1. Calculate the support reactions, calculate internal forces on the designated section C, and plot the internal force diagrams of the H mes shown below. 4.2 武汉理 出版 4.4回 45圈 4.6開 3m 3m 題4.7图 題4.8图
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- 2:04 Laz ו 59% ב Statics assignment 1.pub soln.pdf Problem Number 3: The block shown in the figure below is acted on by its weight, W = 650 KN, a horizontal force H = 450 KN., and the pressure P exerted by the inclined plane. The resultant R of these forces is parallel to the incline. Determine the Forces P and R. Does the block move up or down the incline? Y-axis W = 650 KN P = ? X axis H = 450 KN H=450 KN 35° 40° P= ? W=650 Forces X- Component Y- Component W = 650 KN Wx = -650sin 35 Wy = -650 cos 35 -372.824 -532.449 H = 450 KN Hx = 450 cos 35 Hy = - 450 sin 35 368.618 - 258.109 P = ? Px = - P sin 5 Py = P cos 5 S Rx =R 0.000 PREVIOUS 3/3 NEXT IIA 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 hA load weighing P kN as with AB, AC and AD cables such as are carried. Points A and B are x-axis, C and D points are on the y-z plane. Specific to your name in the data table AB, AC and AD cable using ratings Calculate the forces. P (kn) = 440 L1 (m) = 11 L2 (m) = 10 L3 (m) = 16 %3D %3D %3D L4 (m) = 18 L5 (m) = 16 L6 (m) = 5 %3D %3D %3D
- Q1: In the below drawings, three unbalanced masses and related details are provided. Please answer the following questions. a) a balancing mass is to be located at rp = 50 mm, then what would be the balancing m1 m1 650 mm 200 mm m2 mass (m,) and its angular position (Op) (Please use the r2 m2 A graphical method) r3 m3 15 m3 b) If the system is not balanced what would be the reaction forces at A and B. (FA, FB=?) n = 150 rpm r; = 30 mm m, = 1 kg r2 = 20 mm 40 тm 13= m, =3 kg m; = 2 kg k = please choose an appropriate scaling factor yourself SHOT ON MI 6 MI DUAL CAMERAGiven values: m1 = 10kg, r1=75 cm, m3=20kg, r3=25cm, m4=5kg, had to find the value of r4=? I got r4=50cm m1 = 10 kg at r1 = 75cm m3 = 20 kg at r3 = 25cm m4 = 5 kg at r4 = 50cm 1. Compute the theoretical value of r3 using the following equation: r1m1 = r3m3 + r4m4 2. Compute the % error of r3 using the following equation: % error = |theoretical value - experimental value| / theoretical value x 100%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) = 5
- The linkage shown is used in a vehicle suspension system. Find the forces indicated below when a static force of F = 820 lb is applied to the tire at point C at an angle of 0 = 22° as shown. Assume the connection of the wheel to member AB is rigid. hs hy hi F BY NC SA W4. W, Ws 2013 Michael Swanbom Values for dimensions on the figure are given in the following table. Note the figure may not be to scale. Variable Value Variable Value hi 10 in Wi 8.8 in h2 15.6 in W2 5.6 in h3 18.1 in W3 10.8 in h4 10.9 in WĄ 7.4 in h5 4.9 in W5 9.3 in For all answers, take x as positive to the right and y as positive upward. At point B, member AB exerts a force of B = j lbs on member BHE. At point A, member AD exerts a force of A + j lbs on member AB. At point H, shock GH exerts a force of H = i + i lbs on member ВНЕ. At point E, the magnitude of the force exerted on the car body by member BHE is lbs. At point D, the magnitude of the force exerted on the car body by member AD is lbs. At point G, the…%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)…An eye bolt is used to attach 3 cables to a steel plate. The tension in the three cables create F1=200 lbf, F2=250 lbf, and F3=100 Ibf with 0 = 30 degrees and p=44 degrees. If the eye bolt is in equilibrium, what is the y-component of the sum of other forces on the bolt (force from the nut and plate on the bolt) ? If you add up %3D the three force vectors, the sum other force you are looking for will just be in the opposite direction to put the eye bolt in equilibrium. The y-direction is positive going up. For example, if you find the sum of forces 1, 2, and 3 are 100 Ibf going up, then the other forces in the y-direction must be pointing to the down (-100 lbf) to put the eye bolt in equilibrium. Eye bolt steel plate Nut & Washer esc DOO D00 F4 F3 AA %24 F10
- The cable carrying three 400-N loads has a sag at C of hC = 14 m. Calculate the following: 26.1 The magnitude of the tensile force in segment AB is Blank 1 N. 26.2 The magnitude of the tensile force in segment BC is Blank 2 N. 26.3 The magnitude of the tensile force in segment CD is Blank 3 N. 26.4 The magnitude of the tensile force in segment DE is Blank 4 N.Frames & Machines The truck shown is used to deliver food to aircraft. The elevated unit weighs 2000 Ib with center of gravity at G. 1- Derive the expression for the force in the hydraulic cylinder AB as a function of 0 (the angle that member FD makes with the horizontal). 2- Plot the force over the range 0° < 0 < 45° and discuss your observations. Note: | would suggest that you first solve the problem based on the position (angle) shown. You will need to determine the length of members FD and FB before solving the problem for arbitrary 0. Note that the location of the roller at D and hence the force reactions will change as 0 varies. 40" 17" 57" C 54" E 54" H AB1. - solve the nodal displacements and reaction forces: Al 20 kN/m 20 kN/m 20 kN/m 20 kN/m 5 kN ww Nom