Chapter 64, Problem 28A

Compute the number of cubic centimeters of material in the locating saddle shown. Round the answer to the nearest cubic centimeter.

To determine

The number of cubic centimetres (Volume) of material in the locating saddle.

Answer to Problem 28A

The number of cubic centimetres (Volume) of material in the locating saddle is $151.31{\text{cm}}^{\text{3}}$.

Explanation of Solution

Write the expression of the number of cubic centimetres (Volume) of material in the locating saddle.

 $V=V_1+2V_2+V_3-V_4-V_5$ ........ (I)

Here, the volume of the part 1 is $V_1$, the volume of the part 2 is $V_2$, the volume of the part 3 is $V_3$, the volume of the part 4 is $V_4$ and the volume of the part 4 is $V_4$.

Write the expression of the volume off the part 1 (Square part).

 $V_1=a^2{h}_1$ ........ (II)

Here, the side of the part 1 is $a$ and the height of the part 1 is $h_1$.

Write the expression of the volume off the part 2 (Rectangular part).

 $V_2=l_2{b}_2{h}_2$ ........ (III)

Here, the height of the part 2 is $h_2$, the length of the part 2 is $l_2$ and the width of the part 2 is $b_2$.

Write the expression of the volume off the part 3 (Rectangular part).

 $V_3=l_3{b}_3{h}_3$ ........ (IV)

Here, the height of the part 3 is $h_3$, the length of the part 3 is $l_3$ and the width of the part 3 is $b_3$.

Write the expression of volume of the part 4 (Semi circular part).

 $V_4=\pi r^2{t}_4$ ........ (V)

Here, the radius of the semi circular part is $r$ and the thickness of the semi circular part is $t_4$.

Write the expression of volume of the part 5 (Triangular part).

 $V_5=\frac{1}{2}{b}_5{h}_5{t}_5$ ........ (VI)

Here, the thickness of the triangular part is $t_5$, the height of the part 5 is $h_5$ and the base of the triangular part is $b_5$.

Calculation:

Substitute $6.5\text{cm}$ for $a$ and $2.5\text{cm}$ for $h_1$ in Equation (II).

 $\begin{array}{c}{V}_1={\left(6.5\text{cm}\right)}^2\left(2.5\text{cm}\right)\\ =\left(6.5\text{cm}\right)\left(6.5\text{cm}\right)\left(2.5\text{cm}\right)\\ =105.63{\text{cm}}^{\text{3}}\end{array}$

Substitute $6.5\text{cm}$ for $l_2$, $1\text{cm}$ for $b_2$ and $3.5\text{cm}$ for $h_2$ in Equation (III).

 $\begin{array}{c}{V}_2=\left(6.5\text{cm}\right)\left(1\text{cm}\right)\left(3.5\text{cm}\right)\\ =\left(6.5\text{cm}\right)\left(3.5{\text{cm}}^2\right)\\ =22.75{\text{cm}}^3\end{array}$

Substitute $6.5\text{cm}$ for $l_3$, $1.5\text{cm}$ for $b_3$ and $1\text{cm}$ for $h_3$ in Equation (IV).

 $\begin{array}{c}{V}_3=\left(6.5\text{cm}\right)\left(1.5\text{cm}\right)\left(1\text{cm}\right)\\ =\left(6.5\text{cm}\right)\left(1.5{\text{cm}}^2\right)\\ =9.75{\text{cm}}^{\text{3}}\end{array}$

Substitute $1.5\text{cm}$ for $r$ and $1\text{cm}$ for $t_4$ in Equation (V).

 $\begin{array}{c}{V}_4=\pi {\left(1.5\text{cm}\right)}^2\left(1\text{cm}\right)\\ =\pi \left(2.25{\text{cm}}^2\right)\left(1\text{cm}\right)\\ =7.07{\text{cm}}^{\text{3}}\end{array}$

Substitute $2.5\text{cm}$ for $b_5$, $2\text{cm}$ for $h_5$ and $1\text{cm}$ for $t_5$ in Equation (VI).

 $\begin{array}{c}{V}_5=\frac{1}{2}\left(2.5\text{cm}\right)\left(2\text{cm}\right)\left(1\text{cm}\right)\\ =\left(0.5\right)\left(2.5\text{cm}\right)\left(2\text{cm}\right)\left(1\text{cm}\right)\\ =2.5{\text{cm}}^{\text{3}}\end{array}$

Substitute $105.63{\text{cm}}^{\text{3}}$ for $V_1$, $2.5{\text{cm}}^{\text{3}}$ for $V_5$, $7.07{\text{cm}}^{\text{3}}$ for $V_4$, $9.75{\text{cm}}^{\text{3}}$ for $V_3$ and $22.75{\text{cm}}^3$ for $V_2$ in Equation (I).

 $\begin{array}{c}V=\left(105.63{\text{cm}}^{\text{3}}\right)+2\left(22.75{\text{cm}}^3\right)+\left(9.75{\text{cm}}^{\text{3}}\right)-\left(7.07{\text{cm}}^{\text{3}}\right)-\left(2.5{\text{cm}}^{\text{3}}\right)\\ =\left(105.63{\text{cm}}^{\text{3}}\right)+\left(45.5{\text{cm}}^3\right)+\left(9.75{\text{cm}}^{\text{3}}\right)-\left(7.07{\text{cm}}^{\text{3}}\right)-\left(2.5{\text{cm}}^{\text{3}}\right)\\ =151.31{\text{cm}}^{\text{3}}\end{array}$

Conclusion:

The number of cubic centimetres (Volume) of material in the locating saddle is $151.31{\text{cm}}^{\text{3}}$.