# 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.