Showing posts with label Buoyant Force. Show all posts
Showing posts with label Buoyant Force. Show all posts

Mechanical Properties of Fluids Problems and Solutions Five

We are solving series of problems in Mechanical properties of Fluids.Viscosity is the property of the fluid due to which its relative motion is opposed. When a fluid is moving, it moves like different set of layers, each layer moving in the forward direction opposed the lower layers motion in the forward direction and thus viscous force is generated. It is directly proportional to the area of cross section of the fluid, velocity of the fluid flow and it is inversely proportional to the width of the layer. We can find coefficient of viscosity in this way. When a spherical body is moving in a fluid due to opposing viscous force and upward upthrust and weight in downward direction, some where the resultant force becomes zero and the body acquires a constant velocity called terminal velocity.

Problem

A plate of area 100 square centimeter is placed on the upper surface of castor oil having only 2 mm thickness. Coefficient of thickness is given to us and we need to measure the horizontal force required to move the plate with a certain velocity and the problem is as shown in the diagram below.


Solution

We can solve the problem using the very definition of coefficient of viscosity as shown in the diagram below. We know that the viscous force is directly proportional to the area of cross section,velocity of fluid flow and inversely proportional to the distance between the layers. Solution is as shown in the diagram below.


Problem

A vessel has a height of 40 meter. It has three horizontal tubes each of same diameter and length at different heights from the base as shown in the problem below. We need to find out the length of the single pipe of the same diameter that has to be replaced instead of three pipes so that fluid flow is same.


Solution

The sum of rate of flow through each pipe has to be the fluid flow in the single pipe. We can use Poisellie’s equation and find the length of the new pipe as shown in the diagram below.


Problem

A cylindrical tank has a hole of known area at the bottom. If the water is allowed to flow into the tank from a tube above it with a known rate, we need to find the maximum height of the fluid in the cylinder. Problem is as shown in the diagram below.


Solution

At the maximum possible height of the liquid in the cylinder, the rate of fluid in is equal to the rate of the fluid out. Rate of fluid out can be written in terms of equation of continuity and we also know that the velocity of the fluid coming out of the hole is similar to velocity of the freely falling body. Taking this into consideration, we can solve the problem as shown in the diagram below.



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Mechanical Properties of Fluids Problems and Solutions Four

We are solving series of problems on fluid dynamics. Viscous force acts on all bodies that have relative motion. If a spherical ball is falling through a fluid, it experience a viscous force and it can be measured using a law called Stroke’s law. According to this rule, viscous force is directly proportional to the square of the radius of the ball. There is weight of the ball acting in the downward direction and upthrust acting always in the upward direction. Viscous force always acts opposite to the relative motion of the body. After some travel, the resultant force acting on the body becomes zero and the body acquires a constant velocity and that velocity is called as terminal velocity. 

Problem

An open U tube contains mercury. When certain amount of water of known height is filled in one tube over mercury, we would like to measure the difference between the liquid levels in both the sides and the problem is as shown in the diagram below.


Solution

We know that at the bottom of the system both the pipes has one common point whose height is same and hence the pressure is same at both the points. We know that the pressure can be expressed as the product of height of the liquid, density of the liquid and acceleration due to gravity at the given place. Thus by equating the pressure at both the points we can solve the problem as shown in the diagram below.


Problem

A sphere of known radius is having a cavity of half the radius of the sphere. It is found that the sphere is just floating in the water with its highest point in touch with the water. We need to find the specific gravity of the material and the problem is as shown in the diagram below.


Solution

We know that when the sphere is just floating, its weight is balanced by the upthrust acting on it. We know that upthrust is the product of volume of the fluid displaced, density of the fluid and acceleration due to gravity. As the body is completely immersed in the fluid, volume of the fluid displaced is equal to the volume of the body itself. We know the values of density of water and acceleration due to gravity.

Weight acting in the downward direction is only due to the mass present in the system. We can write mass as volume and density product of the body. Volume of the content means we shall count only the part of mass present in the system. Thus we can measure the weight and equate it ot the upthrust as shown in the diagram below.


Problem

The upthrust force acting on the wing of aeroplane is given to us. Velocity of the air on its lower surface and area of cross section is also given to us and we need to find the velocity of air on its upper surface ad the problem is as shown in the diagram below.


