# AS-Level Maths: Mechanics 1 for Edexcel

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AS-Level Maths: Mechanics 1 for Edexcel
M1.6 Statics of a particle These icons indicate that teacher’s notes or useful web addresses are available in the Notes Page. This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. 1 of 51 © Boardworks Ltd 2005

Contents Types of force Types of force Resolving forces
Particles in equilibrium Friction Examination-style questions Contents 2 of 51 © Boardworks Ltd 2005

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: Weight Normal reaction forces Tension Thrust Friction When an object remains at rest it is said to be in static equilibrium. From Newton’s first law, if the net force on an object is zero then it will remain at rest or move with constant velocity. The first situation is an example of static equilibrium and the second is an example of dynamic equilibrium. A resultant force acting on a body produces an acceleration which is proportional to the resultant force. 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. 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. 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, W = 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. Friction is a force that relates to the molecular structure of the two surfaces in contact. Molecules from one surface link with molecules from the other. As the normal contact force increases, more molecules from the two surfaces are ‘joined’ and so the friction force increases as a result. Approximate values of the coefficient of static friction for various surfaces in contact are Wood/metal 0.4, Wood/brick 0.6, Glass/glass 0.9, Silver/silver 1.4, Brake material/cast iron 0.5.

Thrust and Tension 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. T 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. T

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 This force is essentially electromagnetic in nature.

Contents Resolving forces Types of force Resolving forces
Particles in equilibrium Friction Examination-style questions Contents 8 of 51 © Boardworks Ltd 2005

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° Resolve this force into horizontal and vertical components: 7 N 30° a b 60° a = 7 cos30° = 6.06 to 3 s.f. b = 7 cos60° = 3.50 Therefore, 6.06 N act left and 3.50 N act upwards (to 3 s.f.).

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 = 5 cos35° = 4.10 (to 3 s.f.) Horizontal component = 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 a) Find the values of x, y and z. A fourth force of Newtons acts on the particle. b) Calculate the resultant force, F N, now acting on this particle. c) Find the magnitude of F.

Resultant forces a)  x = 5, y = 1 and z = –4. b) c) Magnitude =
Therefore the magnitude of the force is N = 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 a) 12 So the force acts at an angle of 67.4° (3 s.f.) with the horizontal. 5 b)  the magnitude of the direction vector is 13 N. For a magnitude of 39 N, a force 3 times the direction vector is needed (3 × 13 = 39). Therefore F = –15i + 36j.

Resultant forces Three coplanar forces act at a point in a vertical plane. 6 N 3 N P N 60° a) By resolving vertically, calculate the magnitude of P for which the forces are in equilibrium. Note that, in part a), the system is in equilibrium horizontally as 6cos60o = 3. 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.

Resultant forces a) Resolving vertically, P = 6 cos30° = 5.20
7 50° 40° a) Resolving vertically, P = 6 cos30° = 5.20 b) Resultant vertical force  = 7 – 6 cos40° = 2.40 N Resultant horizontal force  = 6 cos50° – 3 = 0.857 Using the triangle law of addition: R = 0.857 2.40 2.40 0.857 R If the correct direction is not chosen the force will be negative. This means that the force is acting in the opposite direction to that which has been taken. 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.

Particles in equilibrium
Types of force Resolving forces Particles in equilibrium Friction Examination-style questions Contents 20 of 51 © Boardworks Ltd 2005

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. Dynamic equilibrium occurs when the object is moving at constant velocity.

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 Ask students to draw the situation with all forces, then reveal the answer and discuss it. The rod is in thrust. This force is acting upwards.

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 The string is in tension. This force is acting away from the particle. mg

Example 3 An object is resting on a smooth horizontal surface. Draw a diagram showing the forces acting on the object. mg R R is used to denote the Normal Contact Force. (The letter N may sometimes be seen in place of 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. 2g T T = 2g = 2 × 9.8 = 19.6 Therefore the tension in the string is 19.6 N.

Question 2 A particle of mass 3 kg rests on a smooth horizontal surface. Find the force exerted on the particle by the table. 3g R R = 3g = 3 × 9.8 = 29.4 Therefore the force exerted on the particle by the table is 29.4 N.

