Dynamics of particles – problems and solutions

1. Object A with a mass of 6-kg and object B with a mass of 4-kg connected by a cord and pulled by a force of F = 60 N, as shown in the figure below. The coefficient of kinetic friction between the floor and both objects is 0.5 (tan θ = ¾). Acceleration due to gravity is 10 m/s^{2}. What is the magnitude of the tension force?

__Known :__

Mass of object A (m_{A}) = 6 kg

Mass of object B (m_{B}) = 4 kg

Force (F) = 60 Newton

The coefficient of kinetic friction between object and floor (μ_{k}) = 0.5

Acceleration due to gravity (g) = 10 m/s^{2}

Tan θ = 3/4

__Wanted :__ Tension force (T)

__Solution :__

Horizontal component of force F :

F_{x} = F cos θ

F_{x} = (60)(4/5) = (4)(12) = 48 N

Vertical component of force F :

F_{y} = F sin θ

F_{y} = (60)(3/5) = (3)(12) = 36 N

The normal force 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} = w_{B} – F_{y} = m_{B} g – F_{y }= (4)(10) – 36 = 40 – 36 = 4 N

The force of kinetic friction between object A and floor :

f_{kA} = μ_{k }N_{A} = (0.5)(60) = 30 N

The force of kinetic friction between object B and floor :

f_{kB} = μ_{k }N_{B} = (0.5)(4) = 2 N

**Calculate the acceleration of both objects :**

ΣF = m a

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) a

16 = 10 a

a = 16/10

a = 1.6 m/s^{2}

**Calculate the tension force :**

Object A :

ΣF = m a

T_{A} – f_{kA }= m_{A} a

T_{A} – 30 = (6)(1.6)

T_{A} – 30 = 9.6

T_{A }= 9.6 + 30 = 39.6 N

Object B :

ΣF = m a

F_{x }– f_{kB }– T_{B }= m_{B} a

48 – 2 – T_{B }= (4)(1.6)

46 – T_{B }= 6.4

46 – 6.4 = T_{B }

T_{B }= 39.6 N

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

__Known :__

Mass of object A (m_{A}) = 4 kg

Mass of object B (m_{B}) = 2 kg

Force (F) = 30 Newton

The coefficient of kinetic friction between object and floor (μ_{k}) = 0.2

Acceleration due to gravity (g) = 10 m/s^{2}

cos 37^{o} = 0.8

sin 37^{o }= 0.6

__Wanted :__ Acceleration of both objects

__Solution :__

The horizontal component of force F :

F_{x} = F cos θ

F_{x} = (30)(0.8) = 24 N

The vertical component of force F :

F_{y} = 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} + F_{y} = w_{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 N

The force of kinetic friction between object B and floor :

f_{kB} = μ_{k }N_{B} = (0.2)(2) = 0.4 N

**Acceleration of both objects :**

ΣF = m a

F_{x} – f_{kB} – f_{kA} = (m_{A} + m_{B}) a

24 – 0.4 – 8 = (4 + 2) a

15.6 = 6 a

a = 15.6 / 6

a = 2.6 m/s^{2}

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.

Solution :

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 :__

__The tension force (T) :__

Substitute m_{A} in equation 1 with m_{A }in equation 2.

**What is the difference between statics and dynamics in the context of particles?****Answer:**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.

**How does Newton’s Second Law apply to the motion of a particle?****Answer:**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.

**What is the principle of conservation of linear momentum, and how does it apply to particles?****Answer:**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.

**How does the concept of work done on a particle relate to the force applied and the displacement of the particle?****Answer:**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: $W=Fdcosθ$.

**What is the difference between kinetic energy and potential energy in the context of a particle?****Answer:**Kinetic energy is the energy of a particle due to its motion and is given by 1/2 $ mv$, where $m$ is the mass and $v$ 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.

**How is the impulse on a particle related to the change in its momentum?****Answer:**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: $I=FΔt=Δp$.

**What is meant by a “free particle,” and how does it behave?****Answer:**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.

**What role does friction play in the dynamics of particles?****Answer:**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.

**How is the trajectory of a particle influenced by a central force?****Answer:**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.

**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?**

**Answer:**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: $F_{ext}=ddt $, where $P$ is the total linear momentum.