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AS-Level Maths: Mechanics 1for Edexcel

M1.5 Dynamics 2

For more detailed instructions, see the Getting Started presentation.


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Pulleys

Pulleys

Internal and external forces

Towing

Friction on a moving particle

Examination style questions


Connected particles

We talk about connected particles when we are considering situations where two objects are joined together.

1. In the first situation, two masses are connected by a string which passes over a pulley.

There are two types of situations that involve connected particles.

2. In the second situation, is one mass tows another; for example, a car towing a caravan or a train engine pulling a carriage.


Pulleys

Applying Newton’s second law:

Adding: m1g – m2g = m1a + m2am m

a gm m

1 2

1 2

m1g – T = m1a

T – m2g = m2a

Force = mass × accelerationF = ma

T = tension in Newtonsa = accelerationm = mass

This example shows how to find the acceleration of connected masses passing over a pulley.


Pulleys

Applying Newton’s second law:

Adding: m2g – m1gsin = m1a + m2a

sinm ma g

m m

2 1

1 2

m2g – T = m2a

T – m1gcos(90 – ) = m1a

This example shows how to find the acceleration of connected masses passing over a pulley, when the masses are not acting vertically.

90 – θ is the angle between the direction of

the force, and the vertical.


Pulley question 1

Particles of mass 3 kg and 7 kg are attached to each end of a light inextensible string which passes over a smooth fixed pulley.

The system is released from rest.

a) Find the acceleration of the system.

b) Find the tension in the string.


Pulley question 1 solution

a) Consider the 3 kg particle:Consider the 7 kg particle:

Adding:

Therefore the acceleration of the system is 3.92 m s–2.b) T – 3g = 3a

Therefore the tension in the string is 41.2 N.

T – 3g = 3a7g – T = 7a

4g = 10aa = 4g ÷ 10 = 3.92 (to 3 s.f.)

T = (3 × 9.8) + (3 × 3.92) = 41.2 (to 3 s.f.)


Pulley question 2

A particle A of mass 3 kg is resting on a smooth, horizontal table top, which is 2 m above the floor. This particle is connected to a particle B of mass 2 kg by a light inextensible string which hangs freely over a smooth fixed pulley at the edge of the table.

Particle A is held at rest at a point 1 m from the pulley. The system is then released from rest.

a) Find the acceleration of the system before A reaches the pulley.

b) Find the time taken for A to reach the pulley.



a) Consider particle A:

Adding:

Therefore the acceleration of the system is 3.92 m s–2.

Consider particle B:T = 3a

2g – T = 2a

a = 2g ÷ 5 = 3.92 (to 3 s.f.)2g = 5a


Pulley question 2 solution cont.

b) Consider particle A:s = 1u = 0a = 3.92t is unknown

Using s = ut + ½at2:

It takes 0.714 seconds for particle A to reach the pulley.

1 = 0 + 1.96t2

t = 0.714 (to 3 s.f.)

distance (m)

initial velocity (m s–1)

acceleration (m s–2)

time (sec)

Constant acceleration formula


Pulley question 3

A particle A of mass 4 kg is resting on a rough plane inclined at an angle of 30° to the horizontal. The particle is connected by a light inextensible string to a particle B of mass 6 kg, which hangs freely over a smooth fixed pulley at the top of the inclined plane.

The coefficient of friction between the particle and the plane is 0.15.


Find the acceleration of the system.



Resolve perpendicular to the plane: R = 4gcos30° = 33.9 (to 3 s.f.)

F = R = 0.15 × 33.9 = 5.09 (to 3 s.f.)

Consider particle A: T – 5.09 – 4gcos60° = 4a

Consider particle B: 6g – T = 6aAdding: 6g – 5.09 – 4gcos60° = 10a

a = (6g – 5.09 – 4gcos60°) = 3.41 (to 3 s.f.)


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10


Pulley question 4

Example 4: A particle A of mass 3 kg is resting on a sloping surface. It is connected to another particle B of mass 5 kg, which hangs freely by a light inextensible string that passes over a smooth pulley at the bottom of the slope.

The sloping surface is assumed to be smooth and is inclined at an angle of 20° to the horizontal.

Find the acceleration when the system is released from rest.



