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

Electric force and Coulomb's Inverse square law Video Lesson

Electric force of attraction or repulsion exists between every two charges and it follows Coulomb's inverse square law. According to the law, the force between two charges is directly proportional to the product of charges and is inversely proportional to the square of the distance of separation.

Gravitational and electric forces are fundamental in nature. Gravitational force is due to mass of the particle and it exits between every two masses where as electric force exits between the the particles or bodies having only excess charges. Both of them obeys inverse square law but gravitational force is independent of medium but electric force depends on the medium between charges.

Gravitational force is the weakest of fundamental forces but electric force is much stronger than that. Gravitational force is further always attractive force but electric force is either attractive or repulsive. Further classification is explained in the video lesson below.


Coulomb's inverse square law

To identify the relation between the force and charges different experiments were conducted and we got to a conclusion that the electric force between the charges is directly proportional to the product of charges and is inversely proportional to the square of the distance of the separation.  The force also depends on the medium between the charges and its impact is studied basing on the permitivity of the medium. The ratio of permitivity of the medium to the permitivity of the free space is called dielectric constant or relative permitivity. Its value for the vacuum is one and for any non conducting medium it is greater than one. It is explained as shown in the video lesson below.



Effect of the medium on force between the charges

As we have explained earlier, we can understand that the force between the charges gets reduced when the medium is placed instead of the vacuum. It is further explained in the video lesson below. We can find out the effective distance in the presence of medium as shown below.



Impact of force between two charges when third charge is placed 

Electric force between the charges is not effected even if we keep a new charge in between them. What makes a difference is, more forces are acting on each  charge and hence resultant force on each charge is more. But the force between two charges is not affected because of the third charge. To find the resultant charge, we can use vector laws of addition. It is explained in the video lesson below.


Related Posts

Electric Charge and Electric Force

Resultant Force and Coloumb's Law of Electric Force Problems and Solutions
No comments:
Labels: , , ,

Free body diagram and Connected Bodies Video Lesson

Free body diagram is the representation of all the forces acting on a body at a given instant. We shall consider only the forces acting on a body but not the forces applied by the same body. When there are multiple bodies, all of them can be treated as one system if they have same acceleration. When we are solving a system of bodies and we are interested in finding the acceleration and may be tension and contact forces  on a given body. We shall draw free body diagram for each body and we shall further write the equation for the resultant force. We shall consider the force along the direction of motion as positive and vice versa. We are also dealing with contact force in the given video below. Contact force is t he force applied by one body on the other when they are in contact. There will be reaction force also according to Newton's third law of motion.A detailed video lesson is presented below regarding free body diagrams as shown below.



Connected bodies and free body diagram

Here we are discussing about the bodies of different masses connected with the help of string and a force is applied on the system. When force is applied, the string becomes tight and a tension is developed in the wire. We need to identify all the forces acting on each body. We need to be careful and identify only forces acting on the body but not the forces applied by the body. This can be done with the help of free body diagram for each body. Thus we can write equation for resultant force on each body  and we can solve that equations and get the tension in each wire and acceleration of the system as shown in the video lesson below.



Atwood's machine

Two bodies of different masses connected over a smooth pulley with the help of light weighted string could be called as Atwood's machine. We need to find the tension int the wire and the acceleration of the system. We can write free body diagram for each body and equations for resultant force as shown in the video lesson below.


Connected bodies one in vertical mode and other in horizontal position

Let us consider a system of two bodies where one body is hanging vertically from a table and the other body is on a smooth horizontal smooth surface and the two bodies are connected with a light weighted string over a smooth pulley. We need to find the tension in the string and the acceleration of the system. We can draw free body diagram to identify the forces acting on each body and we need to write the equations for resultant force as shown in the video lesson below. We can solve the equations as shown below.




Related Posts

Newton Law of Inertia and Newton law of Force an Introduction


No comments:
Labels: , , ,

Newton's First and Second Law of Motion Video Lesson

Newton's first and second law are helpful in understanding translatory motion. Newton's first law explains the concept of inertia. Any  body has the property of continuing its own state of rest or motion when no external force is acting on the body. This property is called inertia. This itself is called first law of motion. Body can have inertia of rest,motion and direction.

Inertia of rest is the property of body due to which the body continue its state of rest until no external force is acting on the body. We can experience it in daily life. Consider a person in standing position that is in the  in a vehicle that is at rest.When the bus suddenly starts, the connected part of the person with the vehicle that are legs starts moving. But the upper part of the body tries to continue in  the state of rest due to inertia of rest. Thus the lower part of the body moves and the upper part stays there it self. So he fell like he got a jerk and fell like falling back. it is due to inertia of rest. The same can be explained with inertia of rest and direction. A Video lesson about the first law is shown below for your reference.



