Question

Three blocks of mass m, m and 2m are placed at the corners of a triangle having corrdinates (2.5, 1.5), (3.5, 1.5) and (3, 3). Find the center of mass.

Answer 1

Solution:

Given:

⇒ Mass of three blocks are m, m and 2m.

⇒ Coordinates: x₁ = 2.5, y₁ = 1.5, x₂ = 3.5, y₂ = 1.5, x₃ = 3 and y₃ = 3.

To Find:

⇒ Center of mass.

Formula used:

$\sf{\implies x=\dfrac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3}}{m_{1}+m_{2}+m_{3}}}$

$\sf{\implies y=\dfrac{m_{1}y_{1}+m_{2}y_{2}+m_{3}y_{3}}{m_{1}+m_{2}+m_{3}}}$

Now, put the values in the formula to find center of mass.

$\sf{\implies x=\dfrac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3}}{m_{1}+m_{2}+m_{3}}}$

$\sf{\implies x=\dfrac{m\times2.5 + m\times 3.5+2m\times 3}{m+m+2m}}$

$\sf{\implies x=\dfrac{2.5m+3.5m+6m}{4m}}$

$\sf{\implies x=\dfrac{12m}{4m}}$

$\sf{\implies x=3}$

$\sf{\implies y=\dfrac{m_{1}y_{1}+m_{2}y_{2}+m_{3}y_{3}}{m_{1}+m_{2}+m_{3}}}$

$\sf{\implies y=\dfrac{m\times1.5 + m\times 1.5+2m\times 3}{m+m+2m}}$

$\sf{\implies y=\dfrac{1.5m+1.5m+6m}{4m}}$

$\sf{\implies y=\dfrac{10m}{4m}}$

$\sf{\implies y=2.2}$

Hence, center of mass is (3, 2.2).

Answer 2

AnswEr :

⠀

• Mass of three Blocks are m, m and 2m
• Coordinate are (2.5, 1.5), (3.5, 1.5) and (3, 3) respectively.

• Therefore Coordinates will be :

⋆ x₁ = 2.5 and, y₁ = 1.5

⋆ x₂ = 3.5 and, y₂ = 1.5

⋆ x₃ = 3 and, y₃ = 3

_________________________________

Let the Center Masses Be (CMx, CMy)

◑ For Calculation of CMx Formula is :

$\leadsto\sf{ CM_{x}=\dfrac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3}}{m_{1}+m_{2}+m_{3}}}$

$\leadsto\sf{ CM_{x}=\dfrac{m\times2.5 + m\times 3.5+2m\times 3}{m+m+2m}}$

$\leadsto\sf{CM_{x}=\dfrac{2.5m+3.5m+6m}{4m}}$

$\leadsto\sf{CM_{x}=\cancel\dfrac{12m}{4m}}$

$\leadsto\boxed{\sf{CM_{x}=3}}$

◑ For Calculation of CMy Formula is :

$\leadsto\sf{CM_{y}=\dfrac{m_{1}y_{1}+m_{2}y_{2}+m_{3}y_{3}}{m_{1}+m_{2}+m_{3}}}$

$\leadsto\sf{CM_{y}=\dfrac{m\times1.5 + m\times 1.5+2m\times 3}{m+m+2m}}$

$\leadsto\sf{CM_{y}=\dfrac{1.5m+1.5m+6m}{4m}}$

$\leadsto\sf{CM_{y}=\cancel\dfrac{9m}{4m}}$

$\leadsto\boxed{\sf{CM_{y}=2.25}}$

⠀

$\therefore$ Center of Masses is (3 , 2.25)