Question

A square lamina has a side of length l. If an isosceles triangle is removed such that its base lies on one side of the square and its height is h. If the remaining portion is suspended from its apex p of the cut will remain in equilibrium in any position. Find the value of h

Answer 1

Answer on Question#57065 - Physics - Mechanics - Relativity

A square lamina has a side of length $l$ . If an isosceles triangle is removed such that its base lies on one side of the square and its height is $h$ . If the remaining portion is suspended from its apex $P$ of the cut will remain in equilibrium in any position. Find the value of $h$ .

Solution:

Area of the triangle:

$$A _ {t} = \frac {1}{2} h l$$

Area of the lamina:

$$A _ {l} = l ^ {2} - A _ {t} = l \left(l - \frac {1}{2} h\right)$$

The center of mass of the triangle is situated on its altitude at distance $h / 3$ from the base.

From the given information we conclude that the center of mass of the lamina is situated at point $P$ .

Since the center of mass of the square was at point $C$ , then we obtain the following:

$$A _ {t} \cdot \left(\frac {l}{2} - \frac {h}{3}\right) = A _ {l} \cdot \left(\frac {l}{2} - (l - h)\right)$$ $$\frac {1}{2} h l \cdot \left(\frac {l}{2} - \frac {h}{3}\right) = l \left(l - \frac {1}{2} h\right) \cdot \left(\frac {l}{2} - (l - h)\right)$$ $$\frac {2}{3} h ^ {2} - 2 h l + l ^ {2} = 0$$

The only root that meets the requirement $0 < h < l$ is

$$h = \frac {3}{2} \left(1 - \frac {1}{\sqrt {3}}\right) l$$

Answer: $h = \frac{3}{2}\left(1 - \frac{1}{\sqrt{3}}\right)l$ .

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