1. Jul 1, 2004

### daisyi

This one is different, and it is driving me crazy almost as bad as the other two.

A lead sinker is attatched to a round platic sphere. If the sphere floats so that it is half submerged, and has a weight of 1oz. (6.25 * 10^-2lb) and a radius of 1in, (8.33 * 10-2ft) find the volume and weight of the sinker if the specific gravity of lead is 11.3.

2. Jul 1, 2004

### AKG

Treat the sphere and sinker as a single body. If it floats, it's in static equilibrium; the forces down equal the forces up. The forces down are:

$$F_{down} = g(m_{sphere} + m_{sinker}) = g(1oz. + \rho _{sinker}V_{sinker})$$

$\rho _{sinker}$ is the density of the sinker, which you can determine given the specific gravity. So really, you have one equation, two unknowns ($F_{down}$ and $V_{sinker}$). Now, the forces up are:

$$F_{up} = g\rho _{water}(V_{sinker} + V_{sphere}/2)$$

Of course, you know $V_{sphere}$ and $\rho _{water}$, so the only unknowns in this are $V_{sinker}$ again, and $F_{up}$ but $F_{up} = F_{down}$, so you have 2 equations and 2 unknowns (or 3 and 3 if you want to look at $F_{up} = F_{down}$ as it's own equation, and $F_{up}$ and $F_{down}$ as different unknowns). This should be pretty simple to solve.