# Civil Engineering - Applied Mechanics - Discussion

### Discussion :: Applied Mechanics - Section 1 (Q.No.11)

11.

A cable loaded with 0.5 tonne per horizontal metre span is stretched between supports in the same horizontal line 400 m apart. If central dip is 20 m, the minimum tension in the cable, will be

 [A]. 200 tonnes at the centre [B]. 500 tonnes at the centre [C]. 200 tonnes at the right support [D]. 200 tonnes at the left support.

Explanation:

No answer description available for this question.

 Divyesh Patel said: (Mar 7, 2015) I want to know how? Please reply.

 Jay Bermudo said: (May 12, 2015) As you can see, there are tonne in the question so the answer must be the greatest tonne at the center ever in the history of applied mechanics.

 Sujit said: (Jul 17, 2015) Lets l=400m, yc=20m w= 0.5t. Vertical reaction v=wl/2. Horizontal thrust H= WXLXL/8YC. Tension T = Root over of VXV+HXH.

 Salman said: (Aug 5, 2015) The question is about minimum tension not maximum. If it would ask about maximum then 500 is correct.

 Abhay said: (Jan 25, 2016) L = 400 m, δ = 20 m, W = 0.5 t. Vertical reaction at each end = WL/2 = 0.5*400/2 = 100 t. Horizontal thrust H = WL^2/8δ = 0.5*400^2/(8*20) = 500 t. T = Sqrt(V^2+H^2) = sqrt(100^2+500^2) = 509.9 t.

 Amol Bhalerao said: (Jun 21, 2018) (Wl^2)/(8*span). (0.5*400^2)/(8*20) = 500.

 Shobha said: (Oct 18, 2018) Minimum pull in a suspended cable with supports at two ends is equal to HORIZONTAL THRUST.

 Tejas Rahane said: (Apr 19, 2019) Central dip h=(wl^2)/8P. h= dip. l = span. P=pull. w=load.

 Zeba Khan said: (May 14, 2019) What is the maximum pull in the cable?

 Pushkar Jha said: (Jul 3, 2019) Central dip h = (wl^2)/8P.

 Mala said: (Jul 19, 2019) Support reactions V = (0.5x 400)/2 = 100 t. For horizontal trust, equate Mc =0 , moment at centre. 100*200-20*H-0.5*200*100= 0, => H = 500. Tmax = √ (V^2+ H^2). Tmin = H. At center. Hence the answer is 500t at centre.