@MEIConference #MEIConf2019

#MEIConf2019

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#MEIConf2019

Mechanics: A Review of the Principles

(Actually just forces!) Mechanics involves mathematical modelling – representing

the real world with mathematics.

Students need the necessary mathematical skills: solving

equations, using graphs, calculus, geometry and

trigonometry.

They need clarity about the mechanical principles and an

understanding of how they apply to the physical world

appropriate for their level of study.

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The 4 Step Way of Mechanics 1. Diagram:

Shows all the problem’s data, defines the variables fully.

2. Principle(s):

State what it applies to, direction, start and end points.

3. Equation(s):

n unknowns will require n equations.

4. Solution.

Involves simultaneous equations (tricky ones), trigonometry,

quadratic equations etc.

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Forces: Resolving and Addition

A force can be resolved into perpendicular

components.

Forces can be added to give the resultant force, 𝐹.

Parallel forces can be replaced by a single force

using the principle of moments (not usually

required at A-level).

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Types of Forces 1. Weight:

The force between objects and the Earth.

𝑊 = 𝑚𝑔, where 𝑔 is the gravitational field strength, which is

about 9.8 Newtons per kilogram on the Earth’s surface.

2. Contact Force:

The force between solids in contact. Solids resist being pushed into each other – they react by

pushing each other apart – the origin of this is the electrostatic

repulsion between electrons in the atoms.

When rough surfaces are in contact we split the contact force

into two components: the normal reaction (normal to the

surface) and friction (parallel to the surfaces).

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Coulomb’s Model of Friction

𝐹 ≤ 𝜇𝑅

When slipping, or OTPOS,

𝐹 = 𝜇𝑅

Can 𝜇 ever exceed 1?

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⇒ 𝑊 sin 𝜆 = 𝜇𝑊 cos 𝜆 𝜇 = tan𝜆 ⟺ 𝜆 = tan−1 𝜇

The angle of friction

𝑊

𝑅

𝐹

𝜆 𝜆

Resolving parallel to slope: 𝐹 = 𝑊 sin 𝜆

Resolving perpendicular to slope: 𝑅 = 𝑊 cos 𝜆

OTPOS: 𝐹 = 𝜇𝑅

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3. Tension and Thrust

These are forces between bodies connected by ropes, strings,

rods, tow bars, couplings etc. In diagrams these are represented like this:

Students can be confused by what the diagrams mean – the

arrows represent the forces exerted on the objects attached to either

end of the rope in tension. The forces are the same as a consequence

of Newton’s 3rd Law of Motion.

Tension

Thrust

𝑣

𝑣 𝑎

𝑎

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Newton’s Laws of Motion Newton’s 3rd Law (original, poetic version):

“Every action has an equal

and opposite reaction.”

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Newton’s 3rd Law (after complete poetry extraction)

“If body A acts with a force on

body B, then B acts with an equal

and opposite force on body A”

Or

“All forces come in equal and opposite pairs”

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Tug of war

𝑇 𝑇

𝐹𝐴 𝐹𝐵

𝐹𝐴 𝐹𝐵

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Example: The Magdeburg hemispheres were designed by Otto von

Guericke to contain a vacuum, and demonstrate the force of

atmospheric pressure. In 1654, they were placed together and the air

pumped out. In front of Emperor Ferdinand III, two teams of 15 horses

could not pull them apart.

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An easier, more efficient way of doing this would have been to have

used one team of horses and tied the other end to a large tree.

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Add the equal and opposite forces to the diagram

(of a person standing on the ground)

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(not to scale)

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Newton’s 2nd Law of Motion:

𝑭 = 𝒎𝒂

It’s a vector equation – so we need to be clear

about the direction.

We need to be clear what we’re applying it to

(which body has the mass 𝑚?)

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𝐹 = 𝑚𝑎 (car)

𝐷 − 𝑅1 − 𝑇 = 𝑚1𝑎 𝐹 = 𝑚𝑎 (trailer)

𝑇 − 𝑅2 = 𝑚2𝑎

𝐹 = 𝑚𝑎 (whole system)

𝐷 − 𝑅1 − 𝑅2 = 𝑚1 +𝑚2 𝑎

𝑚1 𝑚2

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𝐹 = 𝑚𝑎 (𝑚1 mass)

𝑇 − 𝐹 = 𝑚1𝑎 (1)

𝐹 = 𝑚𝑎 (𝑚2 mass)

𝑚2𝑔 − 𝑇 = 𝑚2𝑎 (2)

𝑚2𝑔 − 𝐹 = 𝑚1 +𝑚2 𝑎 3

Where did equation (3) come from?

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Activity: Weighing yourself in the lift.

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Reading when still=870N; reading when starting upwards = 910N

What was the lift’s acceleration?

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Moments Definition: “The moment of a force about a point* is the found

by multiplying the magnitude of the force by the perpendicular

distance from the point* to the line of action of the force”

The Principle of Moments: If a system of forces is in

equilibrium, the sum of clockwise moments about any point*

equals the sum of anti-clockwise moments.

*strictly speaking the moment is about an axis, not a point.

