NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. A 48-in. boom is held by a ball-and-socket joint at C and by two cables BF and DAE; cable DAE passes around a frictionless pulley at A. 48 in. B W D Y 20 in. E lb) i+( 16 in. C F 30 in. for W = 450 lb, determine the tension in each cable and the reaction at C. The tension in cable DAE is lb. lb. The tension in cable BF is The reaction at Cis-([ 20 in. lb)j + ( lb) k.

An ideal spring, with a pointer attached to its end, hangs next to a scale. With a 100 N weight attached, the pointer indicates "40" on the scale as shown. Using a 200N weight instead results in "60" on the scale. Using an unknown weight X results in "30" on the scale. The weight X is: Hint: Draw the force - Displacement diagram. 100 N O 1.20 N O 2.30 N O 3.40 N O 4.50 N O 5.70 N

5. show all steps in detail, units and formulas please.

2. Three blocks are connected by a weightless cord and rest on an inclined plane (Fig. 2(a)). The following set of simultaneous equations is derived from the free- body diagrams shown in Fig. 2(b): 100a +T + (0)R = 519.72 50a -T +R=216.55 25a + (0)T – R =108.27 Without interchanging the rows and columns, solve for the acceleration a and the tensions T and R, in the two ropes by using Gauss-Seidel iteration method. Stop all calculations after three iterations. 25 kg 50 kg 100 kg 45 1346 48 × 0,375 - 129.93 346.48 173 24 x 0.375 - 64.97 173,24 346,48 692.96 100 x 9.8 = 980 50 9,8 490 25 x 98- 245 a, acceleration 96769 T692.96 x 0.25 = 173.24 173.24

2.22 Calculate the centre distances for a v-belt drive given pulley pitch diameters of 180 mm (motor) and 400 mm (driven) when the belt is (a) an SPÁ 2 360; and (b) an SPB 3 800. [716 mm; 1 441 mm] Calculate the angle of contact on the smaller pulley for both cases. Determine the tension ratio if the groove ang- les are 38° and u is 0,25. (162°; 171

Please write it on paper.

Need help with these  Q2. Identify any zero force members. ( there can be more than one )  Member AB Member AJ Member BCBC Member BH Member BJ Member CD Member CG Member CH Member DE Member DF Member DG Member EF Member FG Member GH Member GJ q3.  You are asked to find the forces in members GHGH and CGCG using the Method of Sections. Which members should you make a cut through?   Member AB Member AJ Member BC Member BH Member BJ Member CD Member CG Member CH Member DE Member DF Member DG Member EF Member FG Member GH Member GJ Q4. Having made a cut through the members in the previous part, you next want to solve for Member GH. Which of the following equilibrium equations is the most efficient? (i.e. which one will allow you to solve for the required member in one equation?) a. ΣFx=0.  b. ΣMC=0. c. ΣMH=0. d. ΣFy=0 e ΣMD=0

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