7.2 Relate external force to mass-centre acceleration | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Systems of Particles

EN PH 131 study topic 7.2

When a system contains several particles, its parts may move differently, but the system as a whole has a mass centre. The motion of that point is governed by the net external force on the complete system. This gives a direct way to predict the mass centre’s acceleration without first finding every particle’s motion. The key is to define the system carefully: forces exerted by objects outside it are external; forces between objects inside it are internal. In this lesson, we use a fixed observer and ordinary Cartesian axes, and we focus on the translational motion of the mass centre.

What you will learn

Define a system and a reference frame before applying the force–acceleration relationship.
Explain why the resultant external force determines the acceleration of a system’s mass centre.
Resolve external forces into components and calculate mass-centre acceleration with consistent SI units.
Distinguish forces acting on the system from internal forces between parts of the system.

1. Define the system and the mass centre

A system is the collection of particles or bodies chosen for analysis. An external force is exerted on that system by something not included in it. An internal force is an interaction between parts that are both included. For example, if two connected components are treated as one system, the force one component exerts on the other is internal.

We use an observer fixed to the ground and a non-accelerating Cartesian reference frame. The origin is chosen for convenience; the positive

x

- and

y

-directions must be stated so that force components and acceleration signs have clear meanings. Position locates a point, displacement is the change in its position, and acceleration describes the time rate of change of velocity. Here the point of interest is the mass centre, denoted by

G

.

For particles with masses

m_i

and positions

\mathbf r_i

, the mass-centre position is the mass-weighted average of their positions. The total mass is

M

. Differentiating this definition twice with respect to time in the fixed frame relates the mass-centre acceleration to the particles’ accelerations.

Newton’s second law applied to each particle gives its net force. When those equations are added, internal interaction forces cancel in pairs for ordinary particle systems, leaving the resultant external force. This is why the complete-system equation does not require the internal forces to be found.

\mathbf r_G=\frac{1}{M}\sum_i m_i\mathbf r_i,\qquad M=\sum_i m_i

The system boundary determines which forces count as external.
Use the same fixed reference frame for positions, velocities, and accelerations.
The system’s total mass is the sum of its particle masses.

2. Relate external force to mass-centre acceleration

The governing rule is that the vector sum of all external forces equals the system’s total mass multiplied by the acceleration of its mass centre. The rule applies to the whole chosen system, not automatically to an individual part. If the system is a single particle, its mass centre is the particle itself and the rule reduces to the familiar particle form of Newton’s second law.

In two dimensions, write the relationship separately along the chosen axes. A positive component of resultant force gives a positive acceleration component; a negative component gives a negative one. A zero resultant component means zero acceleration component, though the mass centre may still have velocity in that direction.

This is an instantaneous relationship: it applies at each time using the external forces and acceleration at that time. If the force is constant, the mass-centre acceleration is constant. If external forces vary, the acceleration can vary too. Do not replace the resultant force with just one applied force when other external forces, such as gravity or contact forces, also act.

A free-body diagram is a picture of external forces on the selected system. It is not a drawing of internal forces or of the mass-centre acceleration. After drawing it, list knowns and unknowns, select axes, resolve every force into components, and apply the equation. Keep mass in kilograms, force in newtons, and acceleration in metres per second squared.

\sum \mathbf F_{\mathrm{ext}}=M\mathbf a_G

Add external force vectors, including weight and contact forces when present.
Internal forces are not included in the complete-system resultant.
The equation predicts mass-centre acceleration, not necessarily the acceleration of every system part.

3. Components, interpretation, and checks

Choose positive axes before assigning signs. A force angled above the positive

x

-axis has a positive vertical component; a force pointing left has a negative horizontal component. Resolve each force using trigonometry, then sum components algebraically. The resulting acceleration components specify both the magnitude and direction of the mass-centre acceleration.

A useful check is dimensional: force divided by mass must have units of acceleration. Another is directional: if the resultant external force points left, the mass-centre acceleration must point left. If the resultant is zero, the acceleration is zero, but the mass centre need not be at rest; it may move at constant velocity.

