Kaplan High Yield- Hydrostatics - help!

[vegan1]

I can't seem to get to the right answers on these questions and am not sure where I am going wrong:

1. A block of mass 5 kg and density 3 g/cm3 is hung from a string while submerged in water. What is the tension in the string? (Answer: 32.66 N)
Would I solve it like this? mg(in water) = mg(in air) - Buoyant Force
and say that Tension = mg(in water)

I keep getting that Buoyant Force is 160 N and mg(air) is 50 N.

2. A cube of side length 3 cm floats in water (p = 1 g/cm3) with 1 cm floating above the water. What is the density of this cube? (Answer: 2/3 g/cm3)
I got 18 cm3 as vol. of submerged part of cube, which gives a Buoy. F of 180 N. Block is floating so is in eqlb. mass of submerged part must then be 18 kg. Since 2/3 is submerged, total mass must be 27 kg. 27 kg / 27 cm3 gives a density of 1.

3. A piece of cork (p = 0.2 g/cm3) with mass 5 grams is held underwater. When the cork is released, what is its initial acceleration? (Answer: 49 m/s2)
Vol. of cube: m/p = 5/.2 = 25 cm3
Buoy. F = ma = pVg = (1 g/cm3) x (25 cm3) x (10) = 250 g = .25 kg = 2.5 N = ma.
a = 2.5 N / .005 kg = 12500 m/s2.

I did this without calculator so I apologize if there's stupid math errors, but I can't find where I'm going wrong. (If you want to ignore my work and just tell me how you get the right answer that would be great, too).

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[Rabolisk]

I could give you the answer to these questions.. but I prefer giving hints so you can work them out.

1. What is the density of water?
2. What is the weight of an object?
3. What are the forces acting on the cork?

 

[Hexon]

hey

i was just about a start a thread on these very questions; my problem is mainly in the second question. i converted everything into metres and kilograms.
My working so far was:

volume of submerged area:
V= Ah= (0.03m)^2(0.02m)^2

Buoyant Force = (rho)(g)(v)
= (1000 kg/m^3)(10m/s^2)(0.0018m^3)
=18N

i drew a free body diagram and reasoned that the buoyant force must equal the weight of the object (since the object is floating), thusly Force (weight) = Force (buoyant) = 18N
Force (weight) = mg = 18N. Taking g as 10m/s^2, i got m= 1.8 kg.

subbing into the density equation for the cube:

rho= m/v = 1.8kg/ 1.8 x 10^-3 m^3 = 1000 kg/m^3.

unfortunately, as we've seen from the OP's answers, it's not the right one. so pretty, please, where did i go wrong?

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

as for the last one, i got this far:
volume of cork= 25 cm^3 as our OP got,
then

Buoyant Force = mg = (0.005 kg)(10 m/s^2) = 0.05N

Buoyant Force = (rho)(g)(v of submerged)

v of submerged = buoyant force/ (rho)(g) = 0.05 N / (1000 kg/m^3) (10m/s) = 5 x 10^ -6 m^3

 

[Ssina]

I can't seem to get to the right answers on these questions and am not sure where I am going wrong:

1. A block of mass 5 kg and density 3 g/cm3 is hung from a string while submerged in water. What is the tension in the string? (Answer: 32.66 N)
Would I solve it like this? mg(in water) = mg(in air) - Buoyant Force
and say that Tension = mg(in water)

I keep getting that Buoyant Force is 160 N and mg(air) is 50 N.

you know that the density of water is 1 g/cm3= 1000 kg/m3 and the density of the block is 3 g/cm3= 3000 g/cm3
also tension doesn't always equal to mg, so just use T (what we are trying to solve for)....
this object is 3x as dense as water so it will sink, but we have attached the sting with the tension. If you draw the diagram, you will have mg down and tension and bouyancy up:
mg= T+ Fb
mg=T+(density of water).g.V(box submerged)
density of block=3000kg/m3=m/V=5/V.... so V=5/3000... now plug in the above equation
5 x 10= T + 1000 x 10 x 5/3000 ... 50 = T + 16.666, so the T = 33.3N

2. A cube of side length 3 cm floats in water (p = 1 g/cm3) with 1 cm floating above the water. What is the density of this cube? (Answer: 2/3 g/cm3)
I got 18 cm3 as vol. of submerged part of cube, which gives a Buoy. F of 180 N. Block is floating so is in eqlb. mass of submerged part must then be 18 kg. Since 2/3 is submerged, total mass must be 27 kg. 27 kg / 27 cm3 gives a density of 1.

for this I wouldn't even bother with bouyancy... you just need this simple formula: (D denoting density)
D-object/D-fluid=V-submerged/V-total
and you don't need to do any conversions here since everything is in g/cm3
D-fluid= 1g/cm2
V-submerged=2x3x3 18cm3 (since 2cm is submerged)
V-total=3x3x3= 27 cm3
so D-object= 18x1/27=2/3
I always think of this formula with icebergs... density of ice is around 0.92 g/cm3, while density of liquid water 1g/cm3 (well we are talking about sea water and it's a bit higher, but 1 is good enough)... so if you have:
D-ice/D-water= 0.92 =V-submerged/V-total
so 92% of an iceberg is under water!

