Vector and Mechanics Calculations, Study notes of Mechanics

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Mechanics 1

Revision Notes

July 2012

• MECHANICS
  • 1. Mathematical Models in Mechanics
    • Assumptions and approximations often used to simplify the mathematics involved:
  • 2. Vectors in Mechanics.
    • Magnitude and direction ←→ components
    • Parallel vectors
    • Adding vectors...........................................................................................................................................................
    • Resolving vectors in two perpendicular components
    • Vector algebra............................................................................................................................................................
    • Vectors in mechanics.................................................................................................................................................
    • Velocity and displacement.
    • Relative displacement vectors
    • Collision of moving particles
    • Closest distance between moving particles
    • Relative velocity
  • 3. Kinematics of a particle moving in a straight line or a plane.
    • Constant acceleration formulae.
    • Vertical motion under gravity
    • Speed-time graphs
  • 4. Statics of a particle.
    • Resultant forces
    • Resultant of three or more forces
    • Equilibrium of a particle under coplanar forces.
    • Types of force
    • Friction
    • Coefficient of friction.
    • Limiting equilibrium
  • 5. Dynamics of a particle moving in a straight line.
    • Newton’s laws of motion.
    • Connected particles
    • Particles connected by pulleys
    • Impulse and Momentum..........................................................................................................................................
    • Internal and External Forces and Impulses.
    • Conservation of momentum.
    • Impulse in string between two particles..................................................................................................................
  • 6. Moments.............................................................................................................................
    • Moment of a Force
    • Sum of moments
    • Moments and Equilibrium
    • Non-uniform rods
    • Tilting rods...............................................................................................................................................................
  • Index

From magnitude and direction form , draw a sketch and use trigonometry to find x and y

components.

Example:

r is a vector of length 7 cm making an angle of -

o with the x -axis.

a = 7 cos 50

o = 4.50 cm ,

b = 7 sin 50

o = 5.36 cm

⇒ r = ⎥ ⎦

cm.

Example: Find a vector of length 20 in the direction of (^) ⎥

⎦

Solution: Any vector in the direction of (^) ⎥

⎦

must be a multiple, k , of (^) ⎥ ⎦

To find the multiple, k , we first find the length of (^) ⎥ ⎦

2 2 − + =

and as 20 = 4 × 5 the multiple k must be 4

so the vector of length 20 is (^) ⎥ ⎦

×

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Parallel vectors

Two vectors are parallel ⇔ one is a multiple of another:

e.g. 6 i – 8 j = 2(3 i – 4 j ) ⇔ 6 i – 8 j and 3 i – 4 j are parallel

Or areparallel 4

and 8

= ×

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7

50

a

b

Adding vectors

Use a vector triangle or a vector parallelogram:

or

1. In component form: ⎥ ⎦

b d

a c

d

c

b

a

2. To add two vectors which are given in magnitude and direction form:

Either a ) convert to component form, add and convert back,

Or b ) sketch a vector triangle and use sine or cosine rule.

Example:

Add together, 3 miles on a bearing of 60

o and 4 miles on a bearing of 140

o. .

Solution:

Using the cosine rule

x

2 = 3

2

• 4

2

• 2 × 3 × 4 × cos 100 = 29.

⇒ x = 5.4006996 = 5.

then, using the sine rule,

sin θ

sin 100

⇒ sin θ = 0.

⇒ θ = 46.83551 ⇒ bearing = 46.8 + 60 = 106.

o

Answer Resultant vector is 5.40 miles on a bearing of 106.

o .

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p

p + q

q

p

p + q

q

θ

x

100

north

north

60

3

140

4

Example: A particle is initially at the point (4, 11) and moves with velocity ⎥ ⎦

m s

• . Find its

position vector after t seconds.

