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.
Showing posts with label Newton laws. Show all posts
Showing posts with label Newton laws. Show all posts
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.
Laws of Motion Problems with Solutions Two
Laws of Motion Problems with Solutions Three
Laws of Motion Problems with Solutions four
Laws of Motion Problems with Solutions Five
Laws of Motion Problems with Solutions Six
Laws of Motion Problems with Solutions Seven
Laws of Motion Problems with Solutions Eight
Linear momentum is a physical quantity that explains about the transfer of kinetic energy of one body with another when there is a collision with other body. It is mathematically defined as the product of mass and velocity of the body. If momentum is more, it transfer kinetic energy to other body more and vice versa. Momentum is a vector quantity and its direction is similar to the direction of velocity.
Linear momentum is a physical quantity that explains the translatory motion. We can define Newton's second law in terms of linear momentum. As per the law, force is defined as the rate of change of linear momentum. Momentum is a physical quantity that gives clarity regarding the ability of transfer of kinetic energy. We can also find the relation between linear momentum and kinetic energy and it is explained in the video lesson below.
Laws of Motion Problems with Solutions Two
Laws of Motion Problems with Solutions Three
Laws of Motion Problems with Solutions four
Laws of Motion Problems with Solutions Five
Laws of Motion Problems with Solutions Six
Laws of Motion Problems with Solutions Seven
Laws of Motion Problems with Solutions Eight
Gravitational force exists between every two particles having some mass and it is directly proportional to the product of their masses and inversely proportional to the square of distance of separation. It is independent of medium between them. If number of bodies are present around any body, the total gravitational force is the vector sum of all the existing forces. We also need to know that the presence of third body is not going to effect the gravitational force between the other two bodies. Acceleration due to gravity varies basing on the position of the body from the surface of the earth and depth of the body from the surface of the earth. It also depends on the location of the body with respect to its location of equator.
Problem
It is given in the problem that the gravitational force between two objects at a separation of one meter is given to us and we need to find the mass of each object.
Solution
We can apply law of gravitational force to know the force between the two bodies and the force is given in the problem and their separation is also given in the problem. By applying them in the formula and by applying the constant value of universal gravitational constant, we can solve the problem as shown in the diagram below.
Problem
It is given in the problem that mass of one ball is four times the mass of the other body and their separation is given as 10 centimeter. The gravitational force between them is given to us and we need to find the mass of each particle.
Solution
We need to apply once again newton's law of gravitation as shown in the diagram below. Universal gravitational constant is a proportional constant and it is constant over the entire universe. Data can be further simplified and problem can be solved as shown in the diagram below.
Problem
It is given in the problem that three spherical balls of mass 1,2 and 3 kilogram are placed at the corners of an equilateral triangle of side one meter. We need to find the effective gravitational force acting on the body of mass one kilogram.
Solution
Let us consider three particles at the three corners of the equilateral triangle as shown in the diagram below. On one kilogram particle there is now gravitational attractive force acting due to other two bodies of masses 2 and 3 kilogram. Their directions are along the line joining them and their resultant can be found using the parallelogram law of vectors.
It can be further simplified as shown in the diagram below. Vector law of addition is applied and we know the angular separation between the bodies and it can be further simplified as shown below.
Problem
We need to know that how a particle of known mass has to be divided into two parts such that the gravitational force between them is maximum and the problem is as shown in the diagram below.
Solution
We can find the gravitational force between any two bodies using the Newton's law of gravitation. We also need to know that if any function has to be maximum, its corresponding differentiation has to become zero as shown in the diagram below. By applying the rules of differentiation and simplifying it further, we can solve the problem as shown in the diagram below.
Problem
It is given that two particles of known mass are separated by a certain distance. We need to know at what distance from the body of mass one kilogram, the gravitational force becomes zero and the problem is as shown in the diagram below.
Solution
We shall imagine any particle of certain mass between the two bodies along the line joining them. On the third body, both the bodies applies force of attraction towards them and hence both of them on the third body are in opposite direction. At any instant, the resultant force becomes zero and the point becomes a neutral point. The problem is solved as shown in the diagram below.
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The concepts are here to deal with laws of motion. We are solving regarding rocket propulsion and tension in the string. Rocket is a variable mass as its fuel gets ejected per each second and the mass of the rocket will be keep on decreasing. We do also solve some problems based on the strings and their connected with the bodies.
Problem
It is given in the problem that fuel of 50 kilogram is consumed per second from the tail of a rocket. We need to measure the thrust on the rocket if the velocity of the exhaust gas is given to us. We also need to measure velocity of the rocket when its mass is reduced to one tenth of its mass. The problem is as shown in the diagram below.
Solution
As mass is variable with time, force is here the product of rate of change of mass with velocity and its value is given to us. The data is substituted as shown in the diagram below.
We know the formula for the final velocity of the rocket and it is simplified as shown in the diagram below.
Problem
This is also a similar problem like the previous one. The speed of the gas is given to us coming from the rocket and we need to measure the upthrust received by the rocket. The problem is as shown below.
Solution
It is solved using the same concept applied in the previous problem and the solution is as shown below.
Problem
The pulleys and the string shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, we need to know the angular separation and the problem is as shown in the diagram below.
Solution
We know that because of weight tension is generated in each wire and it acts away from the weight as shown in the diagram below. This tension can be resolved into components. By equating the forces acting in the vertical direction during the balanced condition, we can solve the problem as shown in the diagram below.
