Ch 11.2 Help
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Ch 11.2 Help
Question 1:
A hydraulic lift has a circular piston of 23.36 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 3.806 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 loadpiston travels during the lift.
How to solve:
Question 2:
A hydraulic lift has a circular piston of 12.34 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 1.435 cm which activates the lift. What is the ratio of the piston travel distance at the input end over the value at the load end ? Assume the shape of the vessel in those parts where the pistons have to move to be cylindrical. Hint: the same fluid volume gets displaced on either end.
How to solve:
Question 3:
A wooden block with a volume of 76.42 cm3 is held under water. What is the buoyant force on the block (see sheet 19)? Indicate with a positive (negative) sign whether the force you calculate depends (does not depend) on the density of the wood.
How to solve:
Question 4:
A wooden block with a volume of 3.912 x 3.912 x 3.912 cm3floats on water. What is the volume of the wood above the water level if the density of the wood is 0.7545 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.
How to solve:
A hydraulic lift has a circular piston of 23.36 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 3.806 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 loadpiston travels during the lift.
How to solve:
Question 2:
A hydraulic lift has a circular piston of 12.34 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 1.435 cm which activates the lift. What is the ratio of the piston travel distance at the input end over the value at the load end ? Assume the shape of the vessel in those parts where the pistons have to move to be cylindrical. Hint: the same fluid volume gets displaced on either end.
How to solve:
Question 3:
A wooden block with a volume of 76.42 cm3 is held under water. What is the buoyant force on the block (see sheet 19)? Indicate with a positive (negative) sign whether the force you calculate depends (does not depend) on the density of the wood.
How to solve:
Question 4:
A wooden block with a volume of 3.912 x 3.912 x 3.912 cm3floats on water. What is the volume of the wood above the water level if the density of the wood is 0.7545 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.
How to solve:
Guest 01 Guest
Re: Ch 11.2 Help
so im pretty sure im doing number three right but the stupid machine says otherwise...
i first convert the volume of the block given into m^3 then i multiply that vaule by the density of water (1E3) and then by 9.8. So what am i doing wrong?
i first convert the volume of the block given into m^3 then i multiply that vaule by the density of water (1E3) and then by 9.8. So what am i doing wrong?
guesto Guest
Re: Ch 11.2 Help
I can't seem to get question 4...
Vsub= the value they gave to the third power.
Vsub=(densityblock/densityh20)Vblock
VblockVsub= volume above water.
I don't think i messed up converting but who knows...
Vsub= the value they gave to the third power.
Vsub=(densityblock/densityh20)Vblock
VblockVsub= volume above water.
I don't think i messed up converting but who knows...
cheers Guest
Re: Ch 11.2 Help
hey kathleen if ur having trouble with question four what about the rest of them...can you pls post equations of question 1,2,and 3...thanks
11.2 que Guest
Re: Ch 11.2 Help
Q1)
Force_a/Force_b = Area_a/Area_b
Q2) Same for question 2
Q3) No idea
Q4) you are right, you mustve messed up converting
Force_a/Force_b = Area_a/Area_b
Q2) Same for question 2
Q3) No idea
Q4) you are right, you mustve messed up converting
yay Guest
Question 2 and 3
I am really confused on 2 and 3 if someone could please post detailed description of how they did the problem would really appreciate it.
sedwards Guest
what?
I never even posted question four, you confused me with the other person. I only posted the signs. I haven't begun working on this quiz yet but I'll let you know when I figure something out. How about try working it on your own first? This website wasn't made to just exchange answers, you guys should try to learn something and work it out before resorting to this.
Kathleen Guest
Re: Ch 11.2 Help
Someone already posted the solutions for questions 1 and 2 guys, try reading the posts before commenting.