↺ 𝐴 = 𝑃𝑑 (clockwise)

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Two methods of taking moments:

A. Using the line of action B. Using components

Both are giving the same vector product 𝑃𝑑 sin 𝜃. Which do we use?

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3𝑚 1𝑚

𝐴

𝐵 2𝑁

4𝑁

Moment about A:

4 × 3 − 1 × 2 = 10𝑁𝑚 c.w.

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A paving slab of weight 𝑊 with sides 𝑎 and 𝑏, held in

equilibrium at an angle of 𝜃 to the ground by a horizontal

force 𝑃. Find 𝑃 in terms of 𝑊 and 𝜃.

↺ 𝐴 12𝑏𝑊cos 𝜃 =12𝑎𝑊sin 𝜃 + 𝑎𝑃 cos 𝜃 + 𝑏𝑃 sin 𝜃

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Moment of P about A:

𝑃𝑑 sin 𝛼 + 𝛽 Moment of P about A:

𝑃𝑑 sin 𝛼 cos 𝛽 + 𝑃𝑑 cos 𝛼 sin 𝛽

Using Moments to derive a compound angle formula

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Ladder Problems e.g. A uniform ladder rests in limiting equilibrium against a smooth

wall and on a rough floor, inclined at an angle 𝜃 to the horizontal. Find

the coefficient of friction.

𝜃 𝜃

How does the angle change as

someone climbs the ladder?

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Jointed Rods e.g. Two identical uniform rods 𝐴𝐵 and 𝐵𝐶 of length 2𝑎 are smoothly

jointed together at 𝐵. End 𝐴 is freely hinged to a wall and rod 𝐵𝐶 rests

on a smooth peg a distance 𝑥 from 𝐵. The system is in equilibrium with

both rods horizontal. Find 𝑥.

𝐴 𝐵 𝐶

2𝑎 𝑥

𝑌 𝑅

𝑊 𝑊

𝑌

𝑌

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The lines of action of 3 non-parallel forces in equilibrium

are concurrent. Although not explicitly on the MEI

syllabus, this is a very useful result,

3 forces NOT in

equilibrium!

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𝑊

𝑅

𝐹

𝑊

𝑅

𝐹

Block on a slope Block on a slope: OTPOT

Toppling on a slope

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𝐹

W

𝑅

𝑃

𝜃

As 𝑃 increases, 𝐹 must increase to maintain equilibrium.

As 𝐹 increases, θ, the angle of the contact force to the

horizontal will decrease. For the lines of action to meet at a

point, R must move to the right.

𝐹

W

𝑅

𝑃

𝜃

OTPOT

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30°

60° 60°

30°

𝑅 𝑇

𝑊

𝑇 = 𝑅

2𝑇 sin 60° = 𝑊

𝑇 =𝑊

3

Question 6 in Handout.

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Dedication and Thanks

I’d like to dedicate this

presentation to my

mechanics teacher, Peter

Atkinson.

He was rigorous, light-

hearted, and a major

inspiration to me.

Mechanics: A Review of the Principles Jon Burr jburr@yorkcollege.ac.uk

Some Problems

1. A block of mass 𝑚𝑚 is to be pulled across a rough floor at a constant velocity by a string inclined at an angle 𝜃𝜃 to the horizontal. The coefficient of friction between the block and floor is 𝜇𝜇 and the tension in the string is 𝑇𝑇. For what value of 𝜃𝜃 is 𝑇𝑇 a minimum?

2. I stood on scales in a lift. When the lift wasn’t moving they read 870N. When the lift started moving up, the reading changed to 910N. What was the acceleration of the lift at that time?

3. The diagram shows a uniform rectangular paving slab 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 of mass 15kg held in equilibrium by a horizontal force 𝐹𝐹 applied at the corner 𝐴𝐴. Corner 𝐴𝐴 rests on the ground; 𝐴𝐴𝐴𝐴 = 30cm and 𝐴𝐴𝐴𝐴 = 40cm. 𝐴𝐴𝐴𝐴 makes an angle of 30° with the horizontal. Find the magnitude of 𝐹𝐹.

[From Sadler & Thorning]

4. The diagram shows a uniform ladder resting in equilibrium with its top end against a smooth vertical wall and its base on a smooth inclined plane. The plane makes an angle 𝜃𝜃 with the horizontal and the ladder makes an angle 𝜙𝜙 with the wall. Prove that tan𝜙𝜙 = 2 tan𝜃𝜃.

𝜃𝜃

𝜙𝜙

𝐹𝐹

𝐴𝐴

𝐴𝐴

𝐴𝐴

𝐴𝐴

𝜃𝜃

𝑇𝑇

5. A uniform smooth rod of mass 𝑀𝑀 and length 2𝑙𝑙 rests partly inside and partly outside a fixed smooth hemispherical bowl of radius 𝑎𝑎. The rim of the bowl is horizontal and one point of the rod is in contact with the rim. Prove that the inclination 𝜃𝜃 of the rod to the horizontal is given by 2𝑎𝑎 cos 2𝜃𝜃 = 𝑙𝑙 cos 𝜃𝜃. Find the reaction between the rod and the bowl at the rim.

(From MEI AS Further Maths 2018)