Kinetics of rigid bodies

Energy is part of the established mechanisms that negotiate with the impact of power and that pushes /pulls couples on the development of mass bodies. Few scientists use the term energy to apply the "dynamic" name to the contemporary mechanics of moving bodies.

Figure 1: Dynamic system

Numerical 1: An application of Newton's law will be illustrated by solving the dynamics of the following system, as illustrated in Figure 1. If you want to calculate the acceleration of 30 kg of mass, pull the rope with the force N. (20 × 9.81).

Sol:

                In this, we do not know the friction and the mass or weight of the pulley and we make the diagram of the free body of 30 kg of mass, as well as the pulley which is attached. Therefore, the tension in all elements of the chain is equal to the force of (20 × 9.81) N.

Figure 2: FBD of Dynamic system 

Using Newton's Second law,

(10 X 9.81) ➗ 30 = 3.27 m/sec

Vertical acceleration 

If instead of applying 20 × 9.81 N of compressive force, 20 kg of weight is suspended at the freeloading. end of the chain, the acceleration will be the same as before?

Figure 3: Dynamic system with vertical acceleration 

We are going to solve this problem. Execution of the FBD of 20 kg of mass.

Now, from the following animation, it will be clear that when the mass of 30 kg increases by a distance of x, the mass of 20 kg will move downwards by a distance of 2x. Therefore, if the acceleration of 30 kg of mass is one (upwards), the acceleration of 20 kg of mass will be 2a (downwards).

Apply Newton's law to a mass of 20 kg.

-T + 20 × 9.81 = 20 × 2a .................. (i) 

or

20 ✕ 9.81 - T = 40. α

Manufacture of a free body of 30 kg of mass with pulley and application of Newton's law.

2T - 30 × 9.81 = 30. α ...........................(ii)

Figure 4: FBD of block 30 Kg 

Solving (i) and (ii)

Now, according to the following diagram, it will be clear that when the mass of 30 kg increases by a distance of x, the mass of 20 kg moves a distance of 2x. So, if the acceleration of 30 kg of mass is equal to one (upwards), the acceleration of 20 kg of mass will be equal to 2a (downwards).

10 × 9.81 = 110. α = 9.81➗ 11 = 0.892 m/sec2 

(ii) Principle of D 'Alembert: 

The second law of Newton is 

                          F =  ma

We can also write it as

                       F  +  (-ma)  = 0 

We know that F is the result of external forces applied to the particle. Considering that (-ma) as a force, we can say that the body is in equilibrium under the action of external forces and force (-ma). This imaginary force is known as an inertial force and the artificial state of created equilibrium is known as dynamic equilibrium. The apparent transformation of a problem into dynamics into one in static is known as the principle of D 'Alembert. D 'Alembert published his work in his "Traite de Dynamique" in 1743

Inertia force is a fictitious force. 

Suppose that a particle is rotated on a horizontal plane by means of a chain.

Figure 5: Inertia force 

For an external observer, the particle moves and has a centripetal acceleration v2 / r. Tension (T) is there, that pulls the particle towards the center. Newton's law can be applied and we get

T= mv2/ r

Q.1. A mass of 10 kg slides on rough ground at a speed of 10 m / s. A mass of 2 kg rests on it. The coefficient of static friction between the mass of 2 kg and the mass of 10 kg is 0.2. The coefficient of kinetic friction between the floor and the mass of 10 kg is 0.1. It is desirable to stop the assembly by applying a horizontal force P. so that the assembly consisting of 10 kg and 2 kg of mass stops at a minimum distance. However, during immobilization, the mass of 2 kg must not slip on a mass of 10 kg. Find the maximum breaking force (P) and the minimum stopping distance.

Try to solve this.................??????????????