query 261

course phy201

Query 26.1026. More Forces (buoyant)

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Question: `q001. Note that this assignment contains 3 questions.

. Water has a density of 1 g per cm^3. If an object is immersed in water, it experiences a buoyant force which is equal to the

weight of the water it displaces.

Suppose that an object of mass 400 grams and volume 300 cm^3 is suspended from a string of negligible mass and volume, and is

submerged in water. If the mass is suspended in equilibrium, what will be the tension in the string?

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Your solution:

The object has a bouyancy of 400 g per cm^3 so this would lessen the tensin from the string.

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Given Solution:

The 400 g mass will experience a downward gravitational force of .4 kg * 9.8 meters/second^2 = 3.92 Newtons. It will also experience

in upward buoyant force equal to the weight of the 300 cm^3 of water it displaces. This volume of water, at 1 g per cm^3, will have a

mass of 300 grams and therefore a weight of .3 kg * 9.8 meters/second^2 = 2.94 Newtons.

The forces acting on the mass are therefore the downward 3.92 Newtons of gravity, the upward 2.94 Newtons of the buoyant force and the

tension, which we will call T, in the string. Since the system is in equilibrium these forces must add up to 0. We thus have

-3.92 Newtons + 2.94 Newtons + T = 0, which has solution

T = .98 Newtons.

In common sense terms, gravity pulls down with 3.92 Newtons of force and the buoyant force pushes of with 2.94 Newtons of force so to

keep all forces balanced the string must pull up with a force equal to the .98 Newton difference.

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Self-critique (if necessary):

ok, so the gravitataion force of the 400g object is

.4kg * 9.8m/s^2 = 3.92N

the volume of the object is 300 cm^3 so it dissplaces 300 g per cm^3 of water giving the water that has been displaced a mass of 300g,

.3kg * 9.8m/s^2 = 2.94N ( this is the force upward opposing direction of gravity)

-3.92N + 2.94N= .98 N total

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Question: `q002. A solid cylinder has a cross-sectional area of 8 cm^2. If this cylinder is held with its axis vertical and is

immersed in water to a depth of 12 cm, what will be the buoyant force on the cylinder?

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Your solution:

not sure

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Given Solution:

At a depth of 12 cm, the volume of the immersed portion will be 12 cm * 8 cm^2 = 96 cm^3. This portion will therefore displace 96

grams of water. The weight of this displace water will be .096 kg * 9.8 meters/second^2 = .94 Newtons. This will be the buoyant force

on the cylinder.

STUDENT COMMENT: oh so the water above the cylander is displaced. i guess the cylander reaches this distance to the surface and maybe

higher.

INSTRUCTOR RESPONSE: The mechanism isn't specified here, but you are told that the cylinder is immersed to depth 12 cm. The cylinder

might be held there by some other force, it might be bobbing up or sinking down at a certain instant, etc.. As long as it displaces 96

cm^3 of water, the buoyant force will be as calculated.

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Self-critique (if necessary):

ok I see

8cm^2 * 12cm = 96cm^3, so 96 grams of water is displaced .096kg * 9.8m/s^2 =.96N with is the bouyant force

I was thinking that i needed to use the formula for a cylinder, but I was not sure.

The key is the volume of water displaced. If the water is displaced by a cylinder, as here, you need the volume of the cylinder. The displaced water volume may of course by spherical, conic, or any other shape. If there's a way to calculate that volume, then that's something you'll probably always want to do.

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Question: `q003. The solid cylinder in the preceding problem has a total length of 18 cm and a mass of 80 grams. If the cylinder is

immersed as before to a depth of 12 cm then released, what will be the net force acting on it at the instant of release?

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Your solution:

The bouyancy = .96N

Force of the cylinder = .08kg * 9.8m/s^2= .78N

ok so .96- .78= .18 in the upward direction

a= f/m

a= .18/ .080= 2.25m/s^2

Ok I see howt his is done I had to look at the solution but I understand.

Gravity is working on the cylinder in the down direction, the buoyancy is acting on the cylinder in the direction opposing that of

gravity. Subtracting the buoyance from the gravity forces we get the net force from that we vcan obtain the acceleration of the object

as is rises the 12 cm to the surface.

confidence rating #$&*

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Given Solution:

The buoyant force on the cylinder is still .94 Newtons, directed upward. Gravity exerts a downward force of .080 kg * 9.8

meters/second^2 = .79 Newtons, approximately. The net force on the cylinder is therefore .94 N - .79 N = .15 N, directed upward. This

will give its 80 gram mass and acceleration a = F / m = .15 N / .080 kg = 1.875 m/s^2.

Note that as the cylinder rises less and less of its volume is submerged, so the buoyant force will decrease while the weight remains

the same. Until the buoyant force has decreased to become equal and opposite to the weight, the net force will continue to be upward

and the cylinder will continue to gain velocity. After this instant the cylinder will continue to rise, but the net force will be

downward so that the cylinder will begin slowing down. Eventually the cylinder will come to rest and the net downward force will

cause it to start descending once more. It will continue descending until the net force is again 0, at which the time it will have a

downward velocity that will carry it beyond this point until it again comes to rest and the cycle will start over again.

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Self-critique (if necessary):

Ok I see how this is done I had to look at the solution but I understand.

Gravity is working on the cylinder in the down direction, the buoyancy is acting on the cylinder in the direction opposing that of

gravity. Subtracting the buoyance from the gravity forces we get the net force from that we vcan obtain the acceleration of the object

as is rises the 12 cm to the surface.

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Self-critique rating #$&*ent:3

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&#Your work looks good. See my notes. Let me know if you have any questions. &#