One is exactly as stated, the ratio of the forces is equal to the ratio of the areas, as in value^2
Convert the values to meters, Do first value listed (larger #) ^2/second value listed(smaller number)^2
It is the same for both questions one and two
However, question one is negative and two is positive
One is exactly as stated, the ratio of the forces is equal to the ratio of the areas, as in value^2
Convert the values to meters, Do first value listed (larger #) ^2/second value listed(smaller number)^2
It is the same for both questions one and two
However, question one is negative and two is positive
Kathleen Guest
question 3
I have attempted to do question 3 I just cant figure it out which is why I am here. If anyone can help me please post. Thank you.
hwilson Guest
Re: Ch 11.2 Help
Seriously is there something wrong with question three or has anyone got it yet?
I converted the given vaue to m^3 then multiplied that by 10^3 and then by 9.8 but no matter how many times i try, it never comes out right...
I did it ont he ractice problems and it marked that correct.... wtf?
I converted the given vaue to m^3 then multiplied that by 10^3 and then by 9.8 but no matter how many times i try, it never comes out right...
I did it ont he ractice problems and it marked that correct.... wtf?
naaaaaa Guest
Question 3
A wooden block with a volume of 76.42 cm3 is held under water. What is the buoyant force on the block (see sheet 19)? Indicate with a positive (negative) sign whether the force you calculate depends (does not depend) on the density of the wood.
What you must do for this question is
1:Take your volume in cm^3 and change it to m^3.
To do this you have to take the cube root of the number they give you so for the example above you need to take the cuberoot(76.42)=4.243612088 this now gives us one side of the wooden block.
Take this number and change it into meters> 4.243612088cm=.04243612088m.
Take this number and then raise this to get a new volume in m^3 so
.04243612088^3=7.642e5. Use this number as the volume
Density=10^3
Density*volume=.07642
This is our mass. Multiply that by g=9.81 to get your final answer.
The answer is negative
What you must do for this question is
1:Take your volume in cm^3 and change it to m^3.
To do this you have to take the cube root of the number they give you so for the example above you need to take the cuberoot(76.42)=4.243612088 this now gives us one side of the wooden block.
Take this number and change it into meters> 4.243612088cm=.04243612088m.
Take this number and then raise this to get a new volume in m^3 so
.04243612088^3=7.642e5. Use this number as the volume
Density=10^3
Density*volume=.07642
This is our mass. Multiply that by g=9.81 to get your final answer.
The answer is negative
Guest24 Guest
Re: Ch 11.2 Help
that still doesn't work for question three, i can't seem to figure it out either.
k Guest
question 4
The question is asking for the volume of the wood ABOVE water. You know how to calculate the volume of the wood under water (submerged) which is Vsub=(ρ of object/ ρ of fluid) * Vobject.
Rho (density) of object = no. given in g/cm^3 * 1000 (keep in mind that 1 g/cm^3=1000 kg/m^3)
Rho (density) of fluid = 1000 kg/m^3
Volume of object = i.e. (3.457/100)^3, where 3.457 was one of the sides of the wooden block in cm. Divide by 100 to convert to meters. Raise it to the third power to calculate volume.
So, Vabove = Vobject  Vsub
Hope this helps.
Rho (density) of object = no. given in g/cm^3 * 1000 (keep in mind that 1 g/cm^3=1000 kg/m^3)
Rho (density) of fluid = 1000 kg/m^3
Volume of object = i.e. (3.457/100)^3, where 3.457 was one of the sides of the wooden block in cm. Divide by 100 to convert to meters. Raise it to the third power to calculate volume.
So, Vabove = Vobject  Vsub
Hope this helps.
guest145 Guest
number 3
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?
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?
student Guest
question 3
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.
sedwards Guest
question 3
the above solution is correct. i just tried mine and it's working. it seems that prof. stephens finally fixed it.
guest145 Guest
Re: Ch 11.2 Help
Yes, as stated everything seems to be working properly so we should all be able to get 4/4's! Great teamwork guys =]
Kathleen Guest
questions 1 and 2
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!
helppp Guest
Re: Ch 11.2 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
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
Daisy Guest
question 4
i need help with #4 also pleaseee..ive tried the way it was explained previously and it came out wrong for me :/
help101 Guest
Question #4
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.
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.
shawty Guest
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