Solve Lead Sinker Problem: Volume & Weight Calculation

[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.

 

[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:

[tex]F_{down} = g(m_{sphere} + m_{sinker}) = g(1oz. + \rho _{sinker}V_{sinker})[/tex]

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

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

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

 

[M98Ranger]

To solve this lead sinker problem, we can use the formula for specific gravity which is defined as the ratio of the density of a substance to the density of water. In this case, we know that the specific gravity of lead is 11.3, which means that it is 11.3 times denser than water.

To find the volume of the lead sinker, we can use the formula for volume of a sphere, V = (4/3)πr^3, where r is the radius of the sphere. Plugging in the given radius of 1 inch, we get V = (4/3)π(1)^3 = (4/3)π cubic inches.

To convert this volume into cubic feet, we can divide it by 1728 (since there are 1728 cubic inches in 1 cubic foot). This gives us a volume of 0.0014 cubic feet.

Next, we can use the formula for specific gravity, SG = (weight of substance/weight of water), to find the weight of the lead sinker. We know that the weight of the sphere is 1oz or 6.25 * 10^-2lb, and since it is half submerged, the weight of the water displaced is also 1oz or 6.25 * 10^-2lb. Plugging these values into the formula, we get:

11.3 = (weight of lead sinker)/(6.25 * 10^-2)

Solving for the weight of the lead sinker, we get 0.708 lb or 11.33 oz.

Therefore, the volume of the lead sinker is 0.0014 cubic feet and its weight is 0.708 lb or 11.33 oz. I hope this helps you solve the problem and alleviate some of your frustration.