Solution

We can assume that the wing is horizontal and hence there is no difference in the gravitational energy. As velocity is different kinetic energy per unit mass and hence pressure energy are going to be different. We can apply Bernoullie’s theorem, we can solve the problem as shown in the diagram below.


Problem

Due to a disease main artery expanded in its area of cross section as per the given data in the problem. Velocity of the blood in non expanded area is given to us and blood density is also given to us. We need to find the excess pressure developed due to this and the problem is as shown in the diagram below.
  

Solution

We can apply equation of continuity and velocity of the other part of artery as shown in the diagram below. Now knowing the velocity, we can apply conservation of energy concept and solve the problem as shown in the diagram below.


Problem

The coefficients of viscosity of two liquids is given to us as 2:3 densities of the fluids is given to us as 4:5. In the equal time we need to know the rate of fluid flow in the tubes and the problem is as shown in the diagram below.


Solution

We need to use a concept called Poisellie’s equation. As per it we can write the equation for the rate of flow as shown in the diagram below and solve the problem. Length and diameter of tubes is given as same and the problem is solved as shown in the diagram below.



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Mechanical Properties of Fluids Problems and Solutions One

Fluid statics is a branch of physics that deals with the fluids in the state of rest and its properties like density,pressure and upthrust. Density is the property of a body which measures how its mass is distributed over its volume. If density is more, its more mass is concentrated in a small volume and vice-versa. Pressure is defined as the force experienced by a body per unit surface area. In this case, force is acting on the body in all directions and hence it changes the volume of the body. Every body in a fluid experience a resultant upward force and it is called upthrust or buoyant force.It depends on the volume of the fluid displaced by the body when it is immersed in the fluid, density of the fluid and acceleration due to gravity. If the body is completely immersed in the fluid, volume of the fluid displaced is equal to the volume of the body itself.

Problem

A body is floating in the water in such a way that six parts out of ten parts of its volume is inside the water. We need to measure the density of the water and the problem is as shown in the diagram below.




Solution

When a body is under water, we know that it experience upthrust. As the body is in the equilibrium state, its weight is balanced by the upthrust and they can be equated mathematically. In the place of weight, we can write the product of volume of the body with its density. Volume of the fluid displaced is six parts of volume of the body as only that much is immersed in the fluid. By applying this data, we can solve the problem as shown in the diagram below.



Problem

A brass sphere weights 100 gram weight in air. It is suspended by a thread in a liquid of known specific density. If the specific density of the brass is also known, we need to find the tension in the thread and the problem is as shown in the diagram below.



Solution

Weight is given in gram weight. We need to convert not only gram into kilogram, we also need to multiply it with acceleration due to gravity so that we are converting it into force that is measured in newton. Tension and upthrust acts in upward direction and weight of the body in the downward direction. As the system is in equilibrium position, we can equate this upward and downward force and solve the problem as shown in the diagram below.


Problem

Ninety grams of sulfuric acid is mixed with ninety grams of water and both the densities are known to us. If the specific gravity of the mixture is given to us, we need to find the loss of volume of the system due to mixing and the problem is as shown in the diagram 
below.



Solution

We need to write the total volume initially is the sum of volumes of the first and second ones and it is to found using the definition as the ratio of mass to the density of the each one as shown in the diagram below.



This much volume is supposed to be there but we can also find out the actual volume available using the data like total mass of the system and the effective density of the mixture and as shown below it is different from the supposed to be the volume measured in the first case. Hence the difference between is the loss of the volume. It is further solved as shown in the diagram below.



Problem

Weight of a solid in air, in water is given to us in the problem as shown in the diagram below and we need to measure the weight of that body when it is completely immersed in a liquid of known specific gravity.



Solution

We can find the loss of the weight of the system as the difference between weight of the body in air to the water. Loss of that weight multiplied with the density of the new system, we can find the loss of the weight of the body in that fluid. Actual mass of the body in the fluid is the difference between actual mass of the body in air to the loss of the weight in the fluid. The solution is as shown in the diagram below.


Problem

A liquid is placed in a cylindrical jar and it is rotating about its axis. If radius and angular velocity of the cylinder is know to us, we need to know the difference between the center of the liquid and rise of the liquid at the sides and the problem is as shown in the diagram below.


