Solve for Ratio of Steel Ax and Aluminum Piston Weights - Buoyancy Problem

[cat_eyes]

[SOLVED] bouyancy problem

Homework Statement

If a steel ax and a aluminum piston have the same apparent weights in the water which can be written as
(Pax)(Vax) - (Pw)(Vax) = (Pp)(Vp) - (Pw)(Vp)
where P=rho(density), p=piston, V=volume, and w=water
Find the ratio (Wax/Wp) of their weights in the air. (Neglect the buoyant force in the air.) Assume densities Pax=7.7g/cm^3 and Pp=2.5g/cm^3

Homework Equations

P=m/v

The Attempt at a Solution

I do not even know how to start it so any help would be greatly appreciated.

 

[chrsr34]

well, if the apparent weight is the same, that means the masses of the 2 will be different because the densities of each are different. Look up the densities for steel and aluminum (should be given) and solve for mass. Look up your equations for density and volume. solve for the unknowns. it should get things rolling.

Chris

 

[Saladsamurai]

With problems like these, I always find it helpful to draw a free-body diagram for each substance I am investigating.

In this case you should, have weight pointing straight down, buoyant force pointing straight up AND Apparent weight acting upward. I always include W_app as a force on my FBD as this will help me determine my masses. I think of as there must be something measuring the W_app such as a spring scale.

I don't know if this is the customary way to account for W_app, but it has always worked for me.

 

[Shooting Star]

I do not even know how to start it so any help would be greatly appreciated.

Write down Archimedes' Principle and we can proceed from there. Try to think how you can apply that in this problem.

 

[cat_eyes]

Ok, so if (Pax)(Vax) - (Pw)(Vax) = (Pp)(Vp) - (Pw)(Vp)
then (770000)Vax - (1000)Vax = (250000)Vp - (1000)Vp
and (769000)Vax = (249000)Vp
right? The only problem is that I don't know how to find the volume now.

 

[cat_eyes]

Oh is Archimedes Principle the one that says that the buoyant force is equal to the weight of the fluid displaced?

 

[chrsr34]

Ok, so if (Pax)(Vax) - (Pw)(Vax) = (Pp)(Vp) - (Pw)(Vp)
then (770000)Vax - (1000)Vax = (250000)Vp - (1000)Vp
and (769000)Vax = (249000)Vp
right? The only problem is that I don't know how to find the volume now.

You have 2 variables in this problem. Solve for one variable and then plug it back into the orginal equation leaving you only 1 variable to solve for. i.e. Vp = 769000Vax/ 249000. Now plug that in for Vp and solve for Vax.

Chris

 

[cat_eyes]

Ok, thanks :)

 

[cat_eyes]

but if (76900)Vax = (249000)(769000Vax/24900) then wouldn't Vax get eliminated because the 249000s cross each other out?

 

[cat_eyes]

anyone?

 

[chrsr34]

Sorry, I think I was thinking a little unclear. Its been a while. I've done a problem like this in the past but just don't remember how. There should be someone here who knows, but ill look into it tomorrow.

Chris

 

[cat_eyes]

Oh, ok thanks

 

[Shooting Star]

Oh is Archimedes Principle the one that says that the buoyant force is equal to the weight of the fluid displaced?

The eqn has been given to you.

(Pax)(Vax) - (Pw)(Vax) = (Pp)(Vp) - (Pw)(Vp), which means (if multiplied with g),

vol*density*g - vol*(density of water)*g is equal for both. Also,

Apparent wt in water = Actual wt - buoyant force. So you can find the actual weights.

 

[cat_eyes]

Oh! Duh, sorry that makes sense, thanks. :)