Consider particle A: T + 3gcos70° = 3aConsider particle B: 5g – T = 5a

Adding: 5g + 3gcos70° = 8a

a = (5g + 3gcos70°)


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8

In this example, gravity and the tension in the string are pulling particle A in the same direction.


Pulley question 5

Example 5: Two particles of mass 2 kg and 5 kg are resting either side of a rough inclined slope. The side that the 2 kg particle rests on is inclined at an angle of 25° to the horizontal and the other side is inclined at an angle of 30° to the horizontal.

The coefficients of friction between the particles and the surface of each side of the slope are equal.

The acceleration when the system is released from rest is 2 m s–2.

a) Find the coefficient of friction between the particles and the slope.

b) Find the tension in the string.


Pulley question 5 Solution

a) Resolve perpendicular to the plane: R1 = 2gcos25°

Resolve perpendicular to the plane: R2 = 5gcos30°Consider the 2 kg particle:

T – 2gcos65° – F1 = 2 × 2

T – 2gcos65° – 2gcos25° = 4Consider the 5 kg particle:

mass × acceleration

Remember, F = R

5gcos60° – F2 – T = 5 × 2

5gcos60° – 5gcos30° – T = 10



Adding: 5gcos60° – 5gcos30° – 2gcos65° – 2gcos25° = 14

5gcos30° + 2gcos25° = 5gcos60° – 2gcos65° – 14

– – ( )

.

° °

°

5 cos60 2 cos65 14to 3 s.f.

5 cos30 2 cos250 036

°8

g g

g g

Therefore,

b) Consider the 2 kg particle: T – 2gcos65° – 2gcos25° = 4

Therefore, T = 2gcos65° + 4 + 2gcos25° × 0.0368

T = 12.9 (to 3 s.f.)

The tension in the string is 12.9 N.


Pulley question 6

Example 6: Two particles A and B are connected by a light inextensible string which passes over a smooth peg. Particle A has mass 3m kg and is resting on a smooth horizontal table. Particle B of mass m kg is hanging freely over the peg which is positioned on the edge of the table.


Find the tension in the string in terms of m.



Consider the 3m kg particle: T = 3maConsider the m kg particle: mg – T = ma

Adding: mg = 4mag = 4aa = ¼g = 2.45

T = 3ma = 3m × 2.45 = 7.35m (to 3 s.f.)

Therefore the tension in the string is 7.35m N.


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Pulleys

Pulleys


Towing





Whether a force is internal or external has nothing to do with the force itself, but rather the system under consideration.

So, whether a force is internal or external is not an intrinsic property of the force.

The next two examples illustrate the distinction between internal and external forces.


Towing example

Example: A car of mass m1 pulls a trailer of mass m2 along a straight horizontal road. The resistance to the motion of the car is F1 and the resistance to the motion of the trailer is F2. The engine of the car produces a driving force of magnitude R.Find, in terms of m1, m2, F1, F2 and R,

a) The acceleration of the system.

b) The tension in the coupling between the car and the trailer.


Towing example solution

a) When considering the motion of the car and trailer as a single entity, the tension in the coupling is an internal force and does not appear on the diagram.

Applying Newton’s second law:R – F1 – F2 = (m1 + m2)a

1 2

21

R F F

m ma



In order to calculate the tension in the coupling, the motion of the car and trailer must be considered separately. In this system, the tension is now an external force. The tension can be calculated by looking at either the motion of the car or the trailer. You could use the unused method to check the answer.

T is now an external force and must appear on the diagram.

Forces acting on the trailer:



Applying Newton’s second law:T – F2 = m2a T = F2 + m2a

1 22 2

1 2

R F FT F m

m m

1 2 2 2 2 2 1 2 2

1 2

m F m F m R m F m FT

m m

1 2 2 2 2

1 2

m F m R m FT

m m

Substituting in answer to part a).



Forces acting on the car:

Applying Newton’s second law:R – T – F1 = m1a T = R – F1 – m1a

1 211

1 2

R F FT R F m

m m

1 1 1 1 21 2 1 1 2 2

1 2

m R m R m F m F m R m F m FT

m m

1 2 2 2 2

1 2

m F m R m FT

m m


Lift example

Example: A man of mass m1 is travelling upwards in a lift of mass m2 which has constant acceleration a. The tension in the cable of the lift is T.Find in terms of m1, m2, g and T,

a) The acceleration of the lift.

b) The normal contact force, R, between the man and the floor of the lift.


