Chapter 62, Problem 27A

To determine

The weight of the steel bearing washer.

Answer to Problem 27A

The weight of the steel bearing washer is $4.62\text{lb}$.

Explanation of Solution

Given information:

The number of rib is $4$ and the weighs of steel is $0.283$

$\text{pound}\text{per}\text{cubic}\text{inch}$.

Write the expression of volume of the rib.

 $V_R=\frac{1}{2}lbh$ ......... (I)

Here, the length of the rib is $l$, the width of the rib is $b$ and the height of the rib is $h$.

Write the expression of the total volume of rib.

 $V_{TR}=n{V}_R$ ......... (II)

Here, the number of rib is $n$.

Write the expression of volume of the larger cylinder.

 $V_{LC}=\frac{\pi }{4}\left(D_1{}^2-d_1{}^2\right)t_1$ ......... (III)

Here, the thickness of the large cylinder is $t_1$, the internal diameter of the large cylinder is $d_1$ and the external diameter of the large cylinder is $D_1$.

Write the expression of volume of the smaller cylinder.

 $V_{SC}=\frac{\pi }{4}\left(D_2{}^2-d_2{}^2\right)t_2$ ......... (IV)

Here, the thickness of the small cylinder is $t_2$, the internal diameter of the small cylinder is $d_2$ and the external diameter of the small cylinder is $D_2$.

Write the expression of the steel bearing washer.

 $W=Vw$ .......... (V)

Here, the weighs is $w$.

Write the expression of the total volume.

 $V=V_{TR}+V_{LC}+V_{SC}$

Substitute $V_{TR}+V_{LC}+V_{SC}$ for $V$ in Equation (V).

 $W=\left(V_{TR}+V_{LC}+V_{SC}\right)w$ ......... (VI)

Calculation:

Substitute $1.625\text{in}$ for $l$, $0.75\text{in}$ for $b$ and $0.375\text{in}$ for $h$ in Equation (I).

 $\begin{array}{c}{V}_R=\frac{1}{2}\left(1.625\text{in}\right)\left(0.75\text{in}\right)\left(0.375\text{in}\right)\\ =\frac{1}{2}\left(1.625\text{in}\right)\left(0.28{\text{in}}^{\text{2}}\right)\\ =0.23{\text{in}}^{\text{3}}\end{array}$

Substitute $4$ for $n$ and $0.23{\text{in}}^{\text{3}}$ for $V_R$ in Equation (II).

 $\begin{array}{c}{V}_{TR}=\left(4\right)\left(0.23{\text{in}}^{\text{3}}\right)\\ =0.92{\text{in}}^{\text{3}}\end{array}$

Substitute $6.375\text{in}$ for $D_1$, $1.625\text{in}$ for $d_1$ and $0.375\text{in}$ for $t_1$ in Equation (III).

 $\begin{array}{c}{V}_{LC}=\frac{\pi }{4}\left\{{\left(6.375\text{in}\right)}^2-{\left(1.625\text{in}\right)}^2\right\}\left(0.375\text{in}\right)\\ =\frac{\pi }{4}\left(0.375\text{in}\right)\left\{\left(40.64{\text{in}}^{\text{2}}\right)-\left(2.64{\text{in}}^{\text{2}}\right)\right\}\\ =\frac{\pi }{4}\left(0.375\text{in}\right)\left(38{\text{in}}^{\text{2}}\right)\\ =11.19{\text{in}}^3\end{array}$

Substitute $3.125\text{in}$ for $D_2$, $1.625\text{in}$ for $d_2$ and $0.75\text{in}$ for $t_2$ in Equation (IV).

 $\begin{array}{c}{V}_{SC}=\frac{\pi }{4}\left\{{\left(3.125\text{in}\right)}^2-{\left(1.625\text{in}\right)}^2\right\}\left(0.75\text{in}\right)\\ =\frac{\pi }{4}\left(0.75\text{in}\right)\left\{\left(9.77{\text{in}}^2\right)-\left(2.64{\text{in}}^2\right)\right\}\\ =\frac{\pi }{4}\left(0.75\text{in}\right)\left(7.13{\text{in}}^2\right)\\ =4.2{\text{in}}^2\end{array}$

Substitute $0.92{\text{in}}^{\text{3}}$ for $V_{TR}$, $4.2{\text{in}}^2$ for $V_{SC}$, $11.19{\text{in}}^3$ for $V_{LC}$ and $0.283$

$\text{pound}\text{per}\text{cubic}\text{inch}$ for $w$ in Equation (VI).

 $\begin{array}{c}W=\left\{\left(0.92{\text{in}}^{\text{3}}\right)+\left(11.19{\text{in}}^3\right)+\left(4.2{\text{in}}^2\right)\right\}\left(0.283\text{pound}\text{per}\text{cubic}\text{inch}\right)\\ =\left\{\left(0.92{\text{in}}^{\text{3}}\right)+\left(11.19{\text{in}}^3\right)+\left(4.2{\text{in}}^2\right)\right\}\left(0.283\text{lb}/{\text{in}}^{\text{3}}\right)\\ =\left(0.283\text{lb}/{\text{in}}^{\text{3}}\right)\left(16.31{\text{in}}^3\right)\\ =4.62\text{lb}\end{array}$

Conclusion:

The weight of the steel bearing washer is $4.62\text{lb}$.