Find the tension in the cables and the reaction force point O
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Q: Find the tension in the cables and reaction force at point O.
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Find the tension in the cables and the reaction force point O
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- 140 lb 12 lb/in. 8 in. Problem 5-54 what are the values of the reaction forces O r1=150KNM r2=177KN O r1=177KN, r2=150KN O r1=177.9167N, r2=157.0833KN O r1=171.9KN, r2=150KNL1 z plane on to L2 L3 (o, 0.0) A tripod shown supports a load of P = 1000 N at point A. Take note that ball and socket joints are used to mount the triopd at points B, D, and C. Consider the following: L1 = 2 L2 = 4 L3 = 5 e = 45° a = 30° Point C coordinates: (2, -2, 3) Determine the following: Force of AB = N Force of AC = N Force of AD =Q3/ Design a helical compression spring made from stainless steel for the following operating conditions: Spring load when the valve is open = 600 N, Spring load when the valve is closed = 250 N, Maximum inside diameter of spring = 25 mm, Length of the spring when the valve is open = 40 mm, Length of the spring when the valve is closed = 50 mm, take into account the Wahl stress factor.
- 5 - note: ignore weight of the bar.A) ( - 6/7)T(CB) + A(y)=0 (A=reaction force at A, A(x), A(y), A(z) are components of force A)B) (2/7)T(CB)+A(x)=392 (A=reaction force at A, A(x), A(y), A(z) are components of force A)C) A(x)=2A(y) (A=reaction force at A, A(x), A(y), A(z) are components of force A)D) A(x)=5A(z) (A=reaction force at A,A(x), A(y), A(z) are components of force A)GIVEN: DISTANCES CO= 7m BO= 7m OD= 8m AD= 8m GIVEN: WEIGHT W1= 195 N W2= 295 N GIVEN: ANGLE X= 55 Degree Y= 48 Degree SOLVE: (a) Tension at CB in NEWTON (b) Reaction at point O in NEWTON PLS PLS SOLVE AND SHOW COMPLETE SOLUTION AND WRITE LEGIBLY. ALSO, DRAW THE FREE-BODY DIAGRAM. ROUND OFF IN 4 DECIMAL PLACES.Q1: In the below drawings, three unbalanced masses and related details are provided. Please answer the following questions. a) a balancing mass is to be located at rp = 50 mm, then what would be the balancing m1 m1 650 mm 200 mm m2 mass (m,) and its angular position (Op) (Please use the r2 m2 A graphical method) r3 m3 15 m3 b) If the system is not balanced what would be the reaction forces at A and B. (FA, FB=?) n = 150 rpm r; = 30 mm m, = 1 kg r2 = 20 mm 40 тm 13= m, =3 kg m; = 2 kg k = please choose an appropriate scaling factor yourself SHOT ON MI 6 MI DUAL CAMERA
- Solve for Problem 2, when RN1 = 0145As seen in the figure, the block on an inclined plane is lifting from 0 0 with two degrees increment in each step. (0, = 0, 0 =2, 02 = 4, ..) Until which degree can we increase the angle without any slipping between the block and the inclined plane? For condition of F > F, the block continues to hang on the inclined plane. Otherwise, it starts to slip. F, = W + sin8(rad), Fy = W+ cosa(rad) F = F, + u, u = 0.50, W = 100 NFind the magnitude of tension in string (or spring scale reading) and the magnitude of vertical FV and horizontal FH components of force applied on meter stick by the hinge. (Project this equation on Ox …this will give you the magnitude of FH force. Project this equation on Oy …this will give you the magnitude of FV force) (Draw a free body diagram) please data: Lruler = 1m mruler= 134 g dstring=80 cm d1= 35 cm d2= 82 cm M1=250g M2=150g θ = 38 degrees α= 62 degrees
- Crank connecting rod in Figure-1crank and connecting rods in the mechanismare the same length; AB = BC = L andBF=FC. One end to point B, the otherfrictionless with the tip horizontally.connected to movable point Dthe stiffness of the spring is k (N/m) when =90the spring is not elongated. spring mechanismforce, R1 and R2 forces and Mstatic under the influence of momentis in balance. The principle of virtual worksstatic balance of the mechanism by applyinggiving the theta angle corresponding to the positionget a relation andthe last of your student number from the table.appropriate values according to the numberCalculate the theta angle using M= -400Nm R1=300N R2=200N L=1,3m k=500N/mPROBLEM A 1) Find the equivalent spring, N/mm 2) Find the force, F k. Kl: =7 K5. =11 0000 K2 =8 X-10 mm k К3 -9 КА -10 К6 -12 ....... ........ .....Member AB (helical spring) and Member AC (axial rod) are carrying a weight at point A. Member AB has the following properties: R = 90 mm, n -4 turns, G- 70 GPa, d- 30 mm, and Talow = 105 MPa while Member AC has the following properties: L= 2.A- 500 mm?, E = 200 GPa, and a low- 30 MPa. Determine the maximum safe %3D value of W (in N). Use simplified formula for the helical spring. A 60°