Lift example solution

When considering the motion of the man and the lift together the normal contact force is an internal force and does not appear on the diagram.

a) Applying Newton’s second law:T – m1g – m2g = (m1 + m2)a

1 2

1 2

T m g m ga

m m



In order to calculate the normal contact force between the man and the floor of the lift, the motion of the man and the lift must be considered separately. The normal contact force can be calculated by looking at either the man or the lift. You could use the unused method to check the answer.

b) Forces acting on the man:

R is now an external force and must appear on the diagram.



Applying Newton’s second law:R – m1g = m1aR = m1a + m1g

1 21 1

1 2

T m g m gR m m g

m m

2 2

1 1 1 2 1 1 2

1 2

m T m g m m g m g m m gR

m m

1

1 2

m TR

m m

Substituting in answer to part a).



Forces acting on the lift:

Applying Newton’s second law:T – N – m2g = m2aN = T – m2g – m2a

Newton’s third law dictates that the force that the man exerts on the floor of the lift is equal and opposite to the force on the man.

1 22 2

1 2

T m g m gN T m g m

m m

2 2

1 1 2 2 1 22 2 2

1 2

m T m T m m g m g m T m m g m gN

m m

1

1 2

m TN

m m

So, N = R


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Pulleys

Pulleys


Towing




Towing question 1

A car is pulling a caravan along a straight horizontal road. The mass of the car is 1200 kg and the mass of the caravan is 900 kg.

The car and caravan are subject to resistive forces of 600 N and 550 N respectively.

There is a constant driving force of 2200 N acting on the car.

a) Find the magnitude of the acceleration of the car

and caravan.

b) Find the tension in the towbar.


Towing question solution

a) Applying Newton’s second law to the whole system:2200 – 600 – 550 = (900 + 1200) × a

2100a = 1050 a = 0.5

Therefore the acceleration of the car and caravan is 0.5 m s–2.



b) Applying Newton’s second Law to the car only,2200 – 600 – T = 1200 × 0.5

T = 2200 – 600 – 600 T = 1000

Therefore the tension in the towbar is 1000 N.


Towing question 2

A train comprises an engine of mass 40,000 kg coupled to two trucks, each of mass 8000 kg. The train moves along a horizontal road. The resistances to the motion of the engine, middle truck and last truck are 15,000 N, 5000 N and 2000 N respectively. The tension in the coupling between the two trucks is 0.

a) Show that the train is decelerating and find the magnitude of this deceleration.

b) Find the tension in the coupling between the engine and the middle truck.

c) Determine whether the engine is exerting a braking or driving force, and find the magnitude of this force.



a) Applying Newton’s second law to the last truck:0 – 2000 = 8000a

a = = –0.25

Therefore the train is decelerating at a rate of 0.25 m s–2.

–2000

8000



b) Applying Newton’s second law to the middle truck:T – 5000 = 8000 × –0.25

T = 5000 – 2000

T = 3000




Use F = ma on the whole train: E – 15,000 – 5000 – 2000 = 56,000 × –0.25

E = 15,000 + 5000 + 2000 – 14,000

E = 8000

Therefore the engine is exerting a driving force of 8000 N.


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Pulleys

Pulleys


Towing




The coefficient of friction

Friction is a very common force that acts on objects moving relative to each other (for example a block sliding along a table) to eventually slow them down.

Many of the examples involving moving objects have involved a resistive force. This is often due to friction.

Friction depends on the roughness of the bodies touching and on the normal contact force. The roughness is characterised by the coefficient of friction, μ, and the frictional force is then F = μR.

As μ gets closer to 1, the rougher the contact between the surfaces. As μ gets closer to 0, the smoother the contact.


Question 5

A children’s slide is inclined at an angle of 25° to the horizontal. A child of mass 30 kg goes down the slide with a constant speed.

a) Draw a diagram showing all the forces.

b) Calculate R, the normal contact force.

c) Calculate the coefficient of friction between the child and the slide.

a)

25°

R

F

30g

a = 0


Question 5 Solution

b) Resolving perpendicular to the plane,

c) Applying Newton’s Second Law down the plane,

R = 30g cos25°

R = 266 (to 3 s.f.)

