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Welcome to our Physics lesson on **Newton's Second Law of Motion for a System of Particles in Coordinates**, this is the second lesson of our suite of physics lessons covering the topic of **Newton's Second Law for System of Particles**, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

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.

You have reached the end of Physics lesson **6.3.2 Newton's Second Law of Motion for a System of Particles in Coordinates**. There are 2 lessons in this physics tutorial covering **Newton's Second Law for System of Particles**, you can access all the lessons from this tutorial below.

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