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In this Physics tutorial, you will learn:

- The definition of particles in physics
- The role of centre of mass in the dynamics of a system of particles
- How to write the Newton's Second Law of Motion for a system of particles
- How to make use of the help of coordinates to study the movement of a system of particles

In the tutorial "Centre of Mass. Types of Equilibrium", we stated that centre of mass of an object helps us to determine its displacement, among other things. This is because an object may swing or rotate, besides moving linearly according a translational trajectory.

Consider a knife thrown towards a target table during a knife throwing activity. In many cases, the person involved holds the knife at its tip instead of its handle. After throwing the knife towards the table, it both rotates and displaces linearly. By the end of process, i.e. when it is stuck on the target table, the knife has rotated by a certain angle to its original direction because it spins several times during the flight.

For the above (and other) reasons, we cannot take the knife as a single object but as a system of points instead. It is obvious that the knife top travels a different distance from the handle. Of course, we can take the centre of mass as a representative of the object, but we know this is a generalization. However, when the object is too small in dimensions compared to the distance travelled by it, the study of the centre of mass is the only method available as if we try to study all points of the system, this would complicate a lot our task.

In this tutorial, we will deal with the motion of a system of particles by considering the forces acting in each individual particle. Obviously, we will deal only with two or three particles, as the method is the same for more particles as well. Increasing the number of particles only makes the calculations longer but the approach doesn't change.

First, we must give the definition of particles in physics. Thus, **"particles are objects whose dimensions are very small compared to their motion"** Therefore, we do not consider the shape of particles at all during the calculation process and as a result, the study of their centre of mass gives the same result as the study of the entire system of particles.

Let's start with the simplest case - a system of two particles. Like in the previous tutorial, "Determining the Centre of Mass in Objects and Systems of Objects", we think them as connected by a very light stick whose mass is neglected. Look at the figure below.

Let's suppose the system moves at another position in t seconds. As a result, we obtain this figure.

It is clear that the small particle (particle 2) is displaced less than the large one (particle 1). Therefore, either we must consider the objects as not related and thus calculating each displacement separately, or we must consider them as a single system and take an average value to represent the displacement just as we did in objects whose motion was irregular during the study of Kinematics.

Obviously, the second method is better, as it shortens the time of calculation and provides a good approximation of the values involved. For this, we need to consider again the centre of mass as a representative of the entire system and relate every kinematic or dynamic approach only to it.

Certainly, all objects move because of the action of any force. Therefore, if there is any motion of a certain object there is a force acting on the object as well. Thus, when a system moves, we can apply the Newton's Second law of Motion which - as discussed in the tutorial "Newton's Second Law of Motion" - outlines the relationship between acceleration, acting force and mass of an object.

The same thing can be said for a system of particles as well. We must only replace the mass of a single object by the mass of the entire system and the force acting on a single object (or better, the resultant force) with the resultant force acting on all particles of the system. Then, the equation

is used to determine the acceleration of the entire system.

The extended Equation of the Newton's Second Law of Motion for a system of particles is

a*⃗*_{sys} = *F**⃗*_{R1} + F*⃗*_{R2} + F*⃗*_{R3} + …*/**m*_{1} + m_{2} + m_{3} + …

where 1, 2, 3, are the indexes for the objects included in the system.

Obviously, the equation (1) is in a vector form, so it is only a general expression of the Newton's Second law of Motion. We must write this equation in components, in order to make it applicable in practice. Therefore, we must write:

a_{x(sys)} = *F*_{Rx(sys)}*/**m*_{sys}

and

a_{y(sys)} = *F*_{Ry(sys)}*/**m*_{sys}

Then, we can use the Pythagorean Theorem to determine the magnitude of acceleration for this system of particles.

Let's explain this point through an example.

Two forces are acting in a system of particles as shown in the figure.

Forces act only at the initial stage, and then the system moves freely.

A light and non-stretchable bar connects the particles to each other. Their respective centres of mass are 3 m away. In the figure, you can also see the values for the masses and the acting force on the particles. The directions of forces and the angles they form to the horizontal direction are also given.

