Three ropes A, B and C are tied together in one single knot K. (See figure.) 11 10 y (m) 6 5 4 3 2 6 00 8 7 K A C 1 B 0 0 1 2 3 4 5 6 x (m) 7 8 9 10 11 If the tension in rope A is 66.1 N, then what is the tension in rope B? (in N) ● A: 21.2 B: 26.4 ○ C: ●D: OE: F: 33.0 41.3 51.6 64.6 G: 80.7 OH: 100.9 Submit Answer Tries 0/12
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- 4.) A 20 kg block is suspended by 2 very thin ropes that meet at a 40° angle. What is the tension on the two ropes? AM ThN < Q口 Formula EQA D Untitled Notebook (2) - Untitled Notebook (2)- Satple Praloom s HI UNSM Ntbok2 Sanple Frirm 3 LnlT Sheet O O & C OBE 63 Atuoods Machine Block Suspended by a Strings 92=-a, (m. - mi). 9% = (mz +m.) Untitled Notebook (2) Tz = sine Block Sliding on a Romp Т. mg w/out friction: 9= gsine w/ friction:A 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 h1. A 100lb weight is suspended by three cables as shown in the figure. The ratios of the stiffness of the cables are: AP:BP:CP = 2k:k:3k, i.e. cable AP and CP are respectively 2x and 3x the stiffness of cable BP. Before the load is applied, all three cables are unloaded and the geometry is as shown. B 45° 60⁰ V 60" P you 45° C 461 Assuming the cables are very stiff and the displacements are therefore small, determine the forces in the three cables in terms of stiffness value k and the weight W(= 100). Assuming the springs are not stiff, and therefore the displacements are not necessarily small, develop a set of equations for the displacement of the mass and the forces in the cables.
- Please include plot/diagram in your solution. Strictly don't use chatgpt.tol F = Fe F= F + F COs # F F sin a. # tan Example (L):Find the two components of the force (100 X) if: 0 = 30, 120 270° as shown in figure. F = 100 N F = 100 N 6%330 6 =120° 0 =270° 0 = 300 0 60 F = 100 N ution:Need help with this one please thank you.
- Q1: Draw just the Free Body Diagram of the following figures. Figure (1) Figure (2) Figure (3) 30° T45° 45° 60° 30 B 1 kN Figure (4) Figure (5) 26 kN 250 N 30 C 13 12 30% 0.15 m B 0.4 m 2 m 4 m 0.2 m ||Required information Stairs leading to a balcony are shown below. |400 lb dimensions in inches 40° 144 600 lb 32 22 16 38 30 30 For the 400 lb and 600 lb forces, and if F= 155 lb and P= 0, determine the reaction at point Dand the force supported by bar BE. The reaction at point Dand the force supported by bar BE are as follows: Dy= 410 Ib TBE = - 701.798 IbA person is pulling a weight as shown below. His shoulder joint is not moving, but his elbow joint is free to rotate. The shoulder joint (S) and the pulley (P) are at the same height (horizontal line). This can be modeled as a diagram shown in the right-hand side. L3 OP S P 30ст L1 L2 = 30cm %D L1 50ст L3 10kg L2 E m = a H H m (a) L2, L3, and 0). (b) moment ME as a function of the elbow flexion angle 0. (c) Find the force directional angle a as a function of L1, L2, L3, and 0 (hint: find PB and BH as functions of L1, Calculate the moment at the elbow joint (E) produced by the weight. In other words, represent the elbow Find the elbow angle Omax where the magnitude of the moment about the elbow joint (||ME||l) is maximized.
- 1st pic is the problem 2nd pic is the solution question to answer: why was cos80 disregarded? can you explain the encircled part of the solution ( i was confused and need to understand how it happened)Find the loading in ED GD and GC. EI, IJ, JK, and KG are equal length. A wire is attached to point G with a tension of 100N.Q1