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Chapter 16: Problem 8

A uniform wire is bent to form the sides of a triangle \(\mathrm{ABC}\). If the centre of gravity of the wire coincides with the centroid of the triangle \(\mathrm{ABC}\), triangle \(\mathrm{ABC}\), must be equilateral.

Short Answer

Expert verified
Triangle \(\text{ABC}\) must be equilateral because a uniform wire's center of gravity coincides with the centroid only if all sides are equal.

Step by step solution

01

Understand the Concept of Centroid

The centroid of a triangle is the point where the three medians intersect. For a uniform wire bent to form a triangle, the center of gravity also coincides with the centroid when all sides are equal in length.
02

Define the Centroid in Context of the Wire

For a uniform wire forming a triangle, the centroid is the balance point, meaning it is equally distributed among its parts. For that to happen, the triangle must distribute weight equally from the centroid.
03

Analyze the Properties of the Triangle

Since the centroid is both the center of gravity and the intersection of medians, and given that the wire is uniform, the only way to ensure the center of gravity coincides with the centroid is to have all sides equal. This is a property of an equilateral triangle.
04

Conclusion

Given that all sides must be the same to keep the center of gravity at the centroid, triangle \(\text{ABC}\) must be equilateral.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

equilateral triangle properties
An equilateral triangle is unique and important in geometry. It has three properties that set it apart from other triangles.

First, **all three sides** are of equal length. This equal length ensures that any line drawn from a vertex to the midpoint of the opposite side will be the same for all vertices.

Second, **all three angles** are equal. Each angle in an equilateral triangle is always 60 degrees, because the total sum of angles in any triangle is 180 degrees.

Third, the equilateral triangle's symmetrical nature ensures that it has excellent balance properties. This is why the centroid, or the point where the medians intersect, will always be the same distance from each vertex. These properties make an equilateral triangle very special in many mathematical contexts.
center of gravity
The center of gravity, also often called the centroid in geometry, is a crucial concept when dealing with triangles and other shapes.

The **center of gravity** is the average location of the weight of an object. In a physical object, this is the single point where weight is evenly distributed, and it is also the balance point.

In the case of a uniform wire bent into a triangle, this center of gravity will coincide with the centroid of the triangle. This means the triangle must be able to balance perfectly at this point.

In an equilateral triangle, due to its symmetry and equal sides, the center of gravity will always coincide with its centroid.
medians of a triangle
A median in a triangle is a line segment that goes from one vertex to the midpoint of the opposite side. Each triangle has **three medians**, and they all intersect at a single point called the centroid.

The point where the medians intersect has special properties. The centroid not only balances the triangle, but **divides each median into a 2:1 ratio**. This means the distance from the centroid to the vertex is twice the distance from the centroid to the midpoint of the opposite side.

Since every triangle has three medians, regardless of the triangle鈥檚 specific type, they always intersect at a single point, hence making the centroid a unique determinant.
triangle geometry
Understanding triangle geometry is essential, as it's a foundation of many geometric principles. In triangles, there are several key concepts:

- **Angles**: The sum of angles in any triangle is always 180 degrees.
- **Sides and Angles**: The longer side of a triangle is opposite the larger angle.
- **Types of Triangles**: Based on sides, triangles can be equilateral, isosceles, or scalene. Based on angles, they can be acute, right, or obtuse.

In the context of this exercise, knowing these foundational principles helps in understanding why an equilateral triangle鈥檚 properties lead to its centroid being a perfect balance point.
Equilateral triangles use their properties to simplify problems related to centroid and medians, making them an excellent example in triangle geometry.

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Most popular questions from this chapter

Show that the centre of mass of a uniform solid right circular cone of height \(h\) is at a distance \(\frac{1}{4} h\) from its base. From a uniform solid right circular cylinder, of radius \(r\) and height \(h\), a right circular cone is bored out. The base of the cone coincides with one end of the cylinder and the vertex \(\mathrm{O}\) is at the centre of the other end. Show that the centre of mass of the remainder of the cylinder is at a distance \(3 h / 8\) from \(O\). The bored-out cylinder is placed with \(\mathrm{O}\) uppermost on a horizontal plane which is rough enough to prevent slipping: the plane is then gradually tilted. Show that the cylinder topples when the inclination of the plane to the horizontal exceeds \(\tan ^{-1}(8 r / 5 h)\)

The area between the curve \(y=f(x)\), the \(x\)-axis and the ordinates 0 and \(x_{1}\) represents a uniform lamina. The \(x\) coordinate of its centre of gravity is given by:(a) \(\frac{\int_{0}^{x_{1}} x y d x}{\int_{0}^{x_{1}} x d x}\) (b) \(\frac{\int_{0}^{x_{1}} x y^{2} d x}{\int_{0}^{x_{1}} y^{2} d x}\) (c) \(\frac{\int_{0}^{x_{1}} y x d x}{\int_{0}^{x_{1}} y d x}\) (d) \(\int_{0}^{x_{1}} x y d x\) (e) \(\frac{x_{1}}{2}\)

A uniform lamina of weight \(W\) is in the shape of a triangle \(A B C\) with \(\mathrm{AB}=\mathrm{AC}=2 a\) and the angle BAC equal to \(2 \alpha\). The side \(\mathrm{AB}\) is fixed along a diameter of uniform solid hemisphere of radius \(a\), the plane of the lamina being perpendicular to the flat surface of the hemisphere. The body rests in equilibrium with a point of the curved surface of the hemisphere in contact with a horizontal table and with \(\mathrm{BC}\) vertical. Show that the wieght of the hemisphere is \(\frac{8}{9} \mathrm{~W} \cot \alpha\). A particle of weight \(W\) is attached to a point \(P\) of \(A B\) where \(A P=3^{2} a\) and the body now settles in equilibrium with the mid-point of \(\mathrm{BC}\) vertically above \(\mathrm{A}\). Prove that \(\tan \alpha=\frac{1}{2}\).

A uniform solid cone has a base radius \(r\) and height \(4 r .\) lt rests with its plane face on an inclined plane which is rough enough to prevent sliding. The cone will topple when the inclination of the plane to the horizontal is greater than: (a) \(45^{\circ}\) (b) arctan \(\frac{1}{4}\) (c) \(\arctan \frac{3}{4}\) (d) \(90^{\circ}\) (e) \(\arctan \frac{1}{2}\).

Prove that the centre of mass of a uniform solid right circular cone, of height \(h\) and semi-vertical angle \(\alpha\) is at a distance \(\frac{3}{4} h\) from its vertex. A frustum is cut from the cone by a plane parallel to the base at a distance \(\frac{1}{2} h\). from the vertex. Show that the distance of the centre of mass of this frustum from its larger plane end is \(11 \mathrm{~h} / 56\). This frustum is placed with its curved surface in contact with a horizontal table. Show that equilibrium is not possible unless \(45 \cos ^{2} \alpha \geqslant 28\).
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