**Calculate:**

- The centre of mass of the system
- The initial acceleration of the system

Take cos 200 = 0.94, sin 200 = 0.34, cos 530 = 0.60 and sin 530 = 0.80.

**a.** Centre of mass of this system is determined by the equation

x_{C} = *x*_{1} × m_{1} + x_{2} × m_{2}*/**m*_{1} + m_{2}

For simplicity, we can take the x-direction (only in this case) according the bar, which connects the two objects. Also we can take the origin at the centre of mass of any object, let's say at centre of the 10 kg object. As a result, we have the following clues:

x_{1} = 0

x_{2} = 3 m

m_{1} = 10 kg

m_{2} = 4 kg

x_{C} = ?

x

m

m

x

Substituting the above values, we obtain

x_{C} = *0 × 10 + 3 × 4**/**10 + 4*

=*12**/**14*m

= 0.86 m

=

= 0.86 m

This result means the centre of gravity of this system is 0.86 m above the lower object as shown below:

**b.** Now, let's calculate the resultant forces for each direction (now we take the horizontal direction as x and the vertical as y as usual). Also we take right as positive and left as negative for the x-direction and up as positive and down as negative for the y-direction as shown in the figure.

If we take the first angle as -530 and the second angle as 200, we have:

F_{1x} = |F_{1}| × cos(-53)^{0} = 40 × 0.60 = 24 N

F_{1y} = -|F_{1}| × sin(-53)^{0} = 40 × (-0.80) = -32 N

F_{2x} = |F_{2}| × cos20^{0} = 25 × 0.94 = 23.5 N

F_{2y} = |F_{2}| × sin20^{0} = 25 × 0.34 = 8.5 N

F

F

F

(The minus sign mean that force is exerted in the negative direction)

Hence, we obtain for the resultant forces in each direction acting on the system

F_{Rx} = F_{1x} + F_{2x} = 24 N + 23.5 N = 47.5 N

F_{Ry} = F_{1y} + F_{2y} = -32 N + 8.5 N = -23.5 N

F

Given that

m_{sys} = m_{1} + m_{2}

= 10 kg + 4 kg

= 14 kg

= 10 kg + 4 kg

= 14 kg

We obtain for the corresponding accelerations of the system in each direction:

a_{x(sys)} = *F*_{Rx}*/**m*_{sys} = *47.5 N**/**14 kg* = 3.39 m/s^{2}

a_{y(sys)} = *F*_{Ry}*/**m*_{sys} = *-23.5 N**/**14 kg* = -1.68 m/s^{2}

a

Therefore,

|a*⃗*_{sys}| = √**a**^{2}_{x(sys)} + a^{2}_{y(sys)}

= √**3.39**^{2} + (-1.68)^{2}

= √**11.49 + 2.82**

= √**14.31**

= 3.78 m/s^{2}

= √

= √

= √

= 3.78 m/s

If we want to find the direction of motion, we consider again the centre of mass of the system. This is because the system's movement may be a combination of many basic motions such as linear, circular, vibrational, etc. Therefore, it would be quite impossible to study what happens with each individual particle of the system.

From Dynamics, we know that objects move in the direction of the resultant force acting on them. The same is true for systems of particles as well. Hence, we can find the direction of motion of the system's centre of mass by finding the direction of resultant force. For this goal, we must calculate the angle θ formed by the resultant force to the horizontal direction. We know that

tan θ = *sin θ**/**cos θ* = *F*_{Ry}*/**F*_{Rx} = *-23.5 N**/**47.5 N* = -0.49

The angle θ, which corresponds to this tangent value, is

θ = arctan (-0.49)

= -26.1^{0}

= 360^{0} - 26.1^{0}

= 333.9^{0}

= -26.1

= 360

= 333.9

This means the acceleration (and the motion as well) is according the vector which is at 333.90 (or -26.10) to the horizontal direction as shown in the figure below.

If particles of the system are given in coordinates, it is much easier to study the system's motion. No angle (and therefore no sines, cosines etc.) are needed. We jump immediately to the second step, i.e. in calculation of the resultant force. Let's illustrate this point through an example.

