what is the weight of the bushing shown below if it is made of steel weighing 0.2833 lb/in.³? (the inner…

Explanation:

Step1: Calculate outer - cylinder volume

The formula for the volume of a cylinder is $V=\pi r^{2}h$. The outer - radius $r_{1}=\frac{2\frac{1}{4}}{2}=\frac{9}{8}$ inches and height $h = 8\frac{1}{2}=\frac{17}{2}$ inches. So, $V_{1}=\pi r_{1}^{2}h=\pi(\frac{9}{8})^{2}\times\frac{17}{2}=\pi\times\frac{81}{64}\times\frac{17}{2}=\frac{1377\pi}{128}$ cubic inches.

Step2: Calculate inner - cylinder volume

The inner - radius $r_{2}=\frac{6/2}{2}=\frac{3}{2}$ inches and height $h=\frac{17}{2}$ inches. So, $V_{2}=\pi r_{2}^{2}h=\pi(\frac{3}{2})^{2}\times\frac{17}{2}=\pi\times\frac{9}{4}\times\frac{17}{2}=\frac{153\pi}{8}$ cubic inches.

Step3: Calculate the volume of the bushing

The volume of the bushing $V = V_{1}-V_{2}=\frac{1377\pi}{128}-\frac{153\pi}{8}=\frac{1377\pi - 2448\pi}{128}=-\frac{1071\pi}{128}$ (There is a mistake above, the outer - diameter should be 6 inches and inner - diameter should be $2\frac{1}{4}$ inches). The correct outer - radius $r_{1}=\frac{6}{2}=3$ inches and inner - radius $r_{2}=\frac{2\frac{1}{4}}{2}=\frac{9}{8}$ inches, height $h = 8\frac{1}{2}=\frac{17}{2}$ inches. $V_{1}=\pi r_{1}^{2}h=\pi\times3^{2}\times\frac{17}{2}=\frac{153\pi}{2}$ cubic inches. $V_{2}=\pi r_{2}^{2}h=\pi(\frac{9}{8})^{2}\times\frac{17}{2}=\frac{1377\pi}{128}$ cubic inches. $V = V_{1}-V_{2}=\frac{153\pi}{2}-\frac{1377\pi}{128}=\frac{9792\pi - 1377\pi}{128}=\frac{8415\pi}{128}$ cubic inches.

Step4: Calculate the weight of the bushing

The weight density of steel is $\rho = 0.2833$ lb/in³. The weight $W=\rho V$. $W = 0.2833\times\frac{8415\pi}{128}\approx0.2833\times\frac{8415\times3.1416}{128}\approx0.2833\times\frac{26439.464}{128}\approx0.2833\times206.5583125\approx58.52$ lb.

Answer:

Approximately $58.52$ lb