Civil Engineering - Surveying - Discussion
Discussion Forum : Surveying - Section 6 (Q.No. 15)
15.
The sag of 50 m tape weighing 4 kg under 5 kg tension is roughly
Discussion:
24 comments Page 2 of 3.
Prity said:
7 years ago
Why you multiplying with P^2 in W? Please explain the formula.
Suhasish Munshi said:
7 years ago
The correct steps:
(W^2 * L)÷(24 *P^2).
W=weight of tape in kg/m,
L=length of tape in m,
P= pull applied or tension in kg,
So, W= 5kg/50m= 0.1kg/m.
L=50m and P=5kg.
(W^2 * L)÷(24 *P^2).
W=weight of tape in kg/m,
L=length of tape in m,
P= pull applied or tension in kg,
So, W= 5kg/50m= 0.1kg/m.
L=50m and P=5kg.
(2)
Munesh kumar meena said:
7 years ago
Tell me the right formula.
Rafi said:
8 years ago
Can you please explain @ Harish, Why you multiplied 10^4 with 24?
JAY PANCHAL said:
8 years ago
In formula why you multiply 24 with 10^4?
Neelam said:
8 years ago
In formula why you multiply 24 with 10^4?
Harish said:
8 years ago
(LW^2/24*10^4)*P^2.
((50*4^2)/(24*10^4))*5^2 = 0.0833.
Ans is right.
((50*4^2)/(24*10^4))*5^2 = 0.0833.
Ans is right.
Rpy said:
8 years ago
The actual formula is LW^2/24p^2.
Where l is the length of tape, W=wl=weightt of tape in kg and p is pull. But it comes 1.33.
Where l is the length of tape, W=wl=weightt of tape in kg and p is pull. But it comes 1.33.
Hasan said:
8 years ago
Please, someone explain it clearly, I am not getting the exact answer.
Raval chaitanya said:
8 years ago
@Adi
Formula = l3w2/24p2.
Formula = l3w2/24p2.
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