Dynamics

SPEC 8.2, 8.6 • FORCES AND NEWTON'S LAWS

Motion under Resultant Force - Applying F = ma to accelerating systems

From Equilibrium to Motion

In previous lessons, we studied equilibrium where forces balance and there is no acceleration. Now we study dynamics - what happens when forces are unbalanced.

Key Principle

When forces are unbalanced, there is a resultant force, which causes acceleration according to Newton's Second Law:

Σ F = ma

where

Σ F

= sum of all forces (resultant force)

The Difference

Equilibrium (Lessons 8.01-8.04)Dynamics (This Lesson)

Σ F = 0

Σ F \neq = 0

Acceleration = 0Acceleration ≠ 0Object at rest or constant velocityObject speeds up or slows down

The Dynamics Method

To solve dynamics problems, follow this systematic approach:

1

Draw a force diagram - Show all forces acting on the object

2

Choose positive direction - Usually the direction of motion

3

Find the resultant force - Add forces in direction of motion, subtract those opposing

4

Apply F = ma - Resultant force = mass × acceleration

5

Use kinematics if needed - Apply suvat equations to find velocity, distance, or time

Key Equations

Dynamics:

F_{res u lt an t} = ma

Kinematics:

v = u + a t

,

s = u t + \frac{1}{2} a t^{2}

,

v^{2} = u^{2} + 2 a s

Worked Example01

A box of mass 5 kg is pulled along a smooth horizontal surface by a horizontal force of 20 N. Find the acceleration of the box.

1

Draw force diagram: Applied force = 20 N (→), Weight = mg (↓), Normal reaction R (↑)

2

Surface is smooth, so no friction force.

3

Resultant horizontal force = 20 N (only horizontal force)

4

Apply F = ma:

20 = 5 \times a

a = \frac{20}{5} = 4 m s^{- 2}

Answer: Acceleration = 4 m s⁻²

Horizontal Motion with Friction

When a surface is rough, friction opposes motion. For a moving object:

Friction When Moving

When an object is moving (not in equilibrium), friction takes its maximum value:

F = μ R

Friction always acts opposite to the direction of motion.

Resultant Force

Resultant = Applied force − Friction =

P - μ R

Then apply:

P - μ R = ma

Worked Example02

A crate of mass 8 kg is pulled along a rough horizontal surface by a horizontal force of 50 N. The coefficient of friction is μ = 0.4. Find the acceleration of the crate.

1

Find the normal reaction. Vertical equilibrium:

R = m g = 8 \times 9.8 = 78.4 N

2

Find friction force (object is moving):

F = μ R = 0.4 \times 78.4 = 31.36 N

3

Find resultant horizontal force:

Resultant = 50 − 31.36 = 18.64 N

4

Apply F = ma:

18.64 = 8 \times a

a = \frac{18.64}{8} = 2.33 m s^{- 2}

Answer: Acceleration = 2.33 m s⁻² (or 2.3 m s⁻²)

Worked Example03

A block of mass 6 kg is pulled along a rough horizontal surface by a force of 40 N acting at 25° above the horizontal. Given μ = 0.35, find the acceleration.

1

Resolve the applied force:

Horizontal:

40 cos 25° = 36.25 N

Vertical:

40 sin 25° = 16.90 N

(upward)

2

Find normal reaction (vertical equilibrium):

R + 16.90 = 6 \times 9.8

R = 58.8 - 16.90 = 41.9 N

3

Find friction force:

F = μ R = 0.35 \times 41.9 = 14.67 N

4

Find resultant horizontal force:

Resultant = 36.25 − 14.67 = 21.58 N

5

Apply F = ma:

21.58 = 6 \times a

a = 3.60 m s^{- 2}

Answer: Acceleration = 3.6 m s⁻²

Inclined Plane Dynamics

When an object moves on an inclined plane, we resolve forces parallel and perpendicular to the plane.

Resolving Forces

R(↖):

R - m g cos θ = 0

→

R = m g cos θ

R(↙):

m g sin θ - F = ma

With friction

F = μ R

:

m g sin θ - μ m g cos θ = ma

Direction of Friction

Object moving DOWN: Friction acts UP the slope

Object moving UP: Friction acts DOWN the slope

Always opposite to the direction of motion!

