Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 3 (Q.No. 36)
36.
A trolley wire weighs 1 kg per metre length. The ends of the wire are attached to two poles 20 m apart. If the horizontal tension is 1000 kg, the central dip of the cable is
Discussion:
8 comments Page 1 of 1.
Konnyei said:
4 years ago
1x100= 8 x h x 1000/20 x 20 (1m=100cm),
h= 5.
h= 5.
Dar said:
5 years ago
10000 = 1000*20^2/8*d = 5.
Rahul said:
6 years ago
For a cable supported at both ends it buckles to resemble shape of catenary but if the dip is small then it can be approximated to a parabola which boils down to equation [(w*l*l)/(8*d)]=T where d is central dip , T is tension generated in cable and w is the UDL of cable and l is length between supports, using this formula we get answer as 5 cm.
Rajinder singh said:
7 years ago
1 * 20 * 20/4 = 8 * h * 1000/20 * 20,
= 5.
= 5.
(1)
Hiwatari said:
7 years ago
Shouldn't the answer be 4 cm?
Formula : T = (W*L*L)/8d.
Where T stands for tension in the wire, W is weight of the wire, L is the length of the wire and d is the dip.
Formula : T = (W*L*L)/8d.
Where T stands for tension in the wire, W is weight of the wire, L is the length of the wire and d is the dip.
Himanahu said:
8 years ago
Please give an explanation in detail.
Prashant said:
8 years ago
Please describe the answer.
Kishor said:
8 years ago
p=8*h*t/l^2
p=w*l^2/4
Solve anh find h is 5.
p=w*l^2/4
Solve anh find h is 5.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers