*3-8. Two electrically charged pith balls, each having a mass of 0.2 g, are suspended from light threads of equal length. Determine the resultant horizontal force of repulsion, F, acting on each ball if the measured distance between them is r = 200 mm. 150 mm + A + + ++ + + F 50 mm r = 200 mm- + + 150 mm + + + + + B
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- 3-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 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 horizontal4. A bead of mass 0.1kg is bound to a linear track which connects the points (1,2,3) and (-1,2,4). The bead moves in response to a force field F = (xy²z, x,z²). Find the work done by the force field in moving the particle using .t₁ dr dt WD = - to F dt
- 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 total weight of the leg and the cast is W=250 N. The distance between the points A and B where the cables are attached to the leg is given as L=100 cm and the angle of the leg with the horizontal is γ=6°. 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 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. ResponseAnswer for the 1st part of the problem. magnitude of force A = 74 N direction of force A = 25o north of west second force = P direction of force P = α south of west arrow_forward The resultant of two forces is horizontal , it means that the y component of resultant = 0 Ry = ASin(25o) - PSin(α) 0 = (74×0.42) - PSin(α) PSin(α) = 31.08 NFor, the smallest force P , the value of Sinα must be maximum which is 1 when the value of αis 90 degree Thus, P×Sin(90o) = 31.08 N P = 31.08 N The magnitude of smallest force P = 31.08 N the direction of force P = 90 degree south of west = along - y axis= arrow pointing vertically downw2:28 Today Edit 2:26 PM Problem Number 2: A horizontal boom 11.5 M in length, AE is supported by guy wires from A to B,C and D. If the tensile load in AD = 360 KN, find the forces in AC and AB so that the resultant force on A will be horizontal. What is the resultant force of the 3-force system? в м Y+ D SM 2.5M 8M 3.OM 1.5M 1.5M 6M 6M A Z+ 10M 1.5M X+
- 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.FY Find the two components of the force (200 N) if a= 60 degrees, as :shown in the figure below F= 200 N α= 60 FX (FX= 100 N) AND (FY= 73.2N) O (FX= 173.2 N) AND (FY= 100N) O (FX= 100 N) AND (FY= 173.2N) OThe 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 distance between the points A and B where the cables are attached to the leg is given as L=100 cm and the angle of the leg with the horizontal is given as γ=8°. 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=50°. 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.
- Q3 (2-104) : The tower is held in place by three cables. If the force of each cable acting on the tower is shown, determine the magnitude and coordinate direction angles a,p,y of the resultant force. Units Used: kN := 1000N Given: x = 20m, a = 16m y = 15m, b = 18m F1 = 600N, c = 6m F2 = 400N, d = 4m F3 = 800N, e = 24mAn 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 WonsherThe 3D rigid truss structure is supported by three connections to the ground in the ry plane at D, E, and F. None of these connections can resist rotation. The connection at D can resist translation in all directions. The connection at F can only resist translation in the +z direction. The connection at E can resist translation in the +x direction and the +z direciton. A force of F = 250 ĝ kN is applied at B. A second force F2 = - 225 î + 140 ĵ - 60.0 k kN is applied at point C. Determine the reactions at connections D, E and F. Enter your answers in Cartesian components. F2 B F E Values for dimensions on the figure are given in the table below. Note the figure may not be to scale. TE Variable Value a 1.84 m 2.13 m 2.76 m The reaction at D is D - k kN. The reaction at E is E k kN. The reaction at F is F k kN. + + II