The value of at any latitude \phi may be obtained from the formula
g=32.09\left(1+0.0053 \sin ^{2} \phi\right) \mathrm{\ ft} / \mathrm{s}^{2}
which takes into account the effect of the rotation of the earth, as well as the fact that the earth is not truly spherical. Determine to four significant figures (a) the weight in pounds, (b) the mass in pounds, (c) the mass in lb ⋅ s² /ft, at the latitudes of 0°, 45°, and 60°, of a silver bar, the mass of which has been officially designated as 5 lb.
\begin{array}{ll} & g=32.09\left(1+0.0053 \sin ^{2} \phi\right) \mathrm{\ ft} / \mathrm{s}^{2} \\\phi=0^{\circ}: & g=32.09 \mathrm{\ ft} / \mathrm{s}^{2} \\\phi=45^{\circ}: & g=32.175 \mathrm{\ ft} / \mathrm{s}^{2} \\\phi=90^{\circ}: & g=32.26 \mathrm{\ ft} / \mathrm{s}^{2}\end{array}
(a) Weight: \quad W=m g
\phi=0^{\circ}: \quad W=\left(0.1554 \mathrm{\ lb} \cdot \mathrm{s}^{2} / \mathrm{ft}\right)\left(32.09 \mathrm{\ ft} / \mathrm{s}^{2}\right)=4.987 \mathrm{\ lb}\blacktriangleleft
\phi=45^{\circ}: \quad W=\left(0.1554 \mathrm{\ lb} \cdot \mathrm{s}^{2} / \mathrm{ft}\right)\left(32.175 \mathrm{\ ft} / \mathrm{s}^{2}\right)=5.000 \mathrm{\ lb}\blacktriangleleft
\phi=90^{\circ}: \quad W=\left(0.1554 \mathrm{\ lb} \cdot \mathrm{s}^{2} / \mathrm{ft}\right)\left(32.26 \mathrm{\ ft} / \mathrm{s}^{2}\right)=5.013 \mathrm{\ lb}\blacktriangleleft
(b) Mass: At all latitudes: \quad\quad m=5.000\mathrm{\ lb}\blacktriangleleft
(c) or \quad m=\frac{5.00 \mathrm{\ lb}}{32.175 \mathrm{\ ft} / \mathrm{s}^{2}}\quad\quad m=0.1554 \mathrm{\ lb} \cdot \mathrm{s}^{2} / \mathrm{ft}\blacktriangleleft