Figure Q1 shows a space truss ABCD and a horizontal load P = 30 kN applied on it at joint D, acting in the x-direction. The rectangular coordinate system (x, y, z) is used here, all linear dimensions are given in metres, and the truss is in static equilibrium.
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- An eye bolt is used to attach 3 cables to a steel plate. The tension in the three cables create F,=200 Ibf, F2=250 lbf, and F3=100 lbf with 0 = 30 degrees and p=27.1 degrees. If the eye bolt is in equilibrium, what is the x-component of the sum of other forces (reaction force) on the bolt? If you add up the three force vectors, the reaction force you are looking for will just be in the opposite direction to put the eye bolt in equilibrium. The x-direction is positive and to the right. Eye bolt rsteel plate Nut & WasherThe figure shows a mechanical model of the Russel fracture traction device and the leg. The leg is held in balance in the position indicated by the two weights attached to the two cables. The total weight of the leg and the cast is W=200 N. The horizontal distance between points A and B where the cables are attached to the leg is L=100 cm and the vertical distance is d=10 cm . Point C is the center of gravity of the cast and leg at three quarters of the L measured from point A ( 3L/4= 75 cm) . The angle that cable 2 makes with the horizontal is measured as β=40 ° . Accordingly, in order for the leg to remain in balance in the position shown; a) Find the tensile force T 1 in cable 1 . (Write your result in N ) b) Find the tensile force T 2 in cable 2 . (Write your result in N ) c) Find the angle α of cable 1 with the horizontalAn 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 lbf with 0 = 30 degrees and p=24.1 degrees. If the eye bolt is in equilibrium, what is the x-component of the sum of other forces on the bolt (force from the nut and plate on the bolt) ? If you add up 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 x-direction is positive to the right. For example, if you find the sum of forces 1, 2, and 3 are 100 Ibf to the right, then the other forces in the x-direction must be pointing to the left (-100 lbf) to put the eye bolt in equilibrium. Eye bolt 2steel 3 plate Nut t Wonsher
- Cable 1 a A Cable 2 В y If the picture does not load, click this link. The shaded shape is supported by Cable 1 and Cable 2 in the position shown (Drawing is NOT to scale). The shape has a uniform weight per unit area of 11kN/m2. Step 1: If A=1.9m, C=0.4m, x=1.2m, y=1.7m, z=3.5m, a = 67°, and B = 18°, find the value of B %3D needed to keep the shape in static equilibrium. Then solve for the tension in Cable 1 (T1) and the tension in Cable 2 (T2). Once you have evaluated your values for B, T1 and T2 it is recommended that you redraw your free body diagram with the correct dimensions, weights, and forces T1x, T1y, T2x, T2y and confirm that you do have your problem correctly in horizontal, vertical, and moment equilibrium. Step 2: Once you have confirmed that you free body diagram is in equilibrium, solve the following equation: Z= T1+T2 Write your final answer for Z (in kN) in the answer box below. Round your answer to two decimal places.4. The access door in Fig. P6-13 is held in the position by cable AB. If the magnitude of tension T in the cable is 50 pound force, find the rectangular components of T acting at A. 10 in. B 10 in. 12 in. 6 in. 20° FIGURE P6-133-In the scheme below, 4 objects are hanging from each of the rings and 5 ropes (blue) keep the entire scheme in balance. The known values are m1=30 kgm1=30 kgα=60ºα=60ºβ=30ºβ=30ºγ=70º a) Attach the force diagram on each of the rings with the graphical decomposition of the forces that are not exclusively on the Cartesian axes. b) Determine the tractions T1, T2, T3, T4 and T5. c) Determine the mass of weights 2, 3, 4 and 5. Tip: If you feel comfortable, use software to solve the systems of equations.
- The figure shows the Russel fracture traction device and a mechanical model of the leg. The leg is held in balance in the position indicated by the two weights attached to the two cables. The combined weight of the leg and the cast is W=210 N. The horizontal distance between points A and B where the cables are attached to the leg is L=100 cm and the vertical distance is d=6 cm. Point C is the center of gravity of the cast and leg at three quarters of the L measured from point A (3L/4= 75 cm). The angle that cable 2 makes with the horizontal is measured as β=33°. Accordingly, in order for the leg to remain in balance in the shown position; a) Find the tensile force T1 in cable 1. (Write your result in N) b) Find the tensile force T2 in cable 2. (Write your result in N) c) Find the angle α of cable 1 with the horizontal.Cartesian coordinat system; A(0,0,0) B(3,4,0) C(3,4,4) F=-20i +10 j+15k N is applied to point C Moment of force F about point A; M, Moment of force F about point B; M, Moment of force F about line AB; M AB Which following one is TRUE? ON MAB(-3i+4j)OM, O A) M = (3i – 4 j)®(-20i +10 j+15k) M, =(j)®(-20i +10 j +15k) O D Mg = 80i – 40j o M, = (3i+4 j+ 4k)® (-20i+10j+15k)ENGINEERING MECHANICS: STATICS (J.L.MERIAM) 2/4: The line of action of the 9.6-kN force F runs through the points A and B as shown in the figure. Determine the x and y scalar components of F. Ans. y, mm B (300, 100) X, mm F 9.6 kN A(-150,-200) 2/5: A cable stretched between the fixed sunnorts A and Bis under a tencien Tef 000 NL
- 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 F10The figure shows the Russel fracture traction device and a mechanical model of the leg. The leg is held in balance in the position indicated by the two weights attached to the two cables. The combined weight of the leg and cast is W=180 N. The horizontal distance between points A and B where the cables are attached to the leg is L=100 cm and the vertical distance is d=5 cm. Point C is the center of gravity of the cast and leg at three quarters of the L measured from point A (3L/4= 75 cm). The angle that cable 2 makes with the horizontal is measured as β=30°. Accordingly, in order for the leg to remain in balance in the shown position; a) Find the tensile force T1 in cable 1. (Write your result in N) Answerb) Find the tensile force T2 in cable 2. (Write your result in N) Answerc) Find the angle α of cable 1 with the horizontal. ResponseExample (2-13): The plate is subjected to the two forces at (A) and (B) as shown. Determine the magnitude of the resultant of these two forces and its direction FA= 8kN measured clockwise from the horizontal. Solution : Parallelogram Law : The parallelogram law of addition is shown in Fig. a. Trigonometry: Using law of cosines (Fig. b), we have: B F8 6 kN