The method concerns the acceleration of the mass centre. It does not assert that every point in an extended or multi-particle system shares that acceleration. Parts can move relative to the mass centre while the centre follows the equation above. For a single rigid body in planar motion, the same relation gives the translational acceleration of its mass centre; additional analysis would be needed to describe rotation, which is not required for the examples here.

a_{Gx}=\frac{\sum F_{\mathrm{ext},x}}{M},\qquad a_{Gy}=\frac{\sum F_{\mathrm{ext},y}}{M}

Check components, signs, units, and the direction of the resultant.
Zero net external force means zero mass-centre acceleration, not necessarily zero velocity.
Do not infer the motion of each part solely from the mass-centre equation.

Worked example

Two-particle system with a horizontal resultant

Two particles of masses

2.0\,\mathrm{kg}

and

3.0\,\mathrm{kg}

are treated as one system. The only horizontal external forces are

18\,\mathrm{N}

to the right and

8\,\mathrm{N}

to the left. Find the horizontal acceleration of the system’s mass centre. Take the ground as the fixed observer, right as positive

x

, and up as positive

y

. The system is considered at the instant the stated forces act.

External horizontal forces

Schematic external-force diagram; the internal interaction is omitted.

Choose the system
Include both particles. Their total mass is the sum of their masses. The stated forces are external; any force the particles exert on each other is internal to this chosen system.
$M=2.0\,\mathrm{kg}+3.0\,\mathrm{kg}=5.0\,\mathrm{kg}$
Find the external resultant
Right is positive, so the leftward force has a negative sign. There is no stated vertical resultant.
$\sum F_{\mathrm{ext},x}=18\,\mathrm{N}-8\,\mathrm{N}=10\,\mathrm{N}$
Apply the mass-centre equation
The complete-system force equation gives the acceleration of the mass centre. A positive result means the acceleration points right.
$a_{Gx}=\frac{10\,\mathrm{N}}{5.0\,\mathrm{kg}}=2.0\,\mathrm{m/s^2}$

Answer: The mass centre accelerates at

2.0\,\mathrm{m/s^2}

to the right.

Check: The resultant points right, matching the acceleration. Also,

\mathrm{N/kg}=(\mathrm{kg\,m/s^2})/\mathrm{kg}=\mathrm{m/s^2}

.

Worked example

A system pulled at an angle

A

12\,\mathrm{kg}

system is acted on by a

50\,\mathrm{N}

pull directed

30^\circ

above the positive horizontal. The ground provides a

20\,\mathrm{N}

friction force to the left. Gravity and the ground’s normal force act vertically and balance. Find the mass-centre acceleration. Use a ground-fixed observer, right as positive

x

, and up as positive

y

. Evaluate the acceleration at the instant these forces are present.

System free-body diagram

Vertical forces balance as stated; the horizontal forces determine the acceleration.

Resolve the pull
The pull has a horizontal component to the right and a vertical component upward. Gravity and the normal force balance the vertical effects, so the vertical resultant is zero.
$F_x=50\cos 30^\circ\,\mathrm{N}=43.3\,\mathrm{N}$
Sum horizontal external forces
Friction acts opposite the positive horizontal direction, so subtract it from the pull’s horizontal component.
$\sum F_{\mathrm{ext},x}=43.3\,\mathrm{N}-20\,\mathrm{N}=23.3\,\mathrm{N}$
Calculate acceleration
Divide the net horizontal external force by the system mass. The positive sign indicates acceleration to the right.
$a_{Gx}=\frac{23.3\,\mathrm{N}}{12\,\mathrm{kg}}=1.94\,\mathrm{m/s^2}$

Answer: The mass-centre acceleration is approximately

1.94\,\mathrm{m/s^2}

to the right, with zero vertical component.

Check: The horizontal resultant is positive, as is the reported acceleration. The balanced vertical forces give

a_{Gy}=0

, consistent with the stated vertical balance.

Worked example

A resultant force with two components

A

4.0\,\mathrm{kg}

system experiences an external resultant of

-6.0\,\mathrm{N}

horizontally and

8.0\,\mathrm{N}

vertically. Find the mass-centre acceleration vector, its magnitude, and its direction. Use a fixed observer with positive

x

to the right and positive

y

upward. The initial state is the instant these forces act; no initial velocity is needed to find acceleration.

