Question

Find the x-coordinate and y-coordinates of the center of gravity (as a multiple of a)

Answer 1

ANSWER

• x = 0.944a

,

• y = 1.333a

EXPLANATION

x- coordinate

The x-coordinate of each pair of piled blocks are:

• x₁ = x₂ = a/2

,

• x₃ = x₄ = 3a/2

The weights of the blocks are:

• W₁ = 20N

,

• W₂ = 80N

,

• W₃ = 70N

,

• W₄ = 10N

The total weight is the sum of the individual weights,

$W=W_1+W_2+W_3+W_4=20N+80N+70N+10N=180N_{}$

By the definition of center of gravity, the x-coordinate is,

$x_{cg}=\frac{\sum ^{}_iW_ix_i}{W}=\frac{(W_1+W_2)x_1+(W_3+W_4)x_3}{W}$

Replace with the values and solve,

$x_{cg}=\frac{(20N+80N)\cdot\frac{a}{2}+(70N+10N)\frac{3a}{2}}{180N}=\frac{50a+120a}{180a}=\frac{170}{180}a=0.944a$

y- coordinate

The same equations apply to the y-coordinate. The y-coordinates of each weight are:

• y₁ = y₄ = a/2

,

• y₂ = y₃ = 3a/2

The y-coordinate of the center of gravity is,

$y_{cg}=\frac{\sum ^{}_iW_iy_i}{W}=\frac{(W_1+W_4)y1_{}+(W_2+W_3)y_2}{W}$

Replace with the values and solve,

$y_{cg}=\frac{(20N+10N)\frac{a}{2}_{}+(80N+70N)\frac{3a}{2}}{180N}=\frac{15a+225a}{180}=\frac{240a}{180}=1.333a$

Hence, the coordinates of the center of gravity are (0.944a, 1.333a).