- #1
Benjamin_harsh
- 211
- 5
- Homework Statement
- A square sheet of metal has a square of one quarter of the original area cut from one corner as shown in the figure. Which of the following statements is true about the position of the centre of gravity of the remaining portion of the sheet?
- Relevant Equations
- original side of the square = ##2x##
A square sheet of metal has a square of one quarter of the original area cut from one corner as shown in the figure. Which of the following statements is true about the position of the centre of gravity of the remaining portion of the sheet?
a) center of gravity lies at a distance of 5/12 of the side of the original square from each uncut side.
b) center of gravity lies at a distance 7/12 of the side of the original square from each uncut side.
c) center of gravity lies at a distance of 63/4 of the side of the original square from each uncut side.
d) None of these.
Solution:
##\overline X = \large \frac{A_{1}\overline{x_{1}} - A_{1}\overline{x_{2}}}{A_{1}-A_{2}} =
\frac{(4x^{2}).(x) - (x^{2})(1.5x)}{4x^{2}-x^{2}}##
##\overline X = \large\frac{5x}{6} = \frac{5}{12}\normalsize(2x)####\overline Y = \large\frac{A_{1}\overline{y_{1}} - A_{1}\overline{y_{2}}}{A_{1}-A_{2}}
= \frac{5x}{6} = \frac{5}{12}\normalsize(2x)##
original side of the square = ##2x##
Why they didn't continue after this step: ##\frac{5}{12}(2x)##?
a) center of gravity lies at a distance of 5/12 of the side of the original square from each uncut side.
b) center of gravity lies at a distance 7/12 of the side of the original square from each uncut side.
c) center of gravity lies at a distance of 63/4 of the side of the original square from each uncut side.
d) None of these.
Solution:
##\overline X = \large \frac{A_{1}\overline{x_{1}} - A_{1}\overline{x_{2}}}{A_{1}-A_{2}} =
\frac{(4x^{2}).(x) - (x^{2})(1.5x)}{4x^{2}-x^{2}}##
##\overline X = \large\frac{5x}{6} = \frac{5}{12}\normalsize(2x)####\overline Y = \large\frac{A_{1}\overline{y_{1}} - A_{1}\overline{y_{2}}}{A_{1}-A_{2}}
= \frac{5x}{6} = \frac{5}{12}\normalsize(2x)##
original side of the square = ##2x##
Why they didn't continue after this step: ##\frac{5}{12}(2x)##?
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