Ch 11.2 Help

Why can't I get 3? Can anyone help? This is what I've tried:

buoyant force = (density of water*Volume)*g
volume=(119.6 cm^3)^(1/3)=4.92694 cm/ 100 cm = (.0492694 m)^3 = .0001196
density of water = 1000
g = 9.81

buoyant force = (1000*.0001196)*9.81

I've been putting the answer as negative because the density of the block shouldn't matter in calculating buoyant force based on the equation on slide 19 in chapter 11.

Can someone please tell me what I'm doing wrong?

I have been working on question 3 for a while I just cant seem to get it. It seems like many people are having problems with this question if anyone has any idea please post it I am sure all of us would really appreciate it.

the above solution is correct. i just tried mine and it's working. it seems that prof. stephens finally fixed it.

Yes, as stated everything seems to be working properly so we should all be able to get 4/4's! Great teamwork guys =]

can someone just explain how they did 1 and 2? I know forcea/force b=area a/areab. when given the diameter do we multiply by 2 to get the area of both and and b? then just divide to get the ratio? i was doing that and it marked me wrong. thanks for your help!

the area of a circle is pi*r^2
so when you solve, divide the diameter by 1/2 and then that equals radius (r). so find the area of a and divide that by area of b. i hope that clarifies your question

can some explain #4. i am totally lost:(

can someone explain how you convert the given volume to kg/m^3 for ques 4? thanks

i need help with #4 also pleaseee..ive tried the way it was explained previously and it came out wrong for me :/

Hey,
I really appreciate everyone's help on this forum, I sometimes post when I can explain the correct solutions, However, I alike many others it seems am having a rough time with questions number 4. Is it possible for someone to readily explain how they arrived at the correct solution. Thanks kindly.

4) A wooden block with a volume of 7.316 x 7.316 x 7.316 cm3 floats on water. What is the volume of the wood above the water level if the density of the wood is 0.6901 g/cm3 (see sheet 20) ? Indicate with a negative (positive) sign whether this fraction is zero (greater than zero) if the density of the wood is greater than the density of water.

Answer is NEGATIVE.

i) First convert the Volume of the object to m^3. The volume given is (7.316)^3=392 cm^3. Now, 1 m^3--->1000000 cm^3, so the volume of the object is 0.000392 m^3.
ii) We know that the density of the fluid (water) is 1000 kg/m^3
iii) Convert the density of the object to SI units: 1 g/cm^3 has 1000 kg/m^3. 0.6901 g/cm^3 * 1000=690.1 kg/m^3
iv) The formula that we need to use is Vsub=(rho_object/rho_fluid)*Vobject. Now we are calculating the volume of the submerged part. So Vsub=0.0002705 m^3
v) The question is asking for the ABOVE volume, so we subtract the submerged volume from the total volume of the object that we have in the first step. So, Vabove=0.000392-0.0002705= 0.0001215 m^3. And this is our final answer.

Hope this helps.

4)

1. Calculate the volume in SI units.
2 volume - density*volume will give the answer.

Just use the given density. After all the unit conversions and such you wind up with the same thing.

For question three just take your cm^3 value and divide by 1000. Then multiply by 9.81. Should work.

5) A hydraulic lift has a circular piston of 41.25 cm diameter at the point where the heavy load is to be lifted (see sheet 15)and a small diameter piston with a diameter of 3.549 cm which activates the lift. What is the ratio of the load over input force ? Indicate with a negative (positive) sign whether the liquid on the input side has to be pushed along a larger (same) distance as the distance the load-piston travels during the lift.

.4125^2/.03549^2 gives me 135 and i put it as negative but i got it wrong...what did i do wrong?

nevermind it worked