Problem
It is given in the problem that a man of mass 60 kilogram is standing on a weighing machine kept in a box of known mass 30 kilogram as shown in the diagram below. If the man manages to keep the box stationary, the reading in the weighing machine is ?
Solution
We need to solve the problem by drawing free body diagrams and further writing the equations of motion. The solution is as shown in the diagram below.
Here we are using the concept that normal reaction is the reading of the scale.
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Motion is the change in the state of the body with respect to time and space. The bodies always tries to keep their state as it is and this property is called inertia. To change the state of the body, we need to apply a force on the body that is sufficient enough to overcome the existing inertia. We are here solving problems about force and momentum. As per the law of conservation of linear momentum, if no external force is acting on the system, the momentum of the system always remains constant and it is conserved.
Problem
It is given in the problem that a body of mass 10 kilogram is being acted upon a force that varies with time. It is also given that a opposing force is also acting on the same body and it is a constant force of 32 newton. The body is also having some initial velocity and we need to know the final velocity of the body after five seconds. The problem is as shown in the diagram below.
Solution
As force is variable with time, to find the work done, we shall integrate the variable force with time and it can be equated to change in momentum. The given data is substituted in the concept and the problem is solved as shown below.
Problem
It is given in the problem that a block of mass m is starting from the state of rest from the top of a smooth inclined plane of known inclination. It has reached the bottom of inclined plane in one part of the time and to travel further on smooth surface some more time is taken. We need to measure the difference in this momentum in the given time difference and the problem is as shown below.
Solution
We can identify the change in momentum and the momentum of each part with specific direction as it is a vector quantity. As the body after coming to the bottom has constant velocity in the further travel, both the momentum are in the direction as shown with the angular separation. We can find the effective value using the parallelogram law and the solution is as shown in the diagram below. The velocity can be found using equations of motion.
Problem
It is given in the problem that two monkeys are holding each other as shown and one monkey is climbing a rope. The mass of the monkeys are given for us and the maximum tension can be bared by upper monkey is given to us and we need to know what force it shall apply on the ope to carry the second money with it with out breaking its tail.
Solution
Tension in the lower part of the rope as the sum of the weight of that rope and acceleration of the money multiplied by the mass of it. We can find the acceleration of the upper monkey as shown below. Similarly the tension in the upper part also can be found as shown below.
Problem
A shall of mass 100 gram is moving with a speed known along straight line inclined at a know angle. It is divided into two parts and we need to measure the velocity and direction of the second part after explosion. The problem is as shown in the diagram below.
Solution
As there is no external force acting on the system, linear momentum is conserved. We can resolve the momentum into components and it can be solved as shown in the diagram below.
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Newton Laws of motion deals with the impact of force on the motion of the body. This force can be defined as the rate of change of momentum. Here in this post we are going to discuss regarding connected bodies. The bodies are connected with the help of string or in contact with each other. The string becomes tight and there is a force is generated in the sting and it is called tension.
Problem
It is given in the problem that two masses of 10 and 20 kilogram are connected by a mass less spring. A force of 200 newton acts on 20 kilogram mass and at a certain instant acceleration of the other body is given. We need to find the acceleration of the other body. The problem is as shown in the diagram below.
Solution
The spring connected between the bodies as if like a string and a tension is generated in it when the force is applied on the connected mass. On 10 kilogram mass only force acting is the tension in the string. On the 20 kilogram mass, applied force is acting on the it towards right and tension acts in opposite direction. Thus we an write equation of motion and solve the problem as shown in the diagram below.
Problem
A block of mass M is pulled by a chain of uniform mass tied to the mass. We need to know the tension in the chain at a particular place. The problem is as shown in the diagram below.
Solution
We know that the chain is uniform and hence its mass is equally distributed. As the point where we need to measure the tension is at one forth of the total length its mass also will be one forth of the total mass of the chain. Thus we can divide the system into two masses. The first one is the mass of the block and the remaining mass of the chain and the second part is one forth of the chain. We can write equations of motion and solve the problem as shown in the diagram below.
Problem
In the given problem two masses are connected with the help of the string as shown in the diagram below. The first mass is on the smooth horizontal surface. The second mass is connected to the firs with the help of a pulley and stand as shown. We need to find the relation between the acceleration of the bodies.
Solution
We can assume that the tension in the horizontal wire is one value and the same will be continued in the connected string. The tension in the bottom wire connected the other bod is the sum of the same two tensions as shown in the diagram below. So we can find the relation as shown here.
Problem
Two blocks of masses are at rest connected with a string over the pulley as shown in the diagram below. A force is applied on the pulley and we need to know the acceleration of the blocks. The problem is as shown in the diagram below.
Solution
The force applied on the pulley is divided across the two strings equally and hence we know the tension in each wire. We know that the weight of each block acts in down ward direction and the tension in the upward direction. By applying laws of motion and resultant force equation, we can solve the problem as shown in the diagram below.
Problem
A particle is at rest at a given position and force is applied along the X direction. We need to measure the velocity of the particle when it is at a particular distance from the origin. The problem is as shown below.
Solution
We know that the work done is stored in the form of change in kinetic energy. As the body is initially at rest, its initial kinetic energy is zero and hence change in kinetic energy is nothing but final kinetic energy. We can write the work done in terms of integration and the problem is solved as shown in the diagram below.
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