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AS-Level Maths: Mechanics 1for Edexcel

M1.6 Statics of a particle

For more detailed instructions, see the Getting Started presentation.


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nte

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Types of force

Types of force

Resolving forces

Particles in equilibrium

Friction

Examination-style questions


Types of force

There are many different types of force that may act on an object.

The most common ones met in mechanics problems are:

When an object remains at rest it is said to be in static equilibrium.

Weight

Normal reaction forces

Tension

Thrust

Friction

This state occurs when the net force acting on the object is zero.

If the net force is not zero, then the object will accelerate.


Weight and acceleration due to gravity

The acceleration due to gravity, g, is the acceleration that a body in free fall experiences if air resistance and other forces are neglected.

The weight, W, of a body is the downward force that the earth exerts on the body due to its mass. This force, by Newton’s Second Law, is equal to the mass multiplied by the acceleration produced.

Therefore,

On or near the surface of the earth g is taken to be approximately 9.8 ms–2 whereas on the moon the acceleration due to lunar gravity would be approximately 1.6 ms–2.

W = mgW = mg


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.

Friction also acts to stop one object moving relative to the other when it would otherwise do so because of a force acting on it. So we need to consider friction when deciding whether a system is in equilibrium.

Friction depends on the roughness of the bodies touching – compare an iron bar sliding over grass to on an ice-hockey puck sliding on ice.


Thrust and Tension

A taut string or rod under extension produces an inward force at each end. This is known as the tension in the string or rod.

A rod under compression produces an outward force at each end. This is known as the thrust in the rod.

Note that a string cannot be compressed and so cannot produce a thrust.

TT

TT


Normal reaction force

When two surfaces are pressed against one another then a force, called the normal reaction force or normal contact force, acts between the two surfaces.

This force acts perpendicular to the area of the surfaces in contact.

Normal reaction force

Weight


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Resolving forces

Types of force

Resolving forces


Friction



Forces as vectors

A force is a vector quantity, so when we deal with forces, we can use vector arithmetic.

One very important skill in mechanics is finding the component of a force in a given direction.

This enables the resultant of several forces acting in different directions to be calculated.


Resolving forces

The component of a force in a given direction is equal to the magnitude of the force multiplied by the cosine of the angle between the force and the given direction.

A B

C

F

The component of the force in the direction AB = F cos.

The component of the force in the direction AC = F cos(90 – )

= F sin.


Resolving forces

A force of 7 N acts on a particle at an angle of 30° to the horizontal. 7 N

30°

a = 7 cos30° = 6.06 to 3 s.f.

Therefore, 6.06 N act left and 3.50 N act upwards (to 3 s.f.).

Resolve this force into horizontal and vertical components:

b = 7 cos60° = 3.50

7 N

30°a

b60°


Resolving forces

A force of 5 N acts on a particle at an angle of 55° to the horizontal. 5 N

55°

Resolve this force into vertical and horizontal components:

Vertical component =

Horizontal component =

5 cos35° = 4.10 (to 3 s.f.)

5 cos55° = 2.87 (to 3 s.f.)

Therefore, 4.10 N act upwards and 2.88 N act left (to 3 s.f.).


Resolving forces

A force of 10 N acts on a particle at an angle of 40° to the vertical.

Calculate the horizontal and vertical components of this force.

10 N 40°

Vertical component = 10 cos40° = 7.66 (to 3 s.f.)

Horizontal component = 10 cos50° = 6.43 (to 3 s.f.)

Therefore, 7.66 N act downwards and 6.43 N act to the left (to 3 s.f.).


Resultant forces

A particle is in equilibrium when acted on by the forces

-6 1

2 , and -3

3 7

x

y

z

a) Find the values of x, y and z.

A fourth force of Newtons acts on the particle.

7

8

1

b) Calculate the resultant force, F N, now acting on this particle.

c) Find the magnitude of F.


Resultant forces

b) 5 6 1 7

2 + 1 + 3 + 8 =

3 7 4 1

c) Magnitude = 22 27 + 8 +1 = 49 + 64 +1 =

a)

6 1

2 3

3 7

x

y

z

0

Therefore the magnitude of the force is N 114

7

8

1

x = 5, y = 1 and z = –4.

114

= 10.7 N (to 3 s.f.)


Resultant forces

A force of magnitude 39 N acts in the direction of –5i + 12j, where i and j are the standard unit vectors.

a) Calculate the angle this force makes with the negative i direction.

b) Find the force in the form ai + bj.


Resultant forces

5

12So the force acts at an angle of 67.4° (3 s.f.) with the horizontal.

For a magnitude of 39 N, a force 3 times the direction vector is needed (3 × 13 = 39). Therefore F = –15i + 36j.

12tan 67.4

5° a)

2 25 12 5 12 13 i jb) the magnitude of the direction vector is 13 N.


Resultant forces

Three coplanar forces act at a point in a vertical plane.

a) By resolving vertically, calculate the magnitude of P for which the forces are in equilibrium.

b) If P = 7 and the 6 N force now acts at 50° to the horizontal, find the magnitude and direction of the resultant force.