Worked Example04

A particle of mass 4 kg is released from rest on a rough plane inclined at 30° to the horizontal. Given μ = 0.2, find the acceleration of the particle.

1

Find the normal reaction:

R = m g cos 30° = 4 \times 9.8 \times 0.866 = 33.95 N

2

Find the friction force:

F = μ R = 0.2 \times 33.95 = 6.79 N

3

Find weight component down slope:

m g sin 30° = 4 \times 9.8 \times 0.5 = 19.6 N

4

Particle moves down, so friction acts up. Resultant down slope:

Resultant = 19.6 − 6.79 = 12.81 N

5

Apply F = ma:

12.81 = 4 \times a

a = 3.20 m s^{- 2}

Answer: Acceleration = 3.2 m s⁻² down the slope

05Worked Example

A box of mass 5 kg is on a rough plane inclined at 25° to the horizontal. A force of 40 N acts at 30° above the slope, pulling the box up the plane. Given μ = 0.3, find the acceleration of the box.

1

Resolve the 40 N force into components:

Parallel to slope:

40 cos 30° = 40 \times 0.866 = 34.64 N

Perpendicular to slope:

40 sin 30° = 40 \times 0.5 = 20 N

2

R(↖):

R + 40 sin 30° - m g cos 25° = 0

R = 5 \times 9.8 \times 0.906 - 20 = 44.39 - 20 = 24.39 N

3

Find friction:

F = μ R = 0.3 \times 24.39 = 7.32 N

4

R(↗): Taking up the slope as positive:

40 cos 30° - m g sin 25° - F = ma

34.64 - 5 \times 9.8 \times 0.423 - 7.32 = 5 a

34.64 - 20.73 - 7.32 = 5 a

5

6.59 = 5 a

a = 1.32 m s^{- 2}

Answer: Acceleration = 1.3 m s⁻² up the slope

Combining Dynamics with Kinematics

Once we find acceleration using F = ma, we can use the suvat equations to find velocity, distance, or time.

The SUVAT Equations

v = u + a t

s = u t + \frac{1}{2} a t^{2}

v^{2} = u^{2} + 2 a s

s = \frac{1}{2} (u + v) t

Two-Step Problems

Step 1: Use F = ma to find acceleration

Step 2: Use suvat equations to find v, s, or t

Choosing the Right Equation

Know u, a, t → find v: Use

v = u + a t

Know u, a, s → find v: Use

v^{2} = u^{2} + 2 a s

Know u, a, t → find s: Use

s = u t + \frac{1}{2} a t^{2}

Worked Example06

A crate of mass 10 kg is pulled from rest along a rough horizontal surface by a horizontal force of 60 N. Given μ = 0.25, find the speed of the crate after it has travelled 8 metres.

1

Find normal reaction:

R = m g = 10 \times 9.8 = 98 N

2

Find friction:

F = μ R = 0.25 \times 98 = 24.5 N

3

Find resultant: 60 − 24.5 = 35.5 N

4

Find acceleration using F = ma:

35.5 = 10 \times a

a = 3.55 m s^{- 2}

5

Use suvat with u = 0, a = 3.55, s = 8:

v^{2} = u^{2} + 2 a s

v^{2} = 0 + 2 \times 3.55 \times 8 = 56.8

v = 7.54 m s^{- 1}

Answer: Speed = 7.5 m s⁻¹

07Worked Example

A block of mass 4 kg is released from rest on a rough plane inclined at 30° to the horizontal. Given μ = 0.3, find the distance travelled in the first 2 seconds.

1

R(↖):

R - m g cos 30° = 0

R = 4 \times 9.8 \times cos 30° = 33.95 N

2

Find friction:

F = μ R = 0.3 \times 33.95 = 10.19 N

3

R(↙): Taking down the slope as positive:

m g sin 30° - F = ma

4 \times 9.8 \times 0.5 - 10.19 = 4 a

19.6 - 10.19 = 4 a

4

a = \frac{9.41}{4} = 2.35 m s^{- 2}

5

Use suvat with u = 0, a = 2.35, t = 2:

s = u t + \frac{1}{2} a t^{2} = 0 + \frac{1}{2} \times 2.35 \times 4 = 4.7 m

Answer: Distance = 4.7 m

Multi-Stage Problems

Some problems involve two or more stages of motion, where forces change partway through.