Newton's Second Law of motion

 Newton's second law of motion is about force. Force is a physical quantity that changes or tries to change the state of the body. Force is defined as the rate of change of momentum. If a body is having a constant mass, then force can be defined as the product of mass and acceleration of the body.  Force is a vector that has both magnitude and direction and it is a vector quantity. The direction of the force is similar to the acceleration. It is explained in the video lesson below.


Resultant force and acceleration

When there is a resultant force acting on a body, then the body velocity gets changed and hence it will have some acceleration. If there are multiple force acting on a body, we shall find the effective force or resultant force using vector laws of addition. If there is no resultant force acting on a body, the body will be either moving with a constant velocity or in the state of rest basing on its previous condition. We can find the resultant force acting on a body as shown in the video lesson below.

Related Posts

Newton Law of Inertia and Newton law of Force an Introduction

No comments:
Labels: , , ,

Oscillations Problems with Solution Two

We are solving series of problems based on  the concept of oscillations. It is also called vibratory or harmonic motion where there  is to and fro motion that gets repeated at regular intervals of time. The vertical projection of uniform circular motion is simple harmonic and we have derived equations for displacement, velocity and acceleration for the body in simple harmonic motion based on that. The time taken to complete one oscillation is called time period and the number of oscillations per one second is called frequency.

Problem

A body of known mass is connected with two springs and they in tern are connected to rigid support as shown in the diagram below. We need to find the effective time period of the system and the problem is as shown in the diagram below.


Solution

When ever the body is slightly disturbed, it starts oscillating and one spring expands and the other spring contracts by the same magnitude. As the force acting on both of them is same, the two springs behaves as if like they are connected in series and we can find the effective spring constant of the system and time period as shown in the diagram below.


Problem

A spring of spring constant K and length L is cut into two parts and the relation between the lengths of two parts and their ratio is given to us in the problem as shown in the diagram below. We need to measure the spring constant of one part of the spring.


Solution

Spring constant is the measure of nature of spring and it depends on the length of spring in the inverse proportional ratio. Taking that into consideration, we need to solve the problem as shown in the diagram below.


Problem

A piece of wood known dimensions is given to us and its density is also known to us as given in the problem below. It is floating in water with one surface vertical to the surface. It is pushed down and released and it starts oscillating. We need to measure the time period of the system.


Solution

When ever a force is applied on the body, there is buoyant force also and the system starts executing oscillatory motion with a certain restoring force. By comparing that with the standard equation, we can solve the problem as shown in the diagram below.


Problem

A cylindrical piston is used to close a cylinder with a certain gas and when it is slightly distributed, it starts oscillating in simple harmonic motion. We need to find the time period of the system and the problem is as shown in the diagram below.


Solution

By comparing it with the standard equation, we can solve the problem as shown in the diagram below. We need to identity the restoring force and the equation for the acceleration of the body in SHM. The problem is solved as shown in the diagram below. Here in the first case, we get the equation for the force.


We need to equate to the product of mass and acceleration and we further need to substitute the value of acceleration for a body in SHM so that we can compare it with the standard equation. Thus we can get the time period of the system as shown in the diagram below.


Problem

A spherical ball of known mass and radius is rolling with out slipping on a concave surface of known radius as shown in the diagram below. If the oscillations are small, we need to find the time period of the system.


Solution

Let us consider the spherical ball is at a particular position and we can find the angle basing on the definition as shown in the diagram below. As we know the equation for the acceleration of a body sliding on a inclined surface, by writing that equation, we can get that and find the time period as shown in the diagram below.



Related Posts


2 comments:
Labels: , , , , ,

Surface Tension Problems with Solutions Three

We are solving series of problems on the property of a liquid called surface tension. This is due to molecular force of attraction among the liquid molecules. If the molecules belongs to same liquid the force is called cohesive force of attraction and if they are due to different liquid molecules then it is called adhesive force of attraction. Which force among this two is dominating depends on the nature of that liquids and the body in contact. Water sticks to our body as adhesive force is dominating and mercury won’t as cohesive force is dominating. This can be even understood in terms of angle of contact and capillary rise of the liquid when a thin tube is placed in it. Surface tension depends on the nature of the liquid as well as temperature of the liquid. If temperature is raised, molecular force of attraction decreases and hence surface tension also decreases.

Problem

A liquid of known density and surface tension is placed in a capillary tube and it has raised to a certain height. We need to find the potential energy of the system and the problem is as shown in the diagram below.