6 N

3 N

P N

60°


Resultant forces

a) Resolving vertically, P = 6 cos30° = 5.20 b) Resultant vertical force =

2.400.857

Rx°

2.40tan 70.3

0.857°x x

Therefore the resultant force is 2.55 N (to 3 s.f.) at an angle of 70.3° (to 3 s.f.) to the negative horizontal.

Using the triangle law of addition:

0.857

2.40

6

3

7

50°40°

7 – 6 cos40° = 2.40 N Resultant horizontal force =

6 cos50° – 3 = 0.857

2 20.857 2.40 2.55 R =


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Types of force

Resolving forces


Friction



Static equilibrium

When an object remains at rest it is said to be in static equilibrium.

This state occurs when an object remains still and the net force acting on it is zero.

If the net force is not zero, then the object will accelerate.

To decide whether an object is in equilibrium, we must resolve all the forces in all the dimensions we are considering to see that no net force is acting.


Example 1

An object is resting on the top of a vertical rod. Draw a diagram showing the forces acting on the object.

T

mg


Example 2

An object is hanging in equilibrium from the bottom of a vertical string. Draw a diagram showing the forces acting on the object.

T

mg


Example 3

An object is resting on a smooth horizontal surface. Draw a diagram showing the forces acting on the object.

mg

R


Question 1

A particle is hanging in equilibrium at the end of a light inextensible string. The mass of the particle is 2 kg.

Find the tension in the string.

T = 2g = 2 × 9.8 = 19.6

Therefore the tension in the string is 19.6 N.

2g

T


Question 2

A particle of mass 3 kg rests on a smooth horizontal surface.

Find the force exerted on the particle by the table.

R = 3g = 3 × 9.8 = 29.4

Therefore the force exerted on the particle by the table is 29.4 N.3g

R


Question 3

A particle of mass 5 kg is held in equilibrium by two light inextensible strings suspended at angles of 15° and 20° respectively to the horizontal.

Find the tension in each string.

15° 20°

T1 T2

5g

Resolving vertically,T1cos75° + T2cos70° = 5g

Resolving horizontally,T1cos15° = T2cos20°

12cos20

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°

°s15

TT


Question 3 solution

22

cos20°cos75° cos70° 5

cos15°

TT g

. ( )2 82 5 to 3 s.f.T

21

cos20°= =

cos15°

TT

Therefore the tensions in the two strings are 82.5 N and 80.3 N respectively.

2cos20°

cos75° cos70° 5cos15°

T g

80.3 (to 3 s.f.)


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Friction

Types of force

Resolving forces


Friction



Friction and the coefficient of friction

If two rough surfaces are in contact then a frictional force may also act to prevent relative motion between the two surfaces.

The size of the frictional force depends on several things.

The maximum size of the frictional force depends on:

The frictional force is never higher than the force required to keep equilibrium.

The surfaces of the objects involved, specifically how rough they are, and

The normal contact force.


Friction and the coefficient of friction

Friction needs to be taken into account when deciding whether a system is in equilibrium.

Therefore Fmax = RFmax = R

The maximum frictional force that can act is proportional to the normal contact force.

where R is the normal contact force.

The constant of this proportionality for any two given surfaces is called the coefficient of friction and is usually written as .

Generally, the rougher the two surfaces, the closer is to 1. The smoother the surfaces, the closer is to 0.


Friction along a rough inclined plane


Question 4

A particle of mass 0.15 kg is resting on a smooth plane inclined at an angle of 30° to the horizontal. The particle is held in equilibrium by a light inextensible string acting up the line of greatest slope.

Find the tension in the string.

Resolving parallel to the plane:

T = 0.15g cos60°

T = 0.735 (to 3 s.f.)



Question 5

A particle of mass 2.5 kg is held in equilibrium on a smooth plane inclined at an angle of 25° to the horizontal by means of a horizontal force H.

Find the magnitude of this force.

Resolving parallel to the plane:H cos25° = 2.5g cos65°

.. ( )

°

°

2 5 cos65= =11 4 to 3 s.f.

cos25

gH

Therefore H has a magnitude of 11.4 N.


Question 6

A particle of mass 2 kg is resting on a rough horizontal surface. The coefficient of friction between the particle and the surface is 0.1. A light inextensible string is attached to the right of the particle at an angle of 30° to the horizontal.

If the particle is on the point of moving to the right, find the tension in the string.


Question 6 solution

T

F

R

2g

30°

Resolving vertically: R + T cos60° = 2g

Resolving horizontally: F = T cos30°

F = RT cos30° = 0.1 × (2g – T cos60°)

T (cos30° + 0.1 cos60°) = 0.2g

T = 2.14 (to 3 s.f.)



Friction along a rough horizontal surface


Question 7

A particle of mass 0.5 kg is resting on a rough horizontal surface. A light inextensible string is attached to the particle at an angle of 15° to the horizontal and exerts a force of 2 N.

