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 1 of 2.
SHIVA KUMAR M N said:
4 years ago
Appy Varignon theorem at A, which states moment produced by the resultant force at any point is equal to the sum of the moment produced by the indivisual force at the same point.
Appy at A, we get 20x = 3 * 2 + 7 * 9 + 10 * 15.
20x = 219, Hence x=10.95cm.
Appy at A, we get 20x = 3 * 2 + 7 * 9 + 10 * 15.
20x = 219, Hence x=10.95cm.
(6)
Anbarasan muthaiah AP/CIVIL STAR LION ENGINEERING said:
8 years ago
RESULTANT* FORCE DISTANCE OF RESULTANT ACT FROM SUPPORT = SUM OF MOMENT
R * x = M.
R = 3+7+10 = 20 KG.
SUM OF M ABOUT B = (3*2)+(7*9)+(10*15) =219 KG.cm.
219= 20x,
x = 219/20 = 10.95cm.
R * x = M.
R = 3+7+10 = 20 KG.
SUM OF M ABOUT B = (3*2)+(7*9)+(10*15) =219 KG.cm.
219= 20x,
x = 219/20 = 10.95cm.
(5)
Habibur Rahman said:
4 years ago
3*2 + 7*9 + 10*15 = 20X.
Then, X = 10.95.
Then, X = 10.95.
(3)
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)
Abdul Khan said:
10 years ago
No one has considered the moment of A reaction, please elaborate I didn't understood.
(1)
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.
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.
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.
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.
Ateeq said:
10 years ago
Eccentric = Total moment at A/Total loads = 3*2+7*9+10*15 = 219/20 = 10.95 cm.
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