A system is composed by three distant particles not connected between them. They are placed on a flat table with negligible mass, as shown in the figure.

Each unit corresponds to 1 m in either direction.

**Calculate:**

- The initial centre of gravity of the system
- At which point will be the centre of mass after 2 s in order to maintain the equilibrium of the system?

Suppose that all objects start moving from rest.

Let's determine the positions of objects first. We will index them from left to right as 1, 2 and 3. Thus, the object 1 is at (5m, 6m), the object 2 is at (14m, 15m) and the object 3 is at (20m, 3m). Therefore, we have the following clues:

m_{1} = 1 kg

m_{2} = 2 kg

m_{3} = 3 kg

x_{1} = 5 m

x_{2} = 14 m

x_{3} = 20 m

y_{1} = 6 m

y_{2} = 15 m

y_{3} = 3m

x_{C} = ?

y_{C} = ?

m

m

x

x

x

y

y

y

x

y

Applying the equations

x_{C} = *x*_{1} × m_{1} + x_{2} × m_{2} + x_{3} × m_{3}*/**m*_{1} + m_{2} + m_{3}

and

y_{C} = *y*_{1} × m_{1} + y_{2} × m_{2} + y_{3} × m_{3}*/**m*_{1} + m_{2} + m_{3}

for each direction, we obtain after the substitutions,

x_{C} = *5 × 1 + 14 × 2 + 20 × 3**/**1 + 2 + 3* = *93**/**6* m = 15.5 m

and

y_{C} = *6 × 1 + 15 × 2 + 3 × 3**/**1 + 2 + 3* = *45**/**6* m = 7.5 m

Therefore, the initial centre of mass is at C^{1}(15.5m, 7.5m)

First, let's work out the components of each force. For this, let's use the figure. You can see that the second force is only vertical; it is 4 units downwards and its magnitude is 40 N. This means each unit corresponds to 10 N of force. Therefore, from the figure we have:

F_{1x} = -3 units × 10 N/unit = -30 N

F_{2x} = 0

F_{3x} = 4 units × 10 N/unit = 40 N

F_{1y} = 5 units × 10 N/unit = 50 N

F_{2y} = - 4 units × 10 N/unit = - 40 N (already given in the figure)

F_{3y} = 3 units × 10 N/unit = 30 N

F

F

F

F

F

Therefore, we obtain for the resultant force in each direction:

F_{Rx} = F_{1x} + F_{2x} + F_{3x}

= -30 N + 0 N + 40 N

= 10 N

= -30 N + 0 N + 40 N

= 10 N

and

F_{Ry} = F_{1y} + F_{2y} + F_{3y}

= 50 N + (-40) N + 30 N

= 40N

= 50 N + (-40) N + 30 N

= 40N

Thus, the magnitude of the resultant force of the system is

|F_{R(sys)}| = √**F**^{2}_{Rx} + F^{2}_{Ry}

= √**10**^{2} + 40^{2}

= √**1700**

≈ 41.2 N

= √

= √

≈ 41.2 N

The acceleration of system is determined by using the Newton's Second Law of Motion. Given that the total mass of the system is

m_{sys} = m_{1} + m_{2} + m_{3}

= 1 kg + 2 kg + 3 kg

= 6 kg

= 1 kg + 2 kg + 3 kg

= 6 kg

we obtain for the system's acceleration

a_{sys} = *F*_{R(sys)}*/**m*_{sys}

=*41.2 N**/**6 kg*

≈ 6.9 m/s^{2}

=

≈ 6.9 m/s

Now, we can determine the components of acceleration, i.e. a_{x(sys)} and a_{y(sys)}. Thus,