Therefore R is 266 N.

30g cos65° – F = 0

F = 124 (to 3 s.f.)

but, F = R

= 124 ÷ 266 = 0.466 (to 3 s.f.)


Question 6

A children’s slide is inclined at an angle to the horizontal. A child of mass 25 kg goes down the slide with a constant speed.

Calculate given that = 0.5.

θ

R

F

25g

a = 0 = 0.5


Question 6 Solution

Resolving perpendicular to the plane, R = 25g cos

Applying Newton’s Second Law down the slope,

We know that cos(90 – )° = sin, so F = 25g sin.

25g cos(90 – )° – F = 0

F = 25g cos(90 – )°

= FR

tan = 0.5

We know that sin ÷ cos = tan, so cancelling gives us:

= 26.6° (to 3 s.f.)

gg

25 sin0.5=

25 cos


Question 7

A small wooden block is projected across a horizontal floor.

Initially it has a speed of 3 ms–1, and it comes to rest after travelling 2 m.

a) Find the acceleration of the block.

b) Find the coefficient of friction between the block and the floor.

m F

R

mg

a


Question 7 Solution

a) Using v2 = u2 + 2as with s = 2, u = 3 and v = 0:

The block is therefore decelerating at 2.25 ms–2.

b) Resolving perpendicularly, R = mg

Applying Newton’s Second Law:

=μ

0 = 9 + 4a a = –2.25

–F = m × –2.25 F = 2.25m

=FR

mmg

2.25 =

m F

R

mg

a = –2.25

0.230 (to 3 s.f.)


Question 8

A lift is accelerating upwards at 2 ms–2. A woman of mass 60 kg is standing in the lift.

Find the normal contact force between the woman and the floor of the lift.

R

60g

a = 2 ms–2

Applying Newton’s Second Law:

R – 60g = 60 × 2

R = 60g + 120

R = 708

Therefore the normal contact force is 708 N.


Question 9

A particle of mass m kg is sliding down a rough plane inclined at 20° to the horizontal. The coefficient of friction between the particle and the surface of the plane is 0.2.

Find the acceleration of the particle down the slope.

20°

R

F

mg

a = 0 μ = 0.2


Question 9 Solution

Applying Newton’s Second Law down the plane:

Resolving perpendicular to the plane, R = mg cos20°

F = R = 0.2mg cos20°

cos70 0.2 cos20= =

mg mga

mg

the particle is accelerating at 0.154 ms–2.

= 0.154 (to 3 s.f.)

mg cos70° – F = mga

F = mg cos70° – mga

cos 70° – 0.2 cos 20°

= 0.2mg cos20°


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Examination-style questions

Pulleys


Towing


Examination-style questions


Examination-style question 1

Two particles, A and B, of masses 1 kg and 2 kg respectively, are connected by a light, inextensible string which passes over a smooth, fixed pulley.This pulley is at the top of a rough plane inclined at an angle of 30° to the horizontal.

A is resting on the plane 2.5 m from the pulley. B is hanging freely over the pulley 1 m above the ground. The coefficient of friction between particle A and the slope is 0.25.

a) Find the acceleration of the system when it is released from rest.

b) Find the time taken for B to reach the ground.

c) Find the distance travelled by A up the slope after B has reached the ground.


Solution 1

a) Resolving perpendicular to the plane, R = gcos30°

Using F = R: F = ¼gcos30° = 2.12 (to 3 s.f.)

Consider A: T – 2.12 – gcos60° = a

Consider B: 2g – T = 2a

Adding: 2g – 2.12 – gcos60° = 3a

a = (2g – 2.12 – gcos60°) = 4.19 (to 3 s.f.)


1

3


Solution 1

For B , u = 0a = 4.19s = 1t = ?

Using s = ut + ½at2

1 = 0 + 0.5 × 4.19 × t2

t2 = 0.477…

t = 0.691 (to 3 s.f.)

Therefore B takes 0.691 seconds to reach the ground.