Resultant external force

The resultant lies in the upper-left quadrant.

Use the force components
The supplied resultant has a negative $x$ -component and positive $y$ -component. Divide each component by the total mass to obtain the corresponding acceleration component.
$a_{Gx}=\frac{-6.0\,\mathrm{N}}{4.0\,\mathrm{kg}}=-1.5\,\mathrm{m/s^2},\qquad a_{Gy}=\frac{8.0\,\mathrm{N}}{4.0\,\mathrm{kg}}=2.0\,\mathrm{m/s^2}$
Find magnitude and direction
The negative horizontal and positive vertical components place the acceleration in the upper-left quadrant. Use the component magnitudes for the angle from the negative horizontal axis.
$|\mathbf a_G|=\sqrt{(-1.5)^2+(2.0)^2}=2.5\,\mathrm{m/s^2},\qquad \theta=\tan^{-1}\!\left(\frac{2.0}{1.5}\right)=53.1^\circ$

Answer: The acceleration vector is

(-1.5\,\mathbf i+2.0\,\mathbf j)\,\mathrm{m/s^2}

. Its magnitude is

2.5\,\mathrm{m/s^2}

, directed

53.1^\circ

above the negative

x

-axis.

Check: The force magnitude is

10\,\mathrm{N}

, so its acceleration magnitude must be

10/4.0=2.5\,\mathrm{m/s^2}

, agreeing with the component calculation. The signs place both vectors in the upper-left quadrant.

Common mistakes and how to avoid them

Using only the largest applied force instead of the vector sum of all external forces.

Correction: Draw the system’s external forces and sum their components, including opposing forces and weight or contact forces when they do not cancel.

Including forces between parts of the chosen complete system in the external resultant.

Correction: Classify a force by the system boundary. Interactions between included parts are internal and cancel in the complete-system equation.

Assuming zero acceleration means the system is at rest.

Correction: Zero resultant external force means constant mass-centre velocity. That velocity may be zero or nonzero.

Treating the mass-centre acceleration as the acceleration of every part.

Correction: The equation describes the translation of the mass centre. Parts of a system may have different accelerations.

Dropping a negative sign when a force points opposite a chosen positive axis.

Correction: State positive directions first, then assign component signs from the actual force directions.

Lesson summary

Define the complete system and the fixed reference frame before listing forces.
The vector sum of external forces equals total system mass times mass-centre acceleration.
Resolve forces along explicitly chosen axes and preserve signs through the calculation.
Check dimensions and direction; zero resultant means zero mass-centre acceleration, not necessarily zero velocity.

Check your understanding

Question 1

A

6.0\,\mathrm{kg}

system has a net external force of

12\,\mathrm{N}

to the left. What is its mass-centre acceleration?

$2.0\,\mathrm{m/s^2}$ to the left
$2.0\,\mathrm{m/s^2}$ to the right
$72\,\mathrm{m/s^2}$ to the left
Zero, because the force is not balanced

Show answer and explanation

2.0\,\mathrm{m/s^2}

to the left

Acceleration equals resultant force divided by mass. The magnitude is

12/6.0=2.0\,\mathrm{m/s^2}

, and its direction matches the leftward resultant.

Question 2

The resultant external force on a system is zero at an instant. Which statement must be true at that instant?

The mass centre has zero acceleration.
The mass centre has zero velocity.
Every part of the system is at rest.
The system has zero mass.

Show answer and explanation

The mass centre has zero acceleration.

The force–acceleration relationship gives zero mass-centre acceleration. It does not require the centre or every part to be at rest.

Key terms

System: The particle or collection of particles selected for analysis.
External force: A force exerted on the selected system by something outside its boundary.
Internal force: A force between parts that are both included in the selected system.
Mass centre: The point whose position is the mass-weighted average of the positions of the system’s particles.
Resultant external force: The vector sum of all external forces acting on the selected system.

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7.2 · Relate external force to mass-centre acceleration

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

EN PH 131 study topic 7.2

What you will learn

1. Define the system and the mass centre

2. Relate external force to mass-centre acceleration

3. Components, interpretation, and checks

Two-particle system with a horizontal resultant

A system pulled at an angle

A resultant force with two components

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through EN PH 131

About this lesson and its review