An object completely submerged in a fluid with a specific gravity of 3 has an apparent weight loss of 40 N. What is the object's specific gravity?
Specific gravity and Bouyant Force
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I've been struggling with these question types and am looking for an answer as well.
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Are you sure you're giving all the information that they gave? I would think you'd need a mass of the object but let's see if we can get through it with proportional reasoning.
You need to think about buoyant force and specific gravity.
SG = density of substance / density of water (1000kg/m^3)
Force Buoyant (Fb) = (rho)(g)(V) where rho is fluid density, g is gravitational acceleration and V is volume of fluid displaced.
We can think of the apparant loss in weight as the magnitude of our Fb. We had only gravity acting upon the object prior to submerging it where we added a buoyant force. That Fb is responsible for the 40N loss in apparent weight.
We can figure rho of the fluid by its specific gravity. SG of 3 = 3000kg/m^3 density.
40N = (3000kg/m^3)(10m/s^2)(V). Solve for V.
V = (4/3000 m^3)
Since we're fully submerged, the volume of fluid displaced is equal to our object's volume.
Where you would calculate your object's density from for its specific gravity would be like this:
rho(object) = mass of object / object density. --> rho = (3000m)/4.
Plug in the rho object to the SG equation.
SGobject = rho object / (1000kg/m^3) = 3m/4
SGobject = 3m/4 where you would need the mass of the object to get the actual specific gravity.
You need to think about buoyant force and specific gravity.
SG = density of substance / density of water (1000kg/m^3)
Force Buoyant (Fb) = (rho)(g)(V) where rho is fluid density, g is gravitational acceleration and V is volume of fluid displaced.
We can think of the apparant loss in weight as the magnitude of our Fb. We had only gravity acting upon the object prior to submerging it where we added a buoyant force. That Fb is responsible for the 40N loss in apparent weight.
We can figure rho of the fluid by its specific gravity. SG of 3 = 3000kg/m^3 density.
40N = (3000kg/m^3)(10m/s^2)(V). Solve for V.
V = (4/3000 m^3)
Since we're fully submerged, the volume of fluid displaced is equal to our object's volume.
Where you would calculate your object's density from for its specific gravity would be like this:
rho(object) = mass of object / object density. --> rho = (3000m)/4.
Plug in the rho object to the SG equation.
SGobject = rho object / (1000kg/m^3) = 3m/4
SGobject = 3m/4 where you would need the mass of the object to get the actual specific gravity.
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What is the source of the question? and is it multiple choice?
From knowing that it is submerged, you know the answer must be >3
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I'm struggling with these as well, the textbook approach to these problems does not teach this quick ratio method being tested on the mcat. I can FBD and sum forces and work out all the math, and then 7 minutes later I just wasted a ton of time getting the answer. The explanations on the practice mcats are not really teaching the method, and i haven't found it in the practice books either.
That's better! I was lurking in this thread and amazed at how difficult it was to solve this one. Anyway, here's a trick I learned from TBR.
For a submerged object, the relative densities = weight / buoyant force. That is:
Density of object / density of fluid = mg / B
Rearranging it gives: density of object = (mg / B) * density of fluid.
Now, the apparent weight loss (40N) is due to the buoyant force, so B = 40.
Solving the formula: density of object = (120/40) * 3 = 9
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Change in force = rho fluid * g * V displaced
Rho object = mass object / V displaced
S object = Rho object / 1000kg/m^3
Magical equation combination activate:
(Mass object * S fluid * g) / Fb = S object
You shouldn't need to draw a FBD if you can see that the change in force is your buoyant force. It's just quick multiplication and division.
I am betting Monopoly money that this question prompt is missing a mass of 12kg in the instructions. Someone feel free to prove me wrong and solve it without a given mass. If you google the question, the first result is an old SDN post from an EK book with that mass listed so...
You can also think of S as the weight change in given fluid divided by the weight change in water
S = (w - w1) / (w - w2)
CAREFUL.
Water has a density of approx. 1000kg/m^3. Specific gravity is calculated as some fluid or object density / density water. The S = 3 for the fluid in the paragraph is not the S of water.
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