Buoyancy is only with the volume submerged so the weight of the box must equal to buoyancy of submerged part in order to float:
Fg=Fb
m.g=D-water.g.V-submerged only
D=m/V-total, so the m of box= D.V-total
plug in and you get:
D-object.V-total.g=D-water.g.V-submerged.... cancel out the g and you get the same equation.

3. A piece of cork (p = 0.2 g/cm3) with mass 5 grams is held underwater. When the cork is released, what is its initial acceleration? (Answer: 49 m/s2)
Vol. of cube: m/p = 5/.2 = 25 cm3
Buoy. F = ma = pVg = (1 g/cm3) x (25 cm3) x (10) = 250 g = .25 kg = 2.5 N = ma.
a = 2.5 N / .005 kg = 12500 m/s2.

I did this without calculator so I apologize if there's stupid math errors, but I can't find where I'm going wrong. (If you want to ignore my work and just tell me how you get the right answer that would be great, too).

for this one I got 40 (so about 39 something if you use g as 9.8) idk how you would get 49!!! but maybe someone can point out where I made any mistake...

as far as the forces go, we have buoyancy up, weight down, and we are accelerating upwards:

Fb - mg = ma (so although buoyancy is causing the acceleration, you still have mg down)
D-water.g.V-cork - mg = ma
Vcork= m/D-cord
D-water.g.m-cork/D-cord - mg = ma
1000x10x 0.005/200 - 0.005x10= 0.005 x a
and a=39 something
I like to think it's 39 and they printed it wrong!! but I'm sure someone will correct me soon!

 

[Hexon]

hey Ssina,

thanks for the help, just wanted to ask you for q2, where did you get this formula?

D-object/D-fluid=V-submerged/V-total

it's the first i've seen of this relationship between specific gravity and the volumes of the submerged and total volume.
It definitely helped me in solving this q that was probably the key piece of information i was missing:S. thanks heaps!

one 'pitfall' i recognise is the conversion of densities from g/cm cubed to kg/m cubed.
because you go from 1 g/cm cubed to 1000 kg/m cubed, it's easy to get different values.

for your last question, should the mass change since the density has changed from 0.2 g/cm cubed to 200kg/m cubed? i mean 5g is the same as 0.005kg

 

[Pisiform]

hey Ssina,

thanks for the help, just wanted to ask you for q2, where did you get this formula?

it's the first i've seen of this relationship between specific gravity and the volumes of the submerged and total volume.
It definitely helped me in solving this q that was probably the key piece of information i was missing:S. thanks heaps!

one 'pitfall' i recognise is the conversion of densities from g/cm cubed to kg/m cubed.
because you go from 1 g/cm cubed to 1000 kg/m cubed, it's easy to get different values.

for your last question, should the mass change since the density has changed from 0.2 g/cm cubed to 200kg/m cubed? i mean 5g is the same as 0.005kg

When anything is floating Fnet = 0, so

F(Buyoant) = F(weight)

(rho_fluid)*g*(V_submerged) = (mass_object)g (replace mass by density*V)

(rho_fluid)*g*(V_submerged) = (rho_object)*(V_entire object)*g (g cancels out)

(V_submerged) / (V_entire object) = (rho_object) / (rho_fluid)



1g/cm^3 = 1000kg/m^3 => For water

Without going into any arithmetic detail, just remeber that when you need to convert
g/cm^3 to kg/m^3 you have to multiply g/cm^3 by 1000. E.g. Mercury has a density of
13.546g/cm^3 ... to convert into kg/m^3 ...multiply by 1000 ... 13546kg/m^3


And yes mass should change. If you have density in g/cm^3 then mass should be in grams, volume
should be in cm^3 and g should be cm/s^2

If density in kg/m^3 then mass should be in kg, Volume in m^3, and g in m/s^2

(btw,Ssina is correct about the answer... Kaplan high yield has a lot of typos)

 

[Ssina]

hey Ssina,

thanks for the help, just wanted to ask you for q2, where did you get this formula?

it's the first i've seen of this relationship between specific gravity and the volumes of the submerged and total volume.
It definitely helped me in solving this q that was probably the key piece of information i was missing:S. thanks heaps!

one 'pitfall' i recognise is the conversion of densities from g/cm cubed to kg/m cubed.
because you go from 1 g/cm cubed to 1000 kg/m cubed, it's easy to get different values.

for your last question, should the mass change since the density has changed from 0.2 g/cm cubed to 200kg/m cubed? i mean 5g is the same as 0.005kg

no problem! I think I came across it actually in the Princeton Review physics book, but you don't really need it... if you follow the above (Pisiform) you can easily come up with it also... I was personally pretty bad with buoyancy so had to cover it a bit extra! also with the densities in different units (I would know water in g/cm3, kg/m3, g/ml, kg/L just in case) plus after you get the same question wrong couple of times, it will bother you enough that you will make sure to know it in your sleep..... and of course it won't be on your MCAT once you know it by heart!

 

[Hexon]

thanks for the explanations, guys