PHYSICISTS OF NEWSCHOOLERS! I need your help...

E

Edward.

Active member
An object is solid throughout. When the object is completely submerged in

ethyl alcohol, its apparent weight is 15.9

N. When completely submerged in water, its apparent weight is 12.1

N. What is the volume of the object?

Here's a hint that my professor provides:

When and object is completely submerged in a fluid, its apparent weight is

equal to its true weight minus the upward acting buoyant force:
image037.gif
.

Write

this equation for each fluid; once for the case that the object is submerged

in ethyl alcohol and once for the case that it is submerged in water. The

unknown volume V will appear in the expression for the buoyant

force FB. Combine the two equations and

solve for V.

Teach me how to solve it please!
 
where it says "click for larger image" we are given this formula:

Apparent Weight = Actual Weight - Buoyant Force
 
i could've done this last year.. but i forgot everything i learned in physics. Im sorry to be no help
 
F_B = roh * V

F_B is the boyant force

roh is the density of the liquid the object is submerged in

V is the volume... What you're looking for.

Write out the two equations like he says. The "actual weight" is the same in each equation so solve each equation for it and set the two equal to each other. The only unknown left is the Volume. Solve for it!
 
I took physics 1 three years ago, but here is what I could find in my old notes.

Fb (water) = (density water)(volume)(gravity)

Fb (alcohol) = (density ethyl alcohol)(volume)(gravity)

SInce Wt = mg-Fb

we get: 15.9N = mg - (1000 kg/m^3)(V)(9.81m/s^2) - water

12.1 N = mg - (806kg/m^3)(V)(9.81m/s^2) - alcohol

subtract water - alcohol

we get: 3.8N = V(g)(Density of water - density of alcohol)

3.8N = V(9.81m/s^2)(1000 - 806)kg/m^3

V = 3.8N/(9.81m/s^2)(194kg/m^3)

V = 1.9967E-3 m^3

 
Thanks everyone I just didn't know that the volume used to calculate the buoyant forces for each liquid was the same as the volume of the object
 
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