Mechanics 1 revision notes, Lecture notes of Mechanics

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Mechanics 1

Revision Notes

April 2016

1. Mathematical Models in Mechanics

Assumptions and approximations often used to simplify the mathematics

involved:

a ) a rigid body is a particle , b ) no air resistance , c ) no wind , d ) force due to gravity remains constant , e ) light pulleys and light strings etc. have no mass , f ) the tension in a light string which remains taut will be constant throughout its length. g ) if a pulley is light or smooth the tensions in the a string going round the pulley will be equal on both sides; the same is true for a smooth peg, h ) if a string is inextensible and remains taut , the accelerations of two particles, one fixed at each end, will be equal , i ) rods are uniform  constant mass per unit length  the centre of mass will be in the middle, j) a lamina is a uniform flat object of negligible thickness , k ) the earth’s surface, although spherical, is usually modelled by a plane, l ) surface is smooth - no friction , m ) forces behave like vectors.

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 .

Adding vectors

Geometrically, use a vector triangle or a vector parallelogram:

or

1. In component form:

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 60o^ and 4 miles on a bearing of 140o.. Solution:

p

p + q

q

p

p + q

q



x

100

north

north

60

3

140

4

Using the cosine rule x^2 = 3^2 + 4^2  2  3  4  cos 100 = 29 16756  x = 54006996 = 5 40 then, using the sine rule,

 sin  = 0 729393

  = 46 83551  bearing = 468 + 60 = 106 8 o

Answer Resultant vector is 540 miles on a bearing of 106 8 o.

Resolving vectors in two perpendicular components

F has components F cos  and F sin 

as shown.

Vector algebra

Notation, = a , = b , etc., but = r , usually! To get from A to B first go A to O using – a then go O to B using b

 = – a + b = b – a.

Vectors in mechanics

Forces behave as vectors (the physicists tell us so) - modelling. Velocity is a vector so must be given either in component form or as magnitude and direction. Speed is the magnitude of the velocity so is a scalar. Acceleration is a vector so must be given either in component form or as magnitude and direction.

x

y

F sin ^ F

 F cos 

a

b

A

O (^) B

Collision of moving particles

Example: Particle A is intially at the point (3, 4) and travels with velocity 9 i – 2 j m s-1. Particle B is intially at the point (6, 7) and travels with velocity 6 i – 5 j m s-1. ( 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 ) rA = (Initial position + displacement)

rB = (Initial position + displacement)

( b ) If the particles collide then their x -coordinates will be equal  x -coords = 3 + 9 t = 6 + 6 t  t = 1

BUT we must also show that the y -coordinates are equal at t = 1. y -coords = 4 – 2t = 7 – 5 t  t = 1.  particles collide when t = 1 at (12, 2).

Closest distance between moving particles

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

rA = and rB =.

(a) Find the vector at time t. (b) Find the distance between A and B at time t in terms of t. (c) Find the minimum distance between the particles and the time at which this occurs.

Solution: (a) = rB  rA =

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

 d^2 = (–1 +2 t )^2 + (1 + t )^2 = 1 – 4 t + 4 t^2 + 1 + 2 t + t^2 = 5 t^2 – 2 t + 2

 d = 5 t^2  2 t  2

(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

 t   dt

d d for max and min  t = 0 2

and the second derivative 10

2

2 2

dt

d d

which is positive for t = 0 2

 d^2 is a minimum at t = 0 2

 minimum value of d = 5  0  22  2  0  2  2  1  8 = 134 to 3 S.F.

Relative velocity

This is similar to relative position in that if C and D are at positions rC and rD

then the position of D relative to C is = rD rel C = rD – rC which leads on to: - if C and D are moving with velocities vC and vD then the velocity of D relative to C is vD rel C = vD – vC.

Example: Particles A and B have velocities vA = (12 t – 3) i + 4 j and vB = (3 t^2 – 1) i + 2 t j. Find the velocity of A relative to B and show that this velocity is parallel to the x - axis for a particular value of t which is to be determined.

Solution: vA = and vB =

 vA rel B = vA – vB =

The y -coordinate = 0 for motion parