Question

Figure p12. 4 shows three uniform objects: a rod, a right triangle, and a square. Their masses and their coordinates in meters are given. Determine the center of gravity for the three-object system.

Answer 1

The center of gravity for the three-objects system is approximately located

at their center of mass.

• The center of gravity of the three-object system is approximately (2.58, 4.75)

Reasons:

The coordinates of the vertices of the objects and their masses are;

The rod:

Vertices, (2, 7), and (9, 7)

Mass, m₁ = 6.00 kg.

The right triangle;

Vertices; (4, 1), (8, 5), (8, 1)

Mass, m₂ = 3.00 kg.

The square

Vertices; (-5, 5), (-2, 5), (-2, 2), (-5, 2)

Mass, m₃ = 5.00 kg

Required;

The center of gravity for the three-object system.

Solution:

The center of gravity is given by the formula;

$\displaystyle X_c =\mathbf{ \frac{X_{c1} \times m_1 +X_{c2} \times m_2 +X_{c3} \times m_3 }{m_1 + m_2 + m_3}}$

$\displaystyle Y_c =\mathbf{ \frac{Y_{c1} \times m_1 +Y_{c2} \times m_2 +Y_{c3} \times m_3 }{m_1 + m_2 + m_3}}$

Where;

$X_{c1}$ = The x-coordinates of the center of the rod = 5.5

$X_{c2}$ = The x-coordinate of the centroid of the triangle  $\displaystyle = \frac{4 + 8 + 8}{3} = \frac{20}{3}$

$X_{c3}$ = The x-coordinate of the centroid of the square = (-5 - (-2)) ÷ 2 - 2 = -3.5

$Y_{c1}$ = The y-coordinate of the center of the rod = 7

$\mathbf{ Y_{c2}}$ = The y-coordinate of the centroid of the triangle  $\displaystyle = \frac{7}{3}$

$Y_{c3}$ = The y-coordinate of the centroid of the square = 3.5

Therefore;

$\displaystyle X_c = \mathbf{ \frac{5.5 \times 6 +\frac{20}{3} \times 3 +(-3.5) \times 5 }{6+ 3 + 5}} \approx 2.58$

$\displaystyle Y_c = \mathbf{ \frac{7 \times 6 +\frac{7}{3} \times 3 +3.5 \times 5 }{6+ 3 + 5}} = 4.75$

The center of gravity for the three-object system is ($X_c$, $Y_c$) ≈ (2.58, 4.75)

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