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- For the ladder in Prob. 6.17, find the internal force system acting on section 2, assuming that x < a/2.Solve the preceding problem for the following data: b = 8.0 in., k = 16 lb/in., a = 45°, and P = 10 lb..17 A mountain-bike rider going uphill applies torque T = Fd(F = l5lb, d = 4 in.) to the end of the handlebars ABCD by pulling on the handlebar extenders DE. Consider the right half of the handlebar assembly only (assume the bars are fixed at the fork at A). Segments AB and CD are prismatic with lengths L, = 2 in.andL3 = 8.5 in, and with outer diameters and thicknesses d01 = 1.25 in. 101 = 0.125 in. and d03 = O.87in.,i03 = 0.ll5in, respectively as shown. Segment BC’ of length L, = 1.2 in. however. is tapered, and outer diameter and thickness vary linearly between dimensions at B and C. Consider torsion effects only. Assume G = 4000 ksi is constant. Derive an integral expression for the angle of twist of half of the handlebar tube when it is subjected to torque T = Fd acting at the end. Evaluate ‘b1-, for the given numerical1ues.
- As shown in the figure, the member is anchored at A and section AB lies in the x–y plane. The dimensions are x1 = 1.6 m, y1 = 1.8 m, and z1 = 1.6 m. The force applied at point C is F=[−180 i+60 j+165 k] N. I need the answers in three components: (I,j,k)N*M There are only two questions I hope anyone can help me to solve it, and show me how to do it. 1) Using the position vector from A to C, calculate the moment about segment AB due to force F. 2) Using the position vector from B to C, calculate the moment about segment AB due to force F. Thank youDetermine the internal forces at point J, with alpha = 0°the horizontal and vertical components of several forces are: (a) Ph = -200lb and Pv = 100lb : (b) Fh = 300lb and Fv = -200lb . determine each force and its angle with respect to the hozirontal (x-axis)
- AB, BC, CD, DE are each 2m long and the plane distance from A-D is 4m. Members are at right angles to each other. a) determine the vertical displacement at E b) determine the rotation at E E=200KN/mm^2 G= 80KN/mm^2 Iabce = constant = 25 x 10^-6 m^4 Jabce = constant = 50 x 10^-6 m^4Determine the algebraic sum of the moments of forces relative to a point in F = 10.0 N; m =9.0 Nm, q =8.0 N/mThe piston of a reciprocating engine exerts a firce of 180 kN on crosshead when the crank is 35 degree (angle beta) past TDC. If the stroke of the piston is 800 mm and the length of the connecting rod is 1.65 m, find, in kN. a) the guide force FG and , b) the firce on the connecting rod Rc .hint: radius AC is 1/2 of the stroke and Angle "phi" may be solved using sine law. Draw FBD
- A sign for a pizza restaurant hangs from a 2.60-m long rod extended out from a building. A cable, attached to the building, is attached to the rod at a point that is 2.10 m from the hidge. (See the figure.)The mass of the rod is 5.90 kg. The mass of the sign is 9.40 kg. The angle between the building and the rod is 74.0 degrees. The angle between the cable and the horizontal is 31.0 degrees.What is the torque on the rod due to the weight of the rod alone? [Take the axis to be at the hinge.]It is common to hang objects on doorknobs and over-the-door hooks. There is a limit to the amount of weight that a door can hold because of the forces exerted on the hinges. A door of height h = 2.5 m and width h/2 has a mass of M = 36 kg. The mass is distributed uniformly, so the center of mass is located at the geometric center of the door. One hinge is located a distance h/4 from the top of the door. The second hinge is a distance h/4 from the bottom of the door. Refer to (a) in the figure. The door’s weight is supported entirely by the two hinges and each hinge supports half of the weight. In other words, the vertical force exerted by each hinge is exactly one half of the total weight, including any additional load. For this problem, take the positive y-direction to be directly upward and the positive x-direction pointing from the hinge side of the door to the knob side. 1.) calculate the force, with its sign in Newtons that the upper hinge exerts on the door in the x axis.…The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5 mm. Suppose that a= 640 mm