FBD help

[mcgill2012]

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what does the free body diagram look like for this situation? ie: how do the forces work? is the weight vector the actual weight of the object, or the apparent weight?

an object with 20 grams of mass is immersed in a fluid and apparently loses 5 grams of mass.

 

[mcgill2012]

sorry i know some of my PS questions may seem trivial, but im learning physics for the first time for the mcat and definately struggling.

 

[fizzgig]

object weighs mg, vector points down from the object's center of mass. if it's hanging around the earth this vector is always there, always draw it, always the full mg value.

object is submerged and 'apparently' loses 5grams of mass. so if you put a scale under it as it sits submerged, part of the force is being counteracted, 5g of weight. so there is a force pointing upwards equal to 5g (g is gravity here not grams btw).

F net = mg - Fbuoy = mg - 5g

note that Fbuoy points up and is located at the object's center of buoyancy, its geometric center (EK says it is where the ctr of mass would be if the object were of uniform density... just something to keep in mind - ctr of mass and ctr of buoyancy do not HAVE to be in the same place). if they're not passing through the same point the object will rotate until they do.

also i didn't convert g to kg i realize bc i'm lazy so the Force you get isn't Newtons at the end. just for what it's worth...

 

[mcgill2012]

thanks you! so if there is a net force of 15g in the down direction, then the object is accelerating downwards and sinking, right?
if an object is floating, does it apparently lose all its weight, since mg must be counterbalanced entirely by Fbuoy for it to float in translational equilibrium?

 

[fizzgig]

yes, if the net force is down the object did not displace enough fluid to counter its weight and it will sink to the bottom.

if an object is floating, yes you can say it apparently lost all its weight. the object sinks into the fluid until the amount of water it displaces is equal to the object's weight. once the 20g object has displaced a volume water that also has mass of 20g, the object stops sinking into the fluid.

 

[mcgill2012]

ok.. so how would you do the following question?
an object with 15 grams of mass is immersed in benzene and suffers an apparent loss of mass of 5 grams. what is the approximate specific gravity of the object if specific gravity of benzene is 0.7?

 

[fizzgig]

the first way i looked at this:
benzene only removed 1/3 of the object's weight. if benzene were to make the object apparently weightless, it would need to triple its Fb, and to do that it would need to triple the weight of fluid the object pushed aside, and to do that it would need to be be 3x as dense. 0.7x3 = 2.1

then i remembered there is a formula. i desperately need to make an equation sheet and take it EVERYWHERE with me for the next month.

Fb = mfluid*g = rhofluid*volfluiddisplaced*g = rhofluid*volobject*g
volobject = massobject/rhoobject
Fb = (rhofluid/rhoobject)*massobject*g
5*10 = 0.7/x *15*10
x/0.7 = 150/50
x/0.7 = 3
x = 0.7*3 = 2.1

hope that's right..

 

[mcgill2012]

thank you! the answer is 2.1. clearly you have nothing to worry about, youre set.

 

[fizzgig]

lol, thanks. i think i'm finally to the point where i can do almost any problems, given 10 minutes and access to the internet. !! O.O !!

sooo all i need to do now is get it all to stay between my ears and become superfast. like a superfast superhero. i'm going to make a cape.