Question

A thin uniform wire is bent to form the two equal sides AB and AC of triangle ABC, where AB = AC = 5 cm. The third side BC, of length 6 cm, is made from a uniform wire of twice the density of the first. The distance of the center of mass from A is

Answer 1

The center of mass from A in the given triangle is 2.5 cm. We can calculate it from specific formula.

Step-by-step explanation:

To find the distance of the center of mass from point A, we need to consider the masses and distances of the segments AB and AC.

Since the wire is uniform, the center of mass of segment AB and AC is located at their midpoints. Therefore, the distance of the center of mass of segment AB from A is 2.5 cm, and the distance of the center of mass of segment AC from A is also 2.5 cm.

Since AB = AC = 5 cm, and BC = 6 cm, we can calculate the combined center of mass using the following formula:

dcm = (mAB * dAB + mAC * dAC) / (mAB + mAC)

Where mAB and mAC are the masses of segments AB and AC respectively, and dAB and dAC are the distances of their center of mass from A.

Since AB and AC have the same length and density, their masses are the same. Therefore, the equation simplifies to:

dcm = (2.5 * 2.5 + 2.5 * 2.5) / (2.5 + 2.5) = 2.5 cm