Solution: The displacement during t seconds will be t × (^) ⎥

⎦

and so the new position vector will be r = ⎥ ⎦

• t × ⎥ ⎦

t

t

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Relative displacement vectors

If you are standing at a point C and X is

standing at a point D then the position vector of

X relative to you is the vector CD

and CD = rD rel C = d – c = rD – rC

Thus if a particle A is at r (^) A = 3 i – 4 j and B is at r (^) B = 7 i + 2 j

then the position of A relative to B is

BA = rA rel B = a – b = r (^) A – rB = (3 i – 4 j ) – (7 i + 2 j ) = –4 i – 6 j.

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Collision of moving particles

Example: Particle A is intially at the point (3, 4) and travels with velocity 9 i – 2 j m s

• .

Particle B is intially at the point (6, 7) and travels with velocity 6 i – 5 j m s

• .

(a) Find the position vectors of A and B at time t.

(b) Show that the particles collide and find the time and position of collision.

Solution:

(a) r (^) A = (^) ⎥ ⎦

⎥+ ×

t

t t (Initial position + displacement)

r (^) B = (^) ⎥ ⎦

⎥+ ×

t

t t (Initial position + displacement)

D

C

O

r C

r D

r D – rC

(b) If the particles collide then their x -coordinates will be equal

⇒ x -coords = 9 t + 3 = 6 t + 6 ⇒ t = 1

BUT we must also show that the y -coordinates are equal at t = 1.

y -coords = –2t + 4 = –5 t + 7 ⇒ t = 1.

⇒ particles collide when t = 1 at (12, 2).

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Closest distance between moving particles

Example: Two particles, A and B , are moving so that their position vectors at time t are

rA = (^) ⎥ ⎦

t

t and rB = (^) ⎥ ⎦

t

t .

(a) Find the position vector of B relative to A.

(b) Find the distance between A and B at time t.

(c) Find the minimum distance between the particles and the time at which this occurs.

Solution:

(a) rB rel A = rB – r (^) A = (^) ⎥ ⎦

t

t

• (^) ⎥ ⎦

t

t = (^) ⎥ ⎦

t

t Answer.

(b) The distance, d , between the particles is the length of

AB = r (^) B rel A = ⎥ ⎦

t

t

⇒ d

2 = (2 t – 1)

2

• ( t + 1)

2 = 4 t

2

• 4 t + 1 + t

2

• 2 t + 1

= 5 t

2

• 2 t + 2

⇒ d = 5 2 2

2 t − t +

(c) The minimum value of d will occur when the minimum value of d

2 occurs

so we differentiate d

2 with respect to t.

d

2 = 5 t

2

• 2 t + 2

2 = t − = dt

d d for max and min ⇒ t = 0.

and the second derivative 10

2

2 2

= dt

d d which is positive for t = 0.

⇒ d

2 is a minimum at t = 0.

⇒ minimum value of d = 5 0. 2 2 0. 2 2 1. 8

2 × − × + = = 1.34 to 3 S .F.

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Vertical motion under gravity

1] The acceleration always acts downwards whatever direction the particle is moving.

2] We assume that there is no air resistance, that the object is not spinning or turning and that

the object can be treated as a particle.

3] We assume that the gravitational acceleration remains constant and is 9.8 m s

• .

4] Always state which direction (up or down) you are taking as positive.

Example: A ball is thrown vertically upwards from O with a speed of 14 m s

• .

(a) Find the greatest height reached.

(b) Find the total time before the ball returns to O.

(c) Find the velocity after 2 seconds.

Solution: Take upwards as the positive direction.

(a) At the greatest height, h , the velocity will be 0 and so we have

u = 14, v = 0, a = –9.8 and s = h (the greatest height).

Using v

2

• u

2 = 2 as we have 0

2

• 14

2 = 2 × (–9.8) × h

⇒ h = 196 ÷ 19.6 = 10.

Answer: Greatest height is 10 m.

(b) When the particle returns to O the distance, s , from O is 0 so we have

s = 0, a = –9.8, u = 14 and t = ?.

