Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 2 (Q.No. 46)
46.
The centre of gravity of an is osceles triangle with base (p) and sides (q) from its base is
Discussion:
14 comments Page 1 of 2.
Uday said:
1 decade ago
Draw a Triangle and than apply hypotenuse theorem to half triangle.
x^2 = q^2-(p/4)^2.
2/3*x = Answer A.
x^2 = q^2-(p/4)^2.
2/3*x = Answer A.
Somesh said:
8 years ago
I think the position of p and q is interchanged in the option A. Otherwise it is correct.
Vishnu said:
5 years ago
Put p is equal to q you should get p/2√3 hence option A is the right one.
Dinesh said:
10 years ago
CG = h/3.
h = √[g^2-(h/2)^2] = (√4g^2-h^2)/2.
h = √[g^2-(h/2)^2] = (√4g^2-h^2)/2.
(1)
Daka said:
9 years ago
In question base (q) and sides (p). Then answer is A.
EREWA said:
7 years ago
Agree @Somesh.
Unless q and p were interchanged.
Unless q and p were interchanged.
PVN said:
6 years ago
A will be correct if p and Q are interchanged.
Subhajit said:
9 years ago
The answer should be root over 4 q^2 - p^2/6.
Anil malara said:
6 years ago
A is correct if p and q are interchanged.
Vinay said:
5 years ago
CG from base BC = (1/6) √(4Q*P^2).
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