Динамикаи зарраҳо - мушкилот ва роҳҳои ҳал

Динамикаи зарраҳо - мушкилот ва роҳҳои ҳал

1. Ҷисми А бо массаи 6 кг ва ҷисми Б бо массаи 4 кг, ки бо ресмон пайваст ва бо қувваи F = 60 Н кашида шудаанд, тавре ки дар расми зер нишон дода шудааст. Коэффитсиенти соиши кинетикӣ байни фарш ва ҳарду ҷисм 0.5 аст (tan θ = ¾). Шитоб аз сабаби қувваи ҷозиба 10 м/с2 аст ^. Бузургии қувваи кашиш чанд аст?

Маълум:

Массаи объекти A (м)_A) = 6 кг

Массаи объекти B (м _B ) = 4 кг

Қувва (F) = 60 Нютон

The coefficient of kinetic friction between object and floor (μ_k) = 0.5

Шитоб аз ҳисоби қувваи ҷозиба (g) = 10 м/с²

Tan θ = 3/4

Дархостшуда: Tension force (T)

Ҳал:

Horizontal component of force F :

F _x = F cos θ

F _x = (60)(4/5) = (4)(12) = 48 N

Vertical component of force F :

Fy ₌ F sin θ

F _y = (60)(3/5) = (3)(12) = 36 N

Дар қувваи муқаррарӣ on object A :

N _A = w _A = m _A g = (6)(10) = 60 N

The normal force on object B :

N _B + F _y – w _B = 0

N _B + F _y = w _B

N_B = в_B - Ф_y = м_B г – Ф_y = (4)(10) – 36 = 40 – 36 = 4 Н

The force of kinetic friction between object A and floor :

f _kA = μ _k N _A = (0.5)(60) = 30 Н

The force of kinetic friction between object B and floor :

f _kB = μ _k N _B = (0.5) (4) = 2 Н

Calculate the acceleration of both objects :

ΣF = ma

F _x – T + T – f _kB – f _kA = (m _A + m _B ) a

F _x – f _kB – f _kA = (m _A + m _B ) a

48 – 2 – 30 = (6 + 4) а

16 = 10 а

а = 16/10

а = 1.6 м/ ^с2

Calculate the tension force :

Object A :

ΣF = ma

T _A – f _kA = m _A a

Т _А – 30 = (6)(1.6)

Т _А – 30 = 9.6

T_A = 9.6 + 30 = 39.6 Н

Object B :

ΣF = ma

F_x - f_kB - Т._B = м_B a

48 – 2 – T _B = (4)(1.6)

46 – T _B = 6.4

46 – 6.4 = T _B

T_B = 39.6 Н

2. If the coefficient of kinetic friction between both blocks and floor is 0.2 , what is the суръатбахшӣ of both objects ? (cos 37^o = 0,8, sin 37^o = 0,6)

Маълум:

Массаи объекти A (м)_A) = 4 кг

Массаи объекти B (м _B ) = 2 кг

Қувва (F) = 30 Нютон

The coefficient of kinetic friction between object and floor (μ_k) = 0.2

Шитоб аз ҳисоби қувваи ҷозиба (g) = 10 м/с²

cos 37 ^o = 0.8

син 37 ^o = 0.6

Дархостшуда: Acceleration of both objects

Ҳал:

The horizontal component of force F :

F _x = F cos θ

F _x = (30)(0.8) = 24 N

The vertical component of force F :

Fy ₌ F sin θ

F _y = (30)(0.6) = 18 N

The normal force on object A :

N _A = w _A = m _A g = (4)(10) = 40 N

The normal force on object B :

N _B + F _y – w _B = 0

N_B + Ф_y = в_B

N _B = w _B – F _y = m _B g – F _y = (2)(10) – 18 = 20 – 18 = 2 N

The force of kinetic friction between object A and floor :

f _kA = μ _k N _A = (0.2)(40) = 8 Н

The force of kinetic friction between object B and floor :

f _kB = μ _k N _B = (0.2) (2) = 0.4 Н

Acceleration of both objects :

ΣF = ma

F _x – f _kB – f _kA = (m _A + m _B ) a

24 – 0.4 – 8 = (4 + 2) а

15.6 = 6 а

а = 15.6 / 6

а = 2.6 м/ ^с2

3. Two objects connected by a cord over a pulley as shown in figure below. Mass of object A = m_A, mass f object B = m_B and the acceleration of block B is a. Acceleration due to gravity is g. What is the tension force on block B.

Ҳал:

The horizontal surface is smooth so there is no friction. The only force that accelerates system is the weight of block B.

System’s acceleration :

Қувваи шиддат (T):

Substitute m_A in equation 1 with m_A дар муодилаи 2.

1. What is the difference between statics and dynamics in the context of particles?
   • ҷавоб: Statics deals with the study of forces on particles in equilibrium, where all the forces are balanced, and there is no acceleration. Dynamics, on the other hand, studies the forces and resulting motion of particles when the forces are unbalanced, causing acceleration.
2. How does Newton’s Second Law apply to the motion of a particle?
   • ҷавоб: Newton’s Second Law states that the force acting on a particle is equal to the mass of the particle times its acceleration: $F=ma$. This law describes how a force will affect the motion of a particle, depending on its mass and the acceleration produced by the force.
3. What is the principle of conservation of linear momentum, and how does it apply to particles?
   • ҷавоб: The principle of conservation of linear momentum states that the total linear momentum of a closed system of particles remains constant if no external forces are acting on the system. This means that the total momentum before and after an interaction between particles within the system will be the same.
4. How does the concept of work done on a particle relate to the force applied and the displacement of the particle?
   • ҷавоб: Work done on a particle is the product of the force applied to the particle, the displacement of the particle, and the cosine of the angle between the force and displacement vectors: $W=Fd\cos \theta$.
5. What is the difference between kinetic energy and potential energy in the context of a particle?
   • ҷавоб: Kinetic energy is the energy of a particle due to its motion and is given by 1/2 ​mv², ки дар он $m$ масса аст ва $v$ is the velocity. Potential energy is the stored energy in a particle due to its position in a force field (such as gravitational or electric fields). The specific expression for potential energy depends on the force acting on the particle.
6. How is the impulse on a particle related to the change in its momentum?
   • ҷавоб: Impulse on a particle is the product of the force applied to the particle and the time duration over which the force is applied. It is equal to the change in the momentum of the particle: $I=F\Delta t=\Delta p$.
7. What is meant by a “free particle,” and how does it behave?
   • ҷавоб: A free particle is one that is not subjected to any forces. According to Newton’s First Law, a free particle will either remain at rest if initially at rest or move with constant velocity if initially in motion.
8. What role does friction play in the dynamics of particles?
   • ҷавоб: Friction is a force that opposes the relative motion between surfaces in contact. In the context of particle dynamics, friction can slow down or stop the motion of a particle, depending on the direction and magnitude of the frictional force.
9. How is the trajectory of a particle influenced by a central force?
   • ҷавоб: A central force is one that acts along the line connecting the particle to a fixed point, called the center. Such forces are common in gravitational and electric interactions and will cause the particle to move in a path that maintains a specific angular momentum about the center, often resulting in elliptical or circular orbits.
10. What is the relationship between the net external force acting on a system of particles and the rate of change of the total linear momentum of the system?

• ҷавоб: According to Newton’s Second Law applied to a system of particles, the net external force acting on the system is equal to the rate of change of the total linear momentum of the system: FСадо Ояндасоз​=dP/dt​, ки дар он $\mathbf{P}$ is the total linear momentum.