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. Resolving vertically, T1cos75° + T2cos70° = 5g 15° 20° T1 T2 5g Resolving horizontally, T1cos15° = T2cos20°

Question 3 solution 80.3 (to 3 s.f.)
Therefore the tensions in the two strings are 82.5 N and 80.3 N respectively.

Contents Friction Types of force Resolving forces
Particles in equilibrium Friction Examination-style questions Contents 29 of 51 © Boardworks Ltd 2005

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 surfaces of the objects involved, specifically how rough they are, and The normal contact force. The frictional force is never higher than the force required to keep equilibrium. Friction is a force that relates to the molecular structure of the two surfaces in contact. Molecules from one surface link with molecules from the other. As the normal contact force increases, more molecules from the two surfaces are ‘joined’ and so the friction force increases as a result. Approximate values of the coefficient of static friction for various surfaces in contact are Wood/metal 0.4, Wood/brick 0.6, Glass/glass 0.9, Silver/silver 1.4, Brake material/cast iron 0.5.

Friction and the coefficient of friction
Friction needs to be taken into account when deciding whether a system is in equilibrium. The maximum frictional force that can act is proportional to 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 . Therefore Fmax = R where R is the normal contact force. Friction is a force that relates to the molecular structure of the two surfaces in contact. Molecules from one surface link with molecules from the other. As the normal contact force increases, more molecules from the two surfaces are ‘joined’ and so the friction force increases as a result. Approximate values of the coefficient of static friction for various surfaces in contact are Wood/metal 0.4, Wood/brick 0.6, Glass/glass 0.9, Silver/silver 1.4, Brake material/cast iron 0.5. 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
Start by adjusting the angle of the plane by dragging on it. Choose values for the coefficient of friction and the mass of the block. Ask students to decide whether the mass will slide when released. The numbers, including the values of R and θ, can be shown on the diagram if required.

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 = (to 3 s.f.) Therefore the tension in the string is N.

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° 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 Resolving vertically: R + T cos60° = 2g
F R 2g 30° Resolving vertically: R + T cos60° = 2g Resolving horizontally: F = T cos30° F = R T cos30° = 0.1 × (2g – T cos60°) T (cos30° cos60°) = 0.2g T = 2.14 (to 3 s.f.) Therefore the tension in the string is 2.14 N.

Friction along a rough horizontal surface
Start by adjusting the angle of the plane by dragging on it. Choose values for the coefficient of friction and the mass of the block. Ask students to decide whether the mass will slide when released. The numbers, including the values of R and θ, can be shown on the diagram if required.

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 2 F R 0.5g 15° Resolving vertically: R + 2 cos75° = 0.5g R = 0.5g – 2 cos75° = 4.38 (to 3 s.f.) Resolving horizontally: F = 2cos15° = 1.93 (to 3 s.f.)  = F/R = = (to 3 s.f.). Therefore the coefficient of friction is 0.441

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° 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 = R Hcos20° – 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° Therefore the horizontal force is 9.06 N.

Examination-style questions
Types of force Resolving forces Particles in equilibrium Friction Examination-style questions Contents 43 of 51 © Boardworks Ltd 2005

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) Resolving parallel to the plane: Tcos10° = 10gcos60°
T = = 49.8 (to 3sf) Therefore the tension in the string is 49.8 N.

Solution 1 Resolving parallel to the plane: Tcos10° + 30cos30° = 10gcos60° T = T = 23.4 (to 3 s.f.) Therefore the tension in the string is 23.4 N.

Examination-style question 2
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 By Pythagoras’ Theorem, AB = 5 cm.
The two angles marked  are alternate as are the two angles marked . Note that  = 90 -  Using trigonometry, sin = and cos = ; sin = and cos = .

Solution 2 Resolving horizontally, T1cos = T2cos
Resolving perpendicularly, T1cos(90 – ) + T2cos(90 – ) = 5g T1sin + T2sin = 5g Therefore T1 = 3g and T2 = 2g.

Examination-style question 3
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 ÷ ø ö ç è æ + 1 3 4 sin cos mg 2 θ

Solution 3 Resolving perpendicular to the plane, R + T cos60 = mgcos
R = mgcos – T Using F = R, F = mgcos – T Resolving parallel to the plane, F + mgcos(90 – ) = Tcos30 as required.