a_{x(sys)} = *F*_{Rx}*/**m*_{sys} = *10 N**/**6 kg* = 1.666..m/s^{2}

and

a_{y(sys)} = *F*_{Ry}*/**m*_{sys} = *40 N**/**6 kg* = 6.666..m/s^{2}

Therefore, the horizontal position of centre of mass after t = 2 s, will be

x(t) = x_{C} + v_{0x} × t + *a*_{x} × t^{2}*/**2*

= 15.5 + 0 × 2 × +*1.67 × 2*^{2}*/**2*

= 15.5 + 3.33 m

= 18.83 m

= 15.5 + 0 × 2 × +

= 15.5 + 3.33 m

= 18.83 m

and the vertical position of centre of mass after t = 2 s, will be

y(t) = y_{C} + v_{0y} × t + *a*_{y} × t^{2}*/**2*

= 7.5 + 0 × 2 +*6.666 × 2*^{2}*/**2*

= 7.5 + 13.33

= 20.83 m

= 7.5 + 0 × 2 +

= 7.5 + 13.33

= 20.83 m

Therefore, the centre of gravity of the system after 2 s will be at (18.83 m, 20,83 m).

As you see, this is a very comprehensive problem, which includes knowledge from at least three sections: Kinematics, Dynamics and Centre of Mass. The solution seems a bit long but it would be much longer if we considered each object separately. Therefore, the concept of centre of mass helps a lot in reducing long mathematical operations during an exercise that involves a system composed by multiple particles.

In physics, particles are objects whose dimensions are very small compared to their motion. Therefore, we do not consider the shape of particles at all during the calculation process and as a result, the study of their centre of mass gives the same result as the study of the entire system of particles.

All objects move because of the action of any force. Therefore, if there is any motion of a certain object, there is a force acting on the object as well. Thus, when a system moves, we can apply the Newton's Second law of Motion, which outlines the relationship between acceleration, acting force and mass of an object.

The same thing can be said for a system of particles as well. We must only replace the mass of a single object by the mass of the entire system and the force acting on a single object (or better, the resultant force) with the resultant force acting on all particles of the system. Then, the equation

a*⃗*_{sys} = *F**⃗*_{R(sys)}*/**m*_{sys}

is used to determine the acceleration of the entire system.

The extended Equation of the Newton's Second Law of Motion for a system of particles is

a*⃗*_{sys} = *F**⃗*_{R}1 + F*⃗*_{R}2 + F*⃗*_{R}3 + …*/**m*_{1} + m_{2} + m_{3} + …

where 1, 2, 3, are the indexes for the objects included in the system.

When written in coordinates, the above equation (which is in vector form), splits in two (or three) similar equations written for each dimension separately:

a_{x(sys)} = *F*_{Rx(sys)}*/**m*_{sys}

and

a_{y(sys)} = *F*_{Ry(sys)}*/**m*_{sys}

Then, we can use the Pythagorean Theorem to determine the magnitude of acceleration for this system of particles.

a_{sys} = √**a**^{2}_{x(sys)} + a^{2}_{y(sys)}

**1)** Two forces are acting horizontally on a system of two particles connected through a very light and non-stretchable stick as shown in the figure.

What is the acceleration of the system (and in which direction does it occur)?

- 3 m/s2 right
- 6.5 m/s2 right
- 6 m/s2 left - down
- 6 m/s2 right

**Correct Answer: D**

**2)** Two particles are initially at the positions shown in the figure.

What is the acceleration of the system written in two decimal places?

- 3.33 m/s2
- 4.57 m/s2
- 1.14 m/s2
- 0.00 m/s2

**Correct Answer: A**

**3)** A system of three particles are shown in the figure below.

Where will the centre of mass be after 3 s if the system starts moving from rest?

- (10.1 m, 10.6 m)
- (12.4 m, 5.25 m)
- (9.1 m, 5.4 m)
- (12.85 m, 1.65 m)

**Correct Answer: B**

We hope you found this Physics tutorial "Newton's Second Law for System of Particles" useful. If you thought the guide useful, it would be great if you could spare the time to rate this tutorial and/or share on social media, this helps us identify popular tutorials and calculators and expand our free learning resources to support our users around the world have free access to expand their knowledge of physics and other disciplines. In our next tutorial, we expand your insight and knowledge of "Newtons Second Law For S System Of Particles" with our Physics tutorial on Moment of Force. Conditions of Equilibrium.

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