Civil Engineering - Highway Engineering - Discussion
Discussion Forum : Highway Engineering - Section 5 (Q.No. 36)
36.
In a braking test, a vehicle travelling at 36 km ph was stopped at a braking distance of 8.0 m. The average value of the vehicle's skid resistance (friction coefficient) is
Discussion:
16 comments Page 1 of 2.
Binny bana said:
5 years ago
A is correct.
Breaking distance =v^2/2gf.
8=10^2/2*9.81*f,
so = 100/156.7.
=.64 answer.
Breaking distance =v^2/2gf.
8=10^2/2*9.81*f,
so = 100/156.7.
=.64 answer.
(6)
Sudhanshu said:
4 years ago
@Shubham.
It's given breaking distance, not SSD.
Braking dist:V^2/254f.
It's given breaking distance, not SSD.
Braking dist:V^2/254f.
(5)
Salim said:
5 years ago
f = v2/254L = 36*36/254*8 = 0.64 but the value of friction factor lies in between 0.35 to 0.40. So, the correct Ans will be (D) none of the above.
(4)
Nabin Adhikari said:
1 decade ago
F = V^2/(254S) = 0.64.
So correct answer might be A.
So correct answer might be A.
(1)
ARUN KUMAR said:
9 years ago
f' = V*2 / 2gL
= (36/3.6)**2/ 2 * 9.81 * 8.0 v = V /3.6
= 36/3.6
= 10m/sec
= 0.64m (say)
= (36/3.6)**2/ 2 * 9.81 * 8.0 v = V /3.6
= 36/3.6
= 10m/sec
= 0.64m (say)
(1)
Gubendhiran said:
8 years ago
SSD=lag distance + braking distance.
SSD=Vt +[(V^2)÷(2g(f+-g))],
Breaking distance=(v^2)÷(254f+-g),
We don't have gradients so leave it.
8=(36^2)÷(254f),
f=0.63779=0.64,
So option A is correct.
SSD=Vt +[(V^2)÷(2g(f+-g))],
Breaking distance=(v^2)÷(254f+-g),
We don't have gradients so leave it.
8=(36^2)÷(254f),
f=0.63779=0.64,
So option A is correct.
(1)
Suman said:
7 years ago
It seems to be option D is correct as 'f' in the breaking distance formula signify longitudinal friction, which varies from 0.4 to 0.35 depending on the speed, while the lateral friction 'f' comes to design the superelevation instead. So option D be the correct answer.
(1)
Shubham gour said:
5 years ago
Options C is correct.
We know that;
SSD= (V*T)+(V^2/2*g*f).
First off ..V= 36*5/18=10 and t=2.5 by IRC.
8 = (10*2.5) +(10^2/2*9.81*f)
8 = 25+100/19.62*f
8 = 25+30.09*f
f = 8/30.09
f = 0.26.
So, take the value is nearest in options is 0.16.
We know that;
SSD= (V*T)+(V^2/2*g*f).
First off ..V= 36*5/18=10 and t=2.5 by IRC.
8 = (10*2.5) +(10^2/2*9.81*f)
8 = 25+100/19.62*f
8 = 25+30.09*f
f = 8/30.09
f = 0.26.
So, take the value is nearest in options is 0.16.
(1)
Roy said:
8 years ago
The correct Answer is A.
Kshitiz said:
8 years ago
Friction coefficient value never exceed 0.4
Longitudinal friction coefficient range 0.3 to o.4
Lateral friction coefficient value generally 0.15
You can use the formula for finding friction coefficient but it exceeds limiting value. So Ans is 0.16 correct ie it took lateral friction value.
Longitudinal friction coefficient range 0.3 to o.4
Lateral friction coefficient value generally 0.15
You can use the formula for finding friction coefficient but it exceeds limiting value. So Ans is 0.16 correct ie it took lateral friction value.
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