Center of Mass

# Center of Mass

389.5k points

A carpenter's square of uniform density has the shape of an L, as shown in the figure. For an $A(x,y)$ coordinate frame with the origin at the lower left corner of the carpenter's square, The x-axis is horizontal and to the right. The y-axis is vertically upward. Because the square is uniform in thickness and has a small thickness, we can assume that the weight of each segment of the square is proportional to its area.

Given in the figure, $B=10 \text{ cm}$, $C=5 \text{ cm}$, $D=4 \text{ cm}$, $E=20 \text{ cm}$

What is the $x$ and $y$ coordinate of the center of gravity?

Center of
lancer

389.5k points

The center of mass of a system of particles is the average position weighted by their masses

$$r_c=\frac{\sum m_i r_i}{\sum m_i}$$

first find the mass of each square,

$$mass=density\times area$$

for the square on the left we have,

$$m_l=\sigma A_l$$ where $$A_l=D \times E = (4)(20)=80\;cm^2$$

and for the square on the bottom,

$$m_b=\sigma A_b$$ where $$A_b=C\times (B-D)=(5)(10-4)=30\; cm^2$$

since the mass is uniformly distributed, then the center of each mass of the square is its geometrical center. Hence the center of the left square is,

$$(x_{cm},y_{cm})=\big( \frac{D}{2},\frac{E}{2}\big)=\big(\frac{4}{2},\frac{20}{2}\big)=(2,10)$$

and the center of mass of the bottom square is,

$$\big(x_{cm},y_{cm})=\big(\frac{B-D}{2}+D,\frac{C}{2}\big)=\big(\frac{10-4}{2}+4,\frac{5}{2}\big)=(7,10)$$

finally, by using the center of mass formula the x and y coordinates are,

$$x_{cm}=\frac{\sigma \big(A_l\frac{D}{2}+A_b\frac{B+D}{2}\big)}{\sigma (A_l+A_b)}=\frac{\big((80)\frac{4}{2}+(30)\frac{10+4}{2}\big)}{(80+30)}=3.36\; cm$$

$$y_{cm}=\frac{\sigma \big(A_l\frac{E}{2}+A_b\frac{C}{2}\big)}{\sigma (A_l+A_b)}=\frac{\big((80)\frac{20}{2}+(30)\frac{5}{2}\big)}{(80+30)}=7.95\; cm$$

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