Problem 3 - 85
A block of wood requires a force of 40 N to keep it immersed in water and a force of 100 N to keep it immersed in glycerin (sp. gr. \( =1.3 \) ). Find the weight and sp. gr. of the wood.
Solution
Let \( V= \) volume of wood
In water:
\[
\begin{array}{l}
{\left[\Sigma F_{V}=0\right]} \\
\begin{array}{l}
B F_{1}-W-F=0 \\
9810 V-W=40 \\
V=\frac{40+W}{9810}
\end{array}
\end{array}
\]
In glycerin:
\[
\begin{array}{l}
{\left[\Sigma F_{V}=0\right]} \\
B F_{2}-W-F=0 \\
\quad(9,810 \times 1.3) V-W=100 \\
(9,810 \times 1.3)\left[\frac{40+W}{9810}\right]-W=100 \\
52+1.3 W-W=100 \\
W=160 \mathrm{~N}
\end{array}
\]
From Eq. (1):
\[
\begin{array}{l}
V=\frac{40+160}{9810} \\
V=0.0204 \mathrm{~m}^{3}
\end{array}
\]
Unit weight, \( \gamma=\frac{W}{V}=\frac{160}{0.0204} \)
Unit weight, \( \gamma=7843 \mathrm{~N} / \mathrm{m}^{3} \)
Sp. gr.. \( s=\frac{\gamma_{\text {wood }}}{\gamma_{\text {water }}}=\frac{7843}{9810} \)
Sp. \( \mathrm{gr}^{2}, \mathrm{~s}=0.8 \)
