Chapter 63, Problem 1A

A rectangular strip of steel 2 ft 4 in. long, 1 ft 6 in. wide, and $\frac{5}{8}$ in. thick has four holes drilled through the strip, each with a diameter of $1\frac{1}{2}$ in. If steel weighs 0.264 lb/cubic inch, what is the weight of this strip? Round the answer to the nearest pound.

To determine

The weight of the rectangular strip of steel.

Answer to Problem 1A

Weight of the rectangular strip $\text{W = 82}\text{lb}\text{.}$

Explanation of Solution

Given:

Length $l=2ft4in.$

Breadth $b=1ft6in.$

Thickness $t=\frac{5}{8}in.$

Diameter of hole $d=1\frac{1}{2}in.$

Specific volume of the steel strip $\nu =0.264\frac{lb}{in{.}^3}$

Concept used:

The weight of the steel strip is

  $W=\nu V$

Here, W is the weight, v is the specific volume and V is the total volume.

Calculation:

The total volume of the steel strip is calculated by the formula

  $V=V_r-4V_c$ _______ (1)

Here, V is the total volume of the steel strip, V_r is the volume of rectangular strip and V_c is the volume of circular holes.

Now, the volume of the rectangular strip V_ris calculated as

  $V_r=lbt$

Here, l is the length, b is the breadth and t is the thickness of rectangle.

By substituting the values we get

  $\begin{array}{l}{V}_r=\left(2ft4in.\right)\left(1ft6in.\right)\left(\frac{5}{8}in.\right)\\ V_r=\left(2\times 12+4in.\right)\left(1\times 12+6in.\right)\left(\frac{5}{8}in.\right)\\ V_r=28\times 18\times \frac{5}{8}in.\\ V_r=315in{.}^3\end{array}$

Again, the volume of the circular holes V_c is calculated as

  $V_c=\pi r^{2}t$

Here, r is the radius and t is the thickness of rectangle.

By substituting the values we get

  $\begin{array}{l}{V}_c=3.14{\left(\frac{1\frac{1}{2}}{2}in.\right)}^2\left(\frac{5}{8}in.\right)\\ V_c=3.14{\left(\frac{3}{4}in.\right)}^2\left(\frac{5}{8}in.\right)\\ V_c=3.14\times \frac{9}{16}\times \frac{5}{8}in.\\ V_c=1.104in{.}^3\end{array}$

Now, substituting the value of volume of rectangular strip and volume of circular holes in equation (1) we get

  $\begin{array}{l}\text{V=315}-\text{4}\left(\text{1}\text{.104}\right)\\ \text{V=315}-\text{4}\text{.416}\\ \text{V=310}\text{.584}\text{in}{\text{.}}^{\text{3}}\end{array}$

The weight of the steel strip is

  $W=\nu V$

Here, W is the weight, v is the specific volume and V is the total volume.

By substituting the values we get

  $\begin{array}{l}W=0.264\times 310.584\\ W=81.99\\ W=82lb\end{array}$

Conclusion:

The weight of the steel strip is $\text{W = 82}\text{lb}\text{.}$