Solution

We know the formula for the gravitational potential energy in terms of the height of the liquid but it is spread over the entire length and not focused at the top. Thus we shall consider the concept of center of mass ans we can assume that the entire mass is consecrated at the center of mass of the liquid which is at geometrically half of the total height. By subsisting the value of the capillary rise in terms of the surface tension, we can solve the problem as shown in the diagram below.


Problem

Several liquid drops of known radius and density are combined to foam a big drop of known radius. If all the energy released in this process is converted into kinetic energy, we need to find the velocity acquired by the drop.


Solution

Energy released can be expressed as the difference between work done for both the radius basing on the definition that surface tension is work done per unit change in the area of cross section of the liquid. We shall equate it to the kinetic energy where mass can be expressed as the product of density and volume. Problem can be further simplified as shown in the diagram below.


Problem

A glass rod of known radius is immersed into a capillary tube of known radius as shown in the problem below. If the arrangement is immersed in the water, we need to find the rise in the liquid level.


Solution

Liquid will rise in the system until the force in the upward direction due to surface tension is balanced by the weight of the raised liquid in the capillary tube system. By writing the mass as the product of volume and density and further volume as the product of area and length, problem can be solved as shown in the diagram below.


Problem

Under isothermal conditions two bubbles of known radius are combined to foam a single drop of know radius. If the external pressure is given to us, we need to measure the surface tension of the system and the problem is as shown in the diagram below.


Solution

We can apply Boyle’s law as the temperature of the system is constant as shown in the diagram below. By simplifying it further we can solve the problem as shown here.


Problem

A soap bubble is blown at the end of very narrow tube as shown in the problem below. Air flows into the tube with a known velocity and it comes to rest inside it. After some time the bubble get some radius and comes out of the tube. We need to find the radius of the bubble so that the air strikes the bubble surface normally.


Solution

We need to equate the force due to pressure to the force due to surface tension and the problem can be solved as shown in the diagram below.




Related Posts

No comments:
Labels: , ,

Surface Tension Problems with Solutions One

We are going to solve series of problems with detailed solutions about a topic called surface tension. Surface tension is the property of liquid due to which the liquid surface experiences a tension and they tend to acquire minimum surface area. It is because of this surface tension, small insects are able to float on the surface of water. It is defined as the force acting on the tangential surface of the liquid normal to the surface of contact per unit length. Surface tension can be explained basing on molecular theory. Every molecule can influence the surrounding and attract the neighboring molecules up to some extend and that distance is called molecular range. Taking the molecule as the center, molecular range as the radius, if we draw a sphere, it is called sphere of influence and within the sphere of influence, core molecule can attract the other molecules.

Problem

The length and thickness of a glass plate is given to us as shown in the diagram below. If this edge is in contact with a liquid of known surface tension, we need to know the force acting on the glass plate due to the surface tension of the liquid.



Solution

We know that surface tension is mathematically force acting on it per unit length. Here length means the length of free surface of the body that is in contact with the liquid. The glass plates both inner and outer surface are in contact with the liquid and hence two lengths has to be taken into count. The problem is solved as shown in the diagram below.



Problem

A drop of water of known volume is pressed between two glass plates so as to spread across a known area. If surface tension of the liquid is known to us, we need to know the force required separating the glass plate and the problem is as shown in the diagram below.


Solution

We can write the volume as the product area of cross section with the length of the liquid. We also know that the surface tension can also be expressed in terms of work done per unit area. Intern work done can be expressed as the product of force and displacement. Taking this into consideration, we can solve the problem as shown in the diagram below.


Problem

A big liquid drop of known radius splits into identical drops of same size in large number and we don’t know the radius of the small drop. We need to measure the work done in this process and the problem is as shown in the diagram below.


Solution

We know that the volume of the liquid is conserved. It means the volume of the big drop is the sum of the volumes of all small drops together and basing on that we can find the relation between smaller and bigger radius as shown in the diagram below. We can write the equation for the work done as the product of surface tension and the change in the area. 


Problem

Work done in blowing a soap bubble of radius R is given to us as W. We need to measure the work done in blowing the same bubble to a different radius and the problem is as shown in the diagram below.


Solution

As discussed in the previous problem, we can define the work done as the product of surface tension and the change in the area of cross section. By applying that data, we can solve problem as shown in the diagram below.




Problem

We need to find the capillary rise of a liquid in a capillary tube when it is dipped in that liquid where surface tension and density of the liquid is given to us. We can treat angle of contact as zero and the problem is as shown in the diagram below.


Solution

We know that when angle of contact is less than ninety degree, the liquid raises above the normal level of the beaker and that property is called capillarity. The capillary rise depends on the radius of the tube, density and surface tension of the liquid. We can apply the formula and solve the problem as shown in the diagram below.



Related Posts

1 comment:
Labels: , , ,
Subscribe to: Comments (Atom)