2. The uniform cable weighing 15N/m is suspended from points A and B. The force in the cable at B is known to be 500 N. a. Find the tension at A, in N b. Find the horizontal distance between A and B, in m, c. Find the length of the cable, in m.

A Fa C P = 900N, a = 42°, B = 40°,0 = 14° A load P is supported by two cables as shown, considering the equilibrium at point-C, estimate the tension in the cables. (Answer upto 1 decimal place) N. Tension in cable AC:

The sign shown has a mass of m = 121 kg and is supported by two cables and a ball and socket at A. Determine the reaction force at A in N. Determine the tension in cable EC in N. Determine the tension in cable BD in N.

A cable supports a load of 50 kg/m uniformly distributed along the horizontal and is suspended from the two fixed points A and B located as shown. a. Calculate the cable tensions at A and B. b. Determine the length of the cable.

1. The figure shows a cable of weight w per unit length. It can be shown that the cable is described by y = ccosh in the indicated system of axes (the so-called catenary curve). The minimum tension in the cable occurs at y = c, and the maximum tension at A and B. These tensions are determined from the equations Tmin Tmax 15 m Determine the maximum- and minimum tensions in a cable over a span of 90m with a sag of 15 m, as shown in the figure. The weight per unit length is 45 Nm. Use the linear interpolation method and work to an accuracy of 1 decimal place. [Hints: (1) The unknown to be found is c. (2) c is a distance, and from the scale of the figure we expect it to be in the order of 100 m.] 90 m- VA wc wyb O

A chain makes angles Y° and Z° at its supports as shown. Determine the following: (a) sag at point C relative to point A; (b) tension support at A; (c) tension at support B and; (d) tension at C. The cable has a mass per length of 0.40 Kg/m.  Given: Y= 18.5° Z= 40° X= 14m W= 0.40 Kg/m

Example 2: A suspended cable is hung between two points, of the same level, and is -> T.cos e Subiected to a horizontal uniformly distributed load, attached to the cable by vertical hangars, as shown in the figure. What is the shape of the cable at equilibrium? and how does the tension vary along the cable? (Neglect self weight of the cable). L. - 32 - 0O000000000OCD

Required information NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. The 6-m long pole ABC is acted upon by a 545-N force as shown in the figure. The pole is held by'a ball-and-socket joint at A and by two cables BD and BE. m F 1.5 m> P The tension in cable BD is The tension in cable BE is The reaction at A is - (I E y B N. N. A N)i + 3 m 3 m 3 m XIS For a 1.5 m, determine the tension in each cable and the reaction at A. 1.5 m 3 m N)j + N)k.

A 100N load is supported by a strong bar. The tension in the cable is 200N. The value of the axial force in the strong bar in N Computer the angle A in degrees Compute the value of angle B if it is twice the value of angle A in degrees

Q4 -The homogeneous bar ABCD shown in the Figure is supported by a Cable that runs from A to B around the smooth peg at E, a vertical cable at C, and a smooth inclined surface at D. Determine the mass of the heaviest bar that can be supported if the stress in each cable is limited to 144 MPa. The area of the cable AB is 260 mm^2 and that of the cable at is 295mm^2 ? 30 3m 3m C 3m Figure P-114