The centre of gravity of an isosceles triangle with base (p) and sides (q) from its base is
A. $$\frac{{\sqrt {4{{\text{p}}^2} - {{\text{q}}^2}} }}{6}$$
B. $$\frac{{4{{\text{p}}^2} - {{\text{q}}^2}}}{6}$$
C. $$\frac{{{{\text{p}}^2} - {{\text{q}}^2}}}{4}$$
D. $$\frac{{{{\text{p}}^2} + {{\text{q}}^2}}}{4}$$
Answer: Option A
This Question Belongs to Mechanical Engineering >> Engineering Mechanics
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Comments ( 2 )
The resultant of two equal forces P making an angle $$\theta ,$$ is given by
A. $$2{\text{P}}\sin \frac{\theta }{2}$$
B. $$2{\text{P}}\cos \frac{\theta }{2}$$
C. $$2{\text{P}}\tan \frac{\theta }{2}$$
D. $$2{\text{P}}\cot \frac{\theta }{2}$$
A. Equal to
B. Less than
C. Greater than
D. None of these
If a number of forces are acting at a point, their resultant is given by
A. $${\left( {\sum {\text{V}} } \right)^2} + {\left( {\sum {\text{H}} } \right)^2}$$
B. $$\sqrt {{{\left( {\sum {\text{V}} } \right)}^2} + {{\left( {\sum {\text{H}} } \right)}^2}} $$
C. $${\left( {\sum {\text{V}} } \right)^2} + {\left( {\sum {\text{H}} } \right)^2} + 2\left( {\sum {\text{V}} } \right)\left( {\sum {\text{H}} } \right)$$
D. $$\sqrt {{{\left( {\sum {\text{V}} } \right)}^2} + {{\left( {\sum {\text{H}} } \right)}^2} + 2\left( {\sum {\text{V}} } \right)\left( {\sum {\text{H}} } \right)} $$
A. $${\text{a}} = \frac{\alpha }{{\text{r}}}$$
B. $${\text{a}} = \alpha {\text{r}}$$
C. $${\text{a}} = \frac{{\text{r}}}{\alpha }$$
D. None of these
Correct answer is:
[√(4q² - p²)]/6
@Bashir Ahmad
We know that COG of triangle is h/3 from its base
Using Pythagorus theorm
h = [√(4q² - p²)]/2
=> COG = h/3 = 1/3 * ( [√(4q² - p²)]/2 )
= [√(4q² - p²)]/6
How