 01: A 105cm Rod is suspended on a wall with a 5N force pulling upwards, 110 degrees away from the horizontal axis and 5cm from a 50kgm hanging object. Find the location of the C.G. away from the wall. Neglect the Mass of the rod and consider it a rigid body. Assume SSLC. 02: Considering a CCW movement is Positive, what is the total value of the Negative Moments at the point with the 50kgm object? Please answer in four decimal places, in Nm. 03: What is the value of C.G. away from the wall? (in meters, two decimal places)A load weighing P kN aswith AB, AC and AD cables such asare carried. Points A and B are x-axis, Cand D points are on the y-z plane.Specific to your name in the data tableAB, AC and AD cable using ratingsCalculate the forces. P (kn) = 440 L1 (m) = 11 L2 (m) = 10 L3 (m) = 16 L4 (m) = 18 L5 (m) = 16 L6 (m) = 5The image below is of an arm in 90 degree elbow flexion holding an implement in the hand. SOLVE FOR THE TORQUE ABOUT THE ELBOW IN ORDER TO MAINTAIN EQUILIBRIUM. Here are the values. d1: 0.024m d2: 0.310m d3: 0.690m Fbarbell: 697N Farm: 74N
- The horizontal force due to water acting on the bolt (Fh): Fh=ρ×A1×V1×V1--V2 Fh=1000×π4×0.0752×1.254×1.254+11.286( Fh)=69.472 This is the only part I did not understand how to get. Can you please explain more? I arrived with this equation, -P1A1 - P2A2 (cancelled since atmos. Pres.) +Fh = rho A1v1 (v2-v1) Why did P1A1 disappear in your solution? How did you come up with (v1 - - v2) ? What is the control volume? Why did you not do a summation of forces is equal to change in momentum but instead go immediately to getting the horizontal force acting on the bolt?The dependence of the tensions t (units: N) in a structure on the external forces f (units: N) follows the linear system Ct=f, where both t and f are 5 by 1 column vectors. The coefficient matrix C has been determined to be [ 11 14 20 -5 -19 ] -5 7 -3 5 -6 4 -5 13 -2 -10 0 0 0 12 -10 0 0 0 0 14 By how many N would the external force f2 need to be decreased (while all other external forces stay the same) in order to decrease the tension t3 by 94 N?Power Engineering - ABEN 3411 Illustrate and discuss the principle of operation for diesel and gasoline engine.