## College Physics (4th Edition)

(a) The dimensions of $V$ are $[L^3]$ (b) The correct interpretation for $V$ would be volume.
(a) A force has the unit newtons which is $kg~m~s^{-2}$. In dimensional analysis, this would be expressed as $[M~L~T^{-2}]$. Let $u$ be the dimensions of $V$ in the equation $F_B = \rho~g~V$. We can find the dimensions of $V$. $[M~L^{-3}]~[L~T^{-2}]~u = [M~L~T^{-2}]$ $[M~L^{-2}~T^{-2}]~u = [M~L~T^{-2}]$ $[L^{-2}]~u = [L]$ $u = \frac{[L]}{[L^{-2}]}$ $u = [L^3]$ The dimensions of $V$ are $[L^3]$. (b) Since volume has the dimensions $[L^3]$, the correct interpretation for $V$ would be volume.