Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 1 (Q.No. 3)
3.
A horizontal rod AB carries three loads of 3.0 kg, 7.0 kg and 10.0 kg at distances of 2.0 cm, 9.0 cm and 15 cm respectively from A where it is hinged. Neglecting the weight of the rod, which is the point at which the rod will be balance ?
Discussion:
16 comments Page 2 of 2.
Ateeq said:
10 years ago
Eccentric = Total moment at A/Total loads = 3*2+7*9+10*15 = 219/20 = 10.95 cm.
Sumit pal said:
1 decade ago
Balance point = Moment from a given point/Total load acting on road.
Balance point = (3*2)+(7*9)+(10*15)/(3+7+10).
= 10.95.
Balance point = (3*2)+(7*9)+(10*15)/(3+7+10).
= 10.95.
MUNIWAR SINGH said:
1 decade ago
Where bending moment is zero; if it is at the x-distance from 10 kg. Then.
3(13-x)+7(6-x) = 10 x.
Then x = 10.95 cm.
3(13-x)+7(6-x) = 10 x.
Then x = 10.95 cm.
PRASHANT DEOKAR said:
1 decade ago
The point of zero moment is in between 7 & 10kg loads .
Assuming plane xx at that portion from B.
Moment @xx+0.
.: 10x = 7(6-X)+3(13-X).
.: X = 10.95CM.
Assuming plane xx at that portion from B.
Moment @xx+0.
.: 10x = 7(6-X)+3(13-X).
.: X = 10.95CM.
Prithviraj suman said:
1 decade ago
cg of a rod = moment from given point/weight of the rod.
cg = (3*2+7*9+10*15)/(10+7+3) = 10.95.
cg = (3*2+7*9+10*15)/(10+7+3) = 10.95.
Varadharajan said:
1 decade ago
Moment about A = (10X15+7X9+3X2) = 219.
Total Load = 10+7+3 = 20.
Eccentricity = 219/20 = 10.95 cm.
Total Load = 10+7+3 = 20.
Eccentricity = 219/20 = 10.95 cm.
(1)
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