Solution 1


Solution 1

Consider A until B reaches the ground:u = 0a = 4.19t = 0.691v = ?

Using v = u + at: v = 0 + 4.19 × 0.691

= 2.90 (to 3 s.f.)

v = velocity

So, the velocity of particle A at the moment that B reaches the ground, is 2.90 m s–1.


Solution 1

When B reaches the ground the string becomes slack and there is no longer a force pulling A up the slope.

Using F = ma up the slope: –2.12 – gcos60° = a a = –7.02 (to 3 s.f.)

For A: u = 2.90a = –7.02v = 0s = ?

Using v2 = u2 + 2as: s =

s = 0.599 (to 3 s.f.)Therefore, A travels a further 0.599 m up the slope after B stops.

( ) .

.

2 2 22 90

2 14 04

v u

a



Two particles, A and B, have masses 3m kg and km kg respectively, where k>3. The particles are connected by a light, inextensible string which passes over a smooth, fixed pulley.

The system has an acceleration of m s–2 when it is released from rest.

a) Find the tension in the string in terms of m and g.

b) Find the value of k.

5

g


Solution 2

a) Consider particle A: T – 3mg =

T =

b) Consider particle B: kmg –

so, 4kmg = 18mg

Therefore, k = 4.5

3

5

mg

18

5

mg

18

5 5

mg kmg

4 18

5 5

kmg mg

4k = 18



A car of mass 850 kg is being towed by a tow-truck of mass 1500 kg. The car and the tow-truck are joined by a light towbar, which is at an angle of 20° to the ground.

The car and the tow-truck experience resistances to motion of 250 N and 600 N respectively.

The driving force exerted by the tow-truck is 3200 N.

a) Find the acceleration of the car and tow-truck.

b) Find the tension in the towbar.


Solution 3

Applying Newton’s second law to the whole system:3200 – 600 – 250 = 2350 × a2350a = 2350a = 1

Therefore the acceleration of the system is 1 m s–2.


Solution 3

Applying Newton’s second law to the car:


T = 1170 (to 3 s.f.)

850 250

cos20

T =

Tcos20° – 250 = 850 × 1



A particle of mass m kg slides down a rough plane inclined at 25° to the horizontal.

The particle passes through a point A with speed 3 ms–1, and 2 seconds later passes a point B with speed 9 ms–1.

a) Find the acceleration of the particle.

b) Find F in terms of m.

c) Find the coefficient of friction between the particle and the surface of the slope.


Solution 4

a) Acceleration = = 3 ms–2

Applying Newton’s Second Law down the plane:

c) Resolving perpendicular to the plane:

FR

= =

9 3

2

mg cos65° – F = m × 3

F = mg cos65° – 3m

R = mg cos25°

b)

25°

R

mg

a = 3 ms–2

F

mg mmg

cos65 3=

cos250.129 (to 3 s.f.)


Examination-style Question 5

A sledge of mass 70 kg is pulled up a slope inclined at an angle to the horizontal where tan = .

The slope is modelled as a rough inclined plane and the rope as a light, inextensible string acting parallel to the line of greatest slope.

The coefficient of friction between the sledge and the slope is 0.3 and the sledge is accelerating up the slope with an acceleration of 0.4 ms–2.

Find the tension in the rope.

740


Solution 5

sin ≈ 7 ÷ 41

From the right-angled triangle with smaller sides of length 7 and 40, which has hypotenuse length ≈ 41, we can deduce the values of sin and cos :

2 240 + 7

= 0.3

θ

R

70g

a = 0.4 ms–2

T

F

Before we resolve the forces, look at tan = .740

and cos ≈ 40 ÷ 41


Solution 5

Resolving perpendicular to the plane,

F = R = 0.3 × 669.3 = 200.8

Applying Newton’s Second Law up the plane,

Therefore the tension in the rope is 346 N.

R = 70g cos R = 669.3

T = 346 (to 3 s.f.)

T = 200.8 + 117.1 + 28

T – 200.8 – 70g sin = 70 × 0.4

T – 200.8 – 70g cos(90 – )° = 70 × 0.4

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© Boardworks Ltd 20051 of 67 These icons indicate that teacher’s notes or useful web addresses are available in the Notes Page. This icon indicates the