Solution

The kinetic energy of the fluid is converted into penitential energy and hence as per the law of conservation of energy, we can equate them. We need to replace linear velocity in terms of angular velocity and radius of the system and the problem can be solved as shown in the diagram below.





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Work Done in lifting mass through Non viscous Medium

To lift a body of certain mass from a certain position to a new position in a liquid, we need to do some work and after the work done, this work done will be stored in the form of potential energy.If the body is in air, it is approximately treated like vacuum and upthrust is ignored.

If the body is in a liquid medium, to lift it from the place, we need to do work against gravitational force. The liquid applies resultant upward force and it is also called as buoyant force.
Here this force is acts in the upward direction and it shall help in lifting the body.

The resultant force acting on the body is the difference between gravitational force and up thrust. This resultant force acts in downward direction as gravitational force is dominated one than buoyant force. As it is acting in the upward .direction, it shall be treated as negative force.

We have ignored viscous force as the medium is assumed to be non viscous.

We can measure the effective acceleration of the system by simplifying the equation.It is shown in detail in the video presented below.

By substituting this effective acceleration value in the potential energy format, we can measure the work done in lifting the body through a liquid as shown below in the video.





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Up thrust or the Buoyant Force and Problems with Solutions

Whenever a body is placed in a fluid, the fluid applies resultant up ward directional force on the body and it is called up thrust. It’s numerical value which is the product of volume of the fluid displaced, density of the fluid and acceleration due to gravity at the given place.Volume of the fluid displaced it depends on portion of the body that is there in the fluid. When the body is completely immersed in the fluid volume of the fluid displaced is equal to the volume of the body itself. When the body is partially immersed in the fluid , volume of the fluid displaced equal to the volume of the body that is there inside the fluid.

According to Archimedes principle ,up thrust is  equal to the weight of the fluid displaced by a static fluid.

Basing on this concept we can find out the relative density of a solid as well as a fluid. Relative density of the fluid is equal to the ratio of the weight of the body in air to the loss of the weight of the body in water.

Relative density of the fluid can be determined as the ratio of loss of the weight of the body in a liquid of the loss of the weight of the body in water.

Up thrust depends on the density of the fluid. As density of the sea water is more than that of ordinary water, upthrust in the case of the sea is more than that of river. Upthrust also depends on acceleration due to gravity. If the arrangement is placed in a lift and the lift is moving in  upward direction than in the place of the acceleration ,we have two consider both acceleration due to gravity as well as the acceleration of the lift.

Under different circumstances we can choose to the equation summed up thrust as shown below.





Determining the volume of the cavity

Let us consider a body having some cavity inside. We can calculate the value of upthrust when it is immersed in water. In the case of the volume of the fluid with can write the volume of the body and in the case of the density of the fluid we can write the density of water. Therefore we can get the volume of the body is the ratio of upthrust to the density of the water and gravity as shown.

Upthrust is nothing but the difference in the weight of the body near to the weight of the body in water. By substituting this terms we can get the volume of the cavity as shown below.




Problems on up thrust

Whenever a body is in a fluid, the fluid applies upthrust, force in the upward direction. The direction of top thrust is always the same. Depending on the volume of the body inside the fluid up thrust will change.

Problem one

A body is floating in water such that 6/10 parts of its volume is under water what is the density of the body?

In solving this problem we have to consider that weight of the body is equal to upthrust as the body is an equilibrium state. In the place of the volume of the fluid displaced, we have  to write only six by 10th of the volume of the body cause only that much of the bodies inside the water.

Problem two

Two solids A and B float on water. It is observed that A floats with half of its volume inside water  and B floats with 2/3  its volume immerse it find the ratio of their densities ?

Even in solving this problem also we are having the same approach. The simple concept is when the body is an equilibrium state the weight acting in the downward direction is equal to the up thrust acting in the upward direction.

Problem three

A cubicle block of wood with each side a 10 cm long floats are at the interface of water and oil. The lower surface is 2 cm below the interface of the liquid. The height of oil and water columns each are 10 cm. Density of the oily is 0.8 g per cc then what is the mass of  block ?

In this problem the weight of the body is equal to the upthrust provided by both water and oil. In the case of the water volume of the fluid displaced is equal to 2 by 10th of the volume of the total body whereas in the case of the oil volume of the fluid displaced is equal to 8 by 10th of the total volume of the body. The problem can be solved as shown below.



It is the study of behavior of liquids in motion.

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