Common Scenarios

Stage 1: Object accelerates under applied force

Stage 2: Force removed → object decelerates due to friction only

Method for Multi-Stage Problems:

1

Analyse Stage 1: Find acceleration using F = ma

2

Use suvat to find final velocity of Stage 1

3

Final velocity of Stage 1 = Initial velocity of Stage 2

4

Analyse Stage 2: Find new acceleration using F = ma

5

Use suvat to find required quantity

Key Connection

The velocity at the end of one stage becomes the initial velocity for the next stage. This links the stages together.

Worked Example08

A box of mass 5 kg is pushed from rest along a rough horizontal floor (μ = 0.4) by a horizontal force of 50 N for 3 seconds. The force is then removed. Find the total distance travelled before the box stops.

1

Stage 1 (Force applied):

R = mg = 49 N, F = μR = 0.4 × 49 = 19.6 N

Resultant = 50 − 19.6 = 30.4 N

a_{1} = 30.4 \div 5 = 6.08 m s^{- 2}

2

Find velocity and distance after 3 s:

v = u + a t = 0 + 6.08 \times 3 = 18.24 m s^{- 1}

s_{1} = u t + \frac{1}{2} a t^{2} = 0 + \frac{1}{2} \times 6.08 \times 9 = 27.36 m

3

Stage 2 (Force removed):

Only friction acts: Resultant = −19.6 N

a_{2} = - 19.6 \div 5 = - 3.92 m s^{- 2}

4

Find distance in Stage 2 (u = 18.24, v = 0):

v^{2} = u^{2} + 2 a s

0 = 18.2 4^{2} + 2 \times (- 3.92) \times s_{2}

s_{2} = \frac{332.70}{7.84} = 42.44 m

5

Total distance =

s_{1} + s_{2} = 27.36 + 42.44 = 69.8 m

Answer: Total distance = 69.8 m

Practice

Question 1: A particle of mass 3 kg is pulled along a rough horizontal surface by a horizontal force of 24 N. The coefficient of friction is 0.45. Find the acceleration of the particle.

1

R = mg = 3 × 9.8 = 29.4 N

2

F = μR = 0.45 × 29.4 = 13.23 N

3

Resultant = 24 − 13.23 = 10.77 N

4

a = 10.77 ÷ 3 = 3.59 m s⁻²

Answer: Acceleration = 3.6 m s⁻²

Practice

Question 2: A box of mass 8 kg is released from rest at the top of a rough plane inclined at 35° to the horizontal. Given μ = 0.25, find the speed of the box after it has travelled 6 metres down the plane.

1

R = mg cos 35° = 8 × 9.8 × 0.819 = 64.21 N

2

F = μR = 0.25 × 64.21 = 16.05 N

3

mg sin 35° = 8 × 9.8 × 0.574 = 44.98 N

4

Resultant down slope = 44.98 − 16.05 = 28.93 N

5

a = 28.93 ÷ 8 = 3.62 m s⁻²

6

Using v² = u² + 2as: v² = 0 + 2 × 3.62 × 6 = 43.44

v = 6.59 m s⁻¹

Answer: Speed = 6.6 m s⁻¹

Summary

Key Formulas

ConceptFormula

Newton's Second Law

F_{res u lt an t} = ma

Friction (moving object)

F = μ R

Horizontal surface

P - μ m g = ma

Incline (down)

m g sin θ - μ m g cos θ = ma

Kinematics

v^{2} = u^{2} + 2 a s

The Dynamics Method

Draw force diagram

Find resultant force in direction of motion

Apply F = ma to find acceleration

Use suvat equations for velocity, distance, time

Exam Tips

Always state the direction of acceleration clearly

Remember: friction = μR when the object is moving

For multi-stage problems, link stages through velocity