If the particle is on the point of moving in the direction of the force, find the coefficient of friction between the particle and the surface.


Question 7 solution

Resolving vertically: R + 2 cos75° = 0.5gR = 0.5g – 2 cos75° = 4.38 (to 3 s.f.)

Resolving horizontally: F = 2cos15° = 1.93 (to 3 s.f.)

Therefore the coefficient of friction is 0.441

2

F

R

0.5g

15°

= F/R = = 0.441 (to 3 s.f.)..

.

1 93

4 38


Question 8

A particle of mass 1.7 kg rests on a rough plane inclined at an angle of 20° to the horizontal. A horizontal force, H, is applied to the particle. The coefficient of friction between the particle and the plane is 0.15.

Find H if

a) The particle is on the point of sliding down the plane.

b) The particle is on the point of sliding up the plane.


Question 8 solution a

Resolving perpendicular to the plane:R = 1.7g cos20° + H cos70°

Resolving parallel to the plane:F + Hcos20° = 1.7gcos70° F = 1.7gcos70° – Hcos20°

F = R 1.7gcos70° – Hcos20° = 0.15(1.7gcos20° + Hcos70°)0.15Hcos70° + Hcos20° = 1.7gcos70° – 0.15 × 1.7gcos20°H(0.15cos70° + cos20°) = 1.7gcos70° – 0.15 × 1.7gcos20°

. . . .

.

1 7 cos70 0 15 1 7 cos203 38 (to 3 s.f.)

0 15cos70 cos

° °

° °20

g gH

Therefore the horizontal force is 3.38 N.


Question 8 Solution b

Resolving perpendicular to the plane:R = 1.7gcos20° + Hcos70°

Resolving parallel to the plane:F + 1.7gcos70° = Hcos20° F = Hcos20° – 1.7gcos70°

F = RHcos20° – 1.7gcos70° = 0.15(1.7gcos20° + Hcos70°)Hcos20° – 0.15Hcos70° = 0.15 × 1.7gcos20° + 1.7gcos70° H(cos20° – 0.15cos70°) = 0.15 × 1.7gcos20° + 1.7gcos70°

. . . . )

.

0 15 1 7 cos20° 1 7 cos70°9 06 (to 3 s.f.

cos20° 0 15cos70°

g gH

Therefore the horizontal force is 9.06 N.


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Types of force

Resolving forces


Friction



Examination-style question 1

A particle of mass 10 kg is held in equilibrium on a smooth plane inclined at an angle of 30o to the horizontal by means of a light inextensible string acting at an angle of 10° to the plane.

a) Draw a diagram showing all the forces acting on the particle.

b) Calculate the tension in the string.

A horizontal force of 30 N is applied to the particle to try to make it slide up the slope. The tension in the string is adjusted so that the particle remains in equilibrium.

c) Calculate the new tension in the string.


Solution 1

a)

b) Resolving parallel to the plane:

Tcos10° = 10gcos60°

T = = 49.8 (to 3sf)


10 cos60°

cos10°

g


Solution 1

c) Resolving parallel to the plane:Tcos10° + 30cos30° = 10gcos60°

T =

T = 23.4 (to 3 s.f.)


10 cos60° – 30cos30°

cos10°

g



A and B are two fixed points on a horizontal line. Two light inextensible strings are attached to A and B and the other ends are attached to a particle C of mass 5 kg. AC = 3 cm, BC = 4 cm and angle C = 90°. The strings are holding particle C in equilibrium.

Find the tensions in the two strings in terms of g.


Solution 2

4

5

3

5

3

5

By Pythagoras’ Theorem, AB = 5 cm.

The two angles marked are alternate as are the two angles marked .

Using trigonometry,

sin = sin = and cos = and cos = 4

5; .


Solution 2

Resolving horizontally,T1cos = T2cos

1 23 4

5 5T T

1 24

3T T

Resolving perpendicularly,T1cos(90 – ) + T2cos(90 – ) = 5gT1sin + T2sin = 5g

1 24 3

55 5

T T g 22

4 4 35

5 3 5T T g

225

515

T g

Therefore T1 = 3g and T2 = 2g.



A particle of mass m kg lies on a rough plane inclined at an angle of ° to the horizontal. The particle is held in equilibrium by means of a light inextensible string held at an angle of 30° to the plane.

The coefficient of friction between the plane and the particle is 0.25 and the particle is about to slide up the plane.

Show that the tension in the string is

134sin4cosmg2 θθ


Solution 3

Resolving perpendicular to the plane,

Resolving parallel to the plane,

R + T cos60 = mgcos1

2R = mgcos – T1

41

8Using F = R, F = mgcos – T

F + mgcos(90 – ) = Tcos301

cos sin cos304

mg T mg T 4 3 1

=8

T

1 3 1cos + sin +

4 2 8mg T mg T T

cos + 4sin= 2

4 3 +1T mg

as required.

© boardworks ltd 2005 1 of 51 these icons indicate that teachers notes or useful web addresses are...

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