Solve for the following: representation of force DE 1. 2. Z-component of force DF IMPLN ww 3. magnitude of the resultant of forces DE, DE, and DG
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- A 10m light rod AB is horizontally placed. One end of this rod is pin supported at point A, and the other end B is connected with a rope BD, which is inclined 30° with respect to x axis. A 10kg box hangs at point C of this rod. Which of the following solutions for the tension force in rope BD is CORRECT? (g=9.8m/s²) O O O A TBD = 117.6 N TBD = 196 N TBD = 169.7 N Don't Know TBD = 98.0 N 6m ► 4m 10kg B D 30°Horizontal cantilever shart with disks is loaded by 2 concentrated forces. 1 \F1 F2 Given: 1-26cm; l,-32cm; n-27cm; r-10cm; F1= 100N; F,= -25ON. Find: absolute value of reactive moment (in N-m) applied at fixed point. 2.The bar AC is supported by a joint at A and a pin B that is free to slide in the same direction (alpha) of the bar. A bar is connected at C and a force is applied in the theta direction. Consider that L1 = 2.7 m, L2 = 1.7 m, alpha = 13 °, theta = 25 °, and P = 68 kN. Determine the magnitude of the force (kN) on pin B.
- As shown, an L-shaped bar is supported by a pin at joint A. The bar's dimensions are aaa = 620 mmmm and bbb = 400 mmmm , and the bar is subjected to a force with magnitude FFF = 5.05 kNkN at joint B. Ignoring the bar's weight, find the actual orientation of the applied force. What is the value of the angle θθtheta?The system shown is subjected to three axial forces P₁ = 250 kN, P₂ = 400 KN and P3 = 275 kN. What is the total change in length due to the loading? (E = 70 GPa). Answers: (a) 0.062 mm. (b) 0.045 mm. (c) 0.073 mm. (d) 0.014 mm. 1 m P1 T 0.6 m A P2 B 1.2 m P3 C D Rigid plate Rigid plate Rigid plate 0.25 m² 2 0.35 m² 0.5 m²Shown below is a leg in a cast and supported by two cables. The combined weight of the leg and cast is 150N acting at the center of mass as shown. The angle of the ankle cable is 60 degrees from the horizontal and the angle of the knee cable is 45 degrees from the horizontal, as shown. Find the cable tensions Tk and TA. Cable 45° 25 cm 15 cm 60° Cast W = 150 N
- F = 200 N F3=300N M. = 100 Nm %3D 0.4 m F, = 300N 0.5 m 0.6 m The pipe system shown in the figure is supported by fixed joint at points O. Calculate the followings: (a) Replace the force-couple system acting on the pipe by a resultant moment and a single force acting on Point O. (b) Replace the force-couple system acting on the pipe by a wrench resultant and determine the coordinates of the point in the xz plane through which the line of action of the wrench passes.A horizontal rod of length L= 9.64 m is attached to an axel in such a way that the axel is located at its left end. Three forces are acting on the rod: F1= 11.6N at the location 2.59 m from the axel, F2 = 15.7 N at the location 5.02 m from the axel, and F3 =15.0 N at the laction 2.31 m from the axel. F1 (pointing up and to the right) makes an angel of 47.7 degrees with the rod, F2 (pointing up and to the left) makes an angel of 45.1 with the rod, and F3 points directly downwards. Calculate the net torque that these three forces exert on the rod. Also, indicate if the torque is clockwise it counter-clockwise.4 m 5m 4 m B 4 m 80 kN 60 kN find the forces in members (BC,BE and DE) ? T or C p 10:36
- A box is suspended by 3 cables as shown. If the weight of the box, W = 83 lb, and cable C has a unit vector, Uc= -0.500 i +-0.766 j+ 0.407 k and cable A has a unit vector Ua= -0.857 i+0.515 j. What is the tension in cable B? BPROBLEMS 2-7.2. In Fig. P-2-7.2, a boom AC is supported by a ball-and-socket joint at C and by the cables BE and AD. If the force multiplier of a force F acting from B to E is F A to D is P = 10 lb/ft and that of a force P acting from = 20 lb/ft, find the component of each force along AC. %3D %3D FAC -37.5 lb; PAC =D100 lb Ans. %3D %3D problem ble indAn 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