- #36
BlackPhysics
- 31
- 8
Work = force * distanceerobz said:Ok, but the units are ##\rm{m}^2## on area. you have ##\rm{m}##, otherwise the values are ok.
Now we can move on to the definition of Work. What is it?
Work = force * distanceerobz said:Ok, but the units are ##\rm{m}^2## on area. you have ##\rm{m}##, otherwise the values are ok.
Now we can move on to the definition of Work. What is it?
Only if the Force is constant over the applied displacement (and in the same direction). Is the Buoyant Force going to be constant over the ##11 \, \rm{cm}## displacment?BlackPhysics said:Work = force * distance
Not necessarily. A formula means nothing unless you can state exactly what the variables mean in relation to each other. For this equation, what must be the relationship between the distance and the force?BlackPhysics said:Work = force * distance
it is not constant...erobz said:Only if the Force is constant over the applied displacement. Is the Buoyant Force going to be constant over the ##11 \, \rm{cm} displacment?
so knowing its not constant now what? how can i calculate the Force.BlackPhysics said:it is not constant...
Have you had any Calculus? If you are given this problem, you should probably be acquainted with the subject.BlackPhysics said:so knowing its not constant now what? how can i calculate the Force.
yes i know calculus. Do i just take the integral of the formula?erobz said:Have you had any Calculus? If you are given this problem, you should be acquainted with the subject.
You could, but there's an easier way. Think of the displaced mass of water. Where was its centre of mass at first, and where is it at the end?BlackPhysics said:yes i know calculus. Do i just take the integral of the formula?
Yes, ##W = \int F \cdot dx ## or you could consider that since the relationship is linear ## \int F \cdot dx = W = F_{avg} \Delta x ## or you could do whatever @haruspex has planned.BlackPhysics said:yes i know calculus. Do i just take the integral of the formula?
F = (0.01963m^2) * 1000 kg. m^3 * 9.8 m/s^2 * (0.11m)erobz said:Yes, ##W = \int F \cdot dx ## or you could consider that since the relationship is linear ## \int F \cdot dx = W = F_{avg} \Delta x ## or you could do whatever @haruspex has planned.
##\pi 2.5^2cm^2## is not 0.0196… m2. Check your conversion from ##cm^2## to ##m^2##.BlackPhysics said:F = (0.01963m^2) * 1000 kg. m^3 * 9.8 m/s^2 * (0.11m)
F = 2.1166W = F * (Delta X)
0.2328 = 2.1166 * 0.11m
Does it look like this?
The average force is just the mass * gravity?
F is the buoyancy forceharuspex said:##\pi 2.5^2## is not 0.0196… Check your conversion from ##cm^2## to ##m^2##.
You still have not said what that F represents. What force is it?
No it isn't. The buoyancy force depends on the weight of the cylinder, which we do not know.BlackPhysics said:F is the buoyancy force
No, the average Force is not ##mg##.BlackPhysics said:F = (0.01963m^2) * 1000 kg. m^3 * 9.8 m/s^2 * (0.11m)
F = 2.1166W = F * (Delta X)
0.2328 = 2.1166 * 0.11m
Does it look like this?
The average force is just the mass * gravity?
##F## is not the Buoyant Force here, it is the force which you push with to counter the Buoyant Force.BlackPhysics said:F is the buoyancy force
F = (0.1963m^2) * 1000 kg. m^3 * 9.8 m/s^2 * (0.11m)
F = 192.42251
Since the force of push is variable, it's not exactly that either.erobz said:F is not the Buoyant Force here, it is the force which you push.
$$dF = \gamma d {V\llap{-}}_D$$haruspex said:Since the force of push is variable, it's not exactly that either.
But the work is the average force times the displacment. You have not calculated the average force, you have calculated the final force at depth ##x= 11 \, \rm{cm}##BlackPhysics said:F = (0.001963M^2) * 1000 kg. m^3 * 9.8 m/s^2 * (0.11m)
F = 2.1166W = F * (Delta X)
0.2328J = 2.1166 * 0.11m
ok so i converted cm^2 to m^2
and the force is not the buoyant force.
So I am guessing its the force that is pushing the cylinder in the water?
How do you calculate that without time? average force = f = m(VI-VF)/Terobz said:But the work is the average force times the displacment. You have not calculated the average force, you have calculated the final force at depth ##x= 11 \, \rm{cm}##
Also, the units of force are missing.
BlackPhysics said:How do you calculate that without time? average force = f = m(VI-VF)/T
Yes, that was my point.erobz said:you have calculated the final force at depth 11cm
That's a fair question. You are right that average force is change in momentum/elapsed time. But here we are dealing with work, not momentum, so you want the force at the average displacement.BlackPhysics said:How do you calculate that without time?
2.327? is that right?haruspex said:Yes, that was my point.
That's a fair question. You are right that average force is change in momentum/elapsed time. But here we are dealing with work, not momentum, so you want the force at the average displacement.
When just starting to push the cylinder down, the slightest force will do. The force needed increases linearly with depth up to the maximum calculated by your equation. So, if you integrate the force wrt depth, what do you get?
W = 2.327 * x?BlackPhysics said:2.327? is that right?
Based on your numbers, it doesn't look right to me.BlackPhysics said:W = 2.327 * x?
0.52889375 is that right?erobz said:Based on your numbers, it doesn't look right to me.
No, I don't get that either.BlackPhysics said:0.52889375 is that right?
You calculated the final (i.e. maximum) force to be 2.116 N (but you keep leaving out the units!).BlackPhysics said:W = 2.327 * x?
1.0577875 is what i got as the avg forceerobz said:Based on your numbers, it doesn't look right to me.
The average Force is nothing more than how you would average two numbers for this problem because the function is linear.
$$ F_{1.05778} = \frac{F(0)+F(2.115575)}{2} $$
Units!BlackPhysics said:1.0577875 is what i got as the avg force
1.0577875 Nerobz said:Units!
Yes i do.erobz said:Now finish it out, and don't forget the units. Do you understand why this works for calculating the area under a line?
Now it's time for the Physicist's to blast you about sig figs! Good Luck!BlackPhysics said:Yes i do.
Avg F = 1.0577875 N
Work = 1.0577875 N * .11M
0.11635 J = 1.0577875 N * .11M
Is this right?