Chapter 8, Problem 8.92P

Small screws are used to fasten a piece of hardwood to the bracket that is formed from 1/20-in.-thick steel sheet metal. For steel, $\gamma =0.283\text{lb/in}{\text{.}}^{\text{3}},$ and for hardwood, $\gamma =0.029\text{lb/in}{\text{.}}^{\text{3}}\text{.}$ Locate the center of gravity of the assembly.

To determine

The centre of gravity of the assembly.

Answer to Problem 8.92P

The centre of gravity of given part is $\overline{x}=3.75\text{ in},\overline{y}=6.67\text{ in},\overline{z}=1.32\text{ in}\text{.}$

Explanation of Solution

Given Information:

Small screws are used to fasten a piece of hardwood to the bracket that is formed from 1/20-in.-thick steel sheet metal as shown in figure below,

  

For steel, ${\gamma }_s=0.283\text{ lb/in}{\text{.}}^3$ and for hardwood, ${\gamma }_w=0.029\text{ lb/in}{\text{.}}^3$.

Calculation:

Consider the following assembly,

  

$i$	Part	$M_i\left(\text{Mass}\text{=}\text{Volume}\text{×}\text{Density}\right)$	${\overline{x}}_i$	${\overline{y}}_i$	${\overline{z}}_i$
1	Steel Plate	$\left(6\times 7.5\times \frac{1}{20}\right)\times 0.283=0.64$	$\frac{7.5}{2}=3.75$	0.00	$\frac{6}{2}=3$
2	Steel Plate	$\left(\left(7.5+3\right)\times 7.5\times \frac{1}{20}\right)\times 0.283=1.11$	$\frac{7.5}{2}=3.75$	$\frac{\left(7.5+3\right)}{2}=5.25$	0.00
3	Steel Plate	$\left(3\times 7.5\times \frac{1}{20}\right)\times 0.283=0.32$	$\frac{7.5}{2}=3.75$	$\left(7.5+3\right)=10.50$	$\frac{3}{2}=1.5$
4	Wood	$\left(3\times 3\times 7.5\right)\times 0.029=1.96$	$\frac{7.5}{2}=3.75$	$7.5+\frac{\left(3\right)}{2}=9$	$\frac{3}{2}=1.5$

Prepare the following table considering $M_i$ as weight of $i^{th}$ part,

$i$	Part	$M_i$	${\overline{x}}_i$	${\overline{y}}_i$	${\overline{z}}_i$	$M_i{\overline{x}}_i$	$M_i{\overline{y}}_i$	$M_i{\overline{z}}_i$
1	Steel Plate	0.64	3.75	0.00	3.00	2.39	0.00	1.91
2	Steel Plate	1.11	3.75	5.25	0.00	4.18	5.85	0
3	Steel Plate	0.32	3.75	10.50	1.50	1.19	3.34	0.47
4	Wood	1.96	3.75	9.00	1.50	7.34	17.62	2.93
	Total	${\displaystyle \sum M_i}=4.03$				$\begin{array}{l}{\displaystyle \sum M_i{\overline{x}}_i}\\ =15.10\end{array}$	$\begin{array}{l}{\displaystyle \sum M_i{\overline{y}}_i}\\ =26.81\end{array}$	$\begin{array}{l}{\displaystyle \sum M_i{\overline{z}}_i}\\ =5.32\end{array}$

Now calculate the centre of gravity as,

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum M_i{\overline{x}}_i}}{{\displaystyle \sum M_i}}=\frac{15.1}{4.03}=3.75\text{ in},\\ \overline{y}=\frac{{\displaystyle \sum M_i{\overline{y}}_i}}{{\displaystyle \sum M_i}}=\frac{26.81}{4.03}=6.67\text{ in},\\ \overline{z}=\frac{{\displaystyle \sum M_i{\overline{z}}_i}}{{\displaystyle \sum M_i}}=\frac{5.32}{4.03}=1.32\text{ in}\text{.}\end{array}$

Conclusion:

Therefore, the centre of gravity of given part is $\overline{x}=3.75\text{ in},\overline{y}=6.67\text{ in},\overline{z}=1.32\text{ in}\text{.}$