Using s = ut + ½ at

2 we have 0 = 14 t – ½ × 9.8 t

2

⇒ t (14 – 4.9 t ) = 0

⇒ t = 0 (at start) or t = 2

6 / 7 seconds.

Answer: The ball takes 2

6 / 7 seconds to return to O.

(c) After 2 seconds, u = 14, a = –9.8, t = 2 and v = ?.

Using v = u + at we have v = 14 – 9.8 × 2

⇒ v = –5.6.

Answer: After 2 seconds the ball is travelling at 5.6 m s

• downwards.

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Speed-time graphs

1] The area under a speed-time graph represents the distance travelled.

2] The gradient of a speed-time graph is the acceleration.

Example: A particle is initially travelling at a speed of 2 m s

• and immediately accelerates at

3 m s

• for 10 seconds; it then travels at a constant speed before decelerating at a 2 m s -

until it stops.

Find the maximum speed and the time spent decelerating: sketch a speed-time graph.

If the total distance travelled is 1130 metres, find the time spent travelling at a constant

speed.

Solution:

For maximum speed: u = 2, a = 3, t = 10, v = u + at ⇒ v = 32 ms

• is maximum

speed.

For deceleration from 32 m s

• at 2 m s - the time taken is 32 ÷ 2 = 16 seconds.

Distance travelled in first 10 secs is area of trapezium = ½(2 + 32) × 10 =

170 metres,

distance travelled in last 16 secs is area of triangle = ½ × 16 × 32 = 256

metres,

⇒ distance travelled at constant speed = 1130 - (170 + 256) = 704 metres

⇒ time taken at speed of 32 m s

• is 704 ÷ 32 = 22 s.

2

32

10 T- 16 t

v

T

Resultant of three or more forces

Reminder:

We can resolve vectors in two perpendicular

components as shown:

F has components F cos θ and F sin θ.

To find the resultant of three forces

1] convert into component form ( i and j ) , add and convert back

or 2] sketch a vector polygon and use sine/cosine rule to find the resultant of two,

then combine this resultant with the third force to find final resultant.

For more than three forces continue with either of the above methods.

Example: Find the resultant of the four

forces shown in the diagram.

Solution: First resolve the 7 N and 4 N

forces horizontally and vertically

Resultant force

is 4 cos 60 + 9 – 7 cos 25 = 4.65585 N

and resultant force

is (7sin 25 + 8) – 4 sin 60 = 7.49423 N

giving this picture

⇒ R = 4. 65585 7. 49423 8. 82 N

2 2

• =

and tan θ =

o

⇒ Answer resultant is 8.82 N at an angle of 58.

0 below the 9 N force.

F sin θ^ F

θ

F cos θ

x

y

60 o

25

o 9 N

8 N

7 N

4 N

7 cos 25 N 4 cos 60 + 9 N

4 sin 60 N

7 sin 25 + 8 N

7.49423 R

Example: Use a vector polygon to find the resultant of the three forces shown in the diagram.

Solution: To sketch the vector polygon, draw the forces end to end. I have started

with the 3 N, then the 4 N and finally the 2 N force.

Combine the 3 N and 4 N forces to find the resultant R 1 = 5 N with θ = 36.

o ,

and now combine R 1 with the 2 N force to find the final resultant R 2 using the

cosine and or sine rule.

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Equilibrium of a particle under coplanar forces.

If the sum of all the forces acting on a particle is zero (or if the resultant force is 0 N) then the

particle is said to be in equilibrium.

Example: Three forces P = (^) ⎥

⎦

N, Q = ⎥

N and R = (^) ⎥ ⎦

b

a N are acting on a

particle which is in equilibrium.

Find the values of a and b.

Solution: As the particle is in equilibrium the sum of the forces will be 0 N.

⇒ P + Q + R = 0

b

a

⇒ Answer: a = –4 and b = –

3 N

2 N

4 N

48 o

4 N

3 N

R 2 N

R 1 N

2 N

θ o

48

o

Example: A particle of mass 2 kg rests in equilibrium on a plane which makes an angle of

25

o with the horizontal.

Find the magnitude of the friction force and the magnitude of the normal reaction.

Solution: First draw a diagram showing all the forces - the weight 2 g N, the friction

F N and the normal reaction R N. Remember that the particle would move down

the slope without friction so friction must act up the slope.

Then draw a second diagram showing forces resolved along and perpendicular

to the slope.

The particle is in equilibrium so

resolving perpendicular to the slope R = 2 g cos 25 = 17.7636,

and resolving parallel to the slope F = 2 g sin 25 = 8.2833.

Answer Friction force is 8.28 N and normal reaction is 17.8 N.

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R

2 g

F

25 o^25 o

2 g cos 25

2 g sin 25

R

F

Coefficient of friction.

There is a maximum value, or limiting value, of the friction force between two surfaces. The ratio

of this maximum friction force to the normal reaction between the surfaces is called the

coefficient of friction.

Fmax = μ N, where μ is the coefficient of friction and N is the normal reaction.

Example: A particle of mass 3 kg lies in equilibrium on a slope of angle 25

o

. If the coefficient

of friction is 0.6, show that the particle is in equilibrium and find the value of the friction

force.

Solution:

Res ⇒ N = 3 g cos 25 = 26.

⇒ Maximum friction force is Fmax = μ N = 0.6 × 26.645 = 16.

Res ⇒ F = 3 g sin 25 = 12.4 < Fmax if the particle is in equilibrium

Thus the friction needed to prevent sliding is 12.4 N and since the maximum possible

value of the friction force is 16.0 N the particle will be in equilibrium and the actual

friction force will be just 12.4 N.

Answer Friction force is 12.4 N.

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N F

3 g sin 25 3 g cos 25

25 o

N F

3 g

5. Dynamics of a particle moving in a straight line.

Newton’s laws of motion.

1. A particle will remain at rest or will continue to move with constant velocity in a straight

line unless acted on by a resultant force.

2. For a particle with constant mass, m kg , the resultant force F N acting on the particle and

its acceleration a m s

• satisfy the equation F = m a.

3. If a body A exerts a force on a body B then body B exerts an equal force on body A but in

the opposite direction.

Example: A box of mass 30 kg is being pulled along the ground by a horizontal force of

60 N. If the acceleration of the trolley is 1.5 m s

• find the magnitude of the friction force.

Solution: First draw a picture !!

No need to resolve as forces

are already at 90

o to each other.

Resolve horizontally

⇒ 60 – F = 30 × 1.

⇒ F = 60 – 45 = 15.

Answer Friction force is 15 N.

Example: A ball of mass 2 kg tied to the end of a string. The tension in the string is 30 N.

Find the acceleration of the ball and state in which direction it is acting.

Solution: First draw a picture!!

Resolve upwards ⇒ 30 – 2 g = 2 a

⇒ a = 5.

Answer Acceleration is 5.2 m s

• upwards.

R

30 g

60 F

30

a

2 g

Example: A particle of mass 25 kg is being pulled up a slope at angle of 25

o above the

horizontal by a rope which makes an angle of 15

o with the slope. If the tension in the rope

is 300 N and if the coefficient of friction between the particle and the slope is 0.25 find the

acceleration of the particle.

Solution:

Res ⇒ N + 300 sin 15 = 25 g cos 25 ⇒ N = 144.

and, since moving, friction is maximum ⇒ F = μ N = 0.25 × 144.39969 =

Res ⇒ 300 cos 15 - ( F + 25 g sin 25) = 25 a

⇒ a = 6.

Answer the acceleration is 6.01 m s

• .

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N + 300 sin 15

a

F + 25 g sin 25

300 cos 15

25 g cos 25

25 o

15 o

25 g

300 N

F

a

25 o