Question

Find the weight of the steel rivet shown in the figure. (Steel weighs 0.0173 |b/cu cm.) (Round to the nearest tenth as needed.)

Answer 1

Given that the density of steel, we can find the weight using the formula:

$\rho=\frac{m}{V}$

We want the weight (m), then

$m=\rho V$

We know that

$\rho=0.0173\text{ lb/cm}^3$

Then we need to calculate the volume of the rivet.

As we can see it's compounded by a cylinder and a truncated cone, the formula to calculate the volume for these figures are

$V=\pi r^2h$

For the cylinder

h → height

r → radius

and for the truncated cone

$V=\frac{\pi h}{3}(R^2+Rr+r^2)$

R → Big radius

r → Small radius

h → height

Now we have the volume formulas, the total volume will be the sum of these two volumes, then we can say that

$m=\rho(V_1+V_2)$

Where V1 is the cylinder volume and V2 is the truncated cone volume.

Let's find V1:

$\begin{gathered} V_1=\pi r^2h \\ \\ V_1=\pi(1.7)^2\cdot10.7 \\ \\ V_1=97.1475_{} \end{gathered}$

Result in cubic centimeters.

And V2 is

$\begin{gathered} V_2=\frac{\pi h}{3}(R^2+Rr+r^2) \\ \\ V_2=\frac{\pi(2.1)}{3}((3.4)^2+3.4\cdot1.7+(1.7)^2) \\ \\ V_2=44.4880 \end{gathered}$

Result also in cubic centimeters.

Therefore, the total volume is

$\begin{gathered} V=V_1+V_2 \\ V= 141.6356 \end{gathered}$

In cubic centimeters as well.

Now we have the volume we can just apply

$m=\rho V$

Therefore

$\begin{gathered} m=0.0173\cdot141.6356 \\ \\ m=2.4503\text{ lb} \end{gathered}$

Then, the weight of the steel rivet is 2.45lb.

If we round it to the nearest tenth it will be 2.5lb

Therefore, the final answer, rounded to the nearest tenth is

$m=2.5\text{ lb}$