Given that the density of steel, we can find the weight using the formula:
[tex]\rho=\frac{m}{V}[/tex]We want the weight (m), then
[tex]m=\rho V[/tex]We know that
[tex]\rho=0.0173\text{ lb/cm}^3[/tex]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
[tex]V=\pi r^2h[/tex]For the cylinder
h → height
r → radius
and for the truncated cone
[tex]V=\frac{\pi h}{3}(R^2+Rr+r^2)[/tex]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
[tex]m=\rho(V_1+V_2)[/tex]Where V1 is the cylinder volume and V2 is the truncated cone volume.
Let's find V1:
[tex]\begin{gathered} V_1=\pi r^2h \\ \\ V_1=\pi(1.7)^2\cdot10.7 \\ \\ V_1=97.1475_{} \end{gathered}[/tex]Result in cubic centimeters.
And V2 is
[tex]\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}[/tex]Result also in cubic centimeters.
Therefore, the total volume is
[tex]\begin{gathered} V=V_1+V_2 \\ V=$141.6356$ \end{gathered}[/tex]In cubic centimeters as well.
Now we have the volume we can just apply
[tex]m=\rho V[/tex]Therefore
[tex]\begin{gathered} m=0.0173\cdot141.6356 \\ \\ m=2.4503\text{ lb} \end{gathered}[/tex]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
[tex]m=2.5\text{ lb}[/tex]