Discussion :: Applied Mechanics - Section 1 (Q.No.11)
|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)|
(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.
l = span.
|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.
=> H = 500.
Tmax = √ (V^2+ H^2).
Tmin = H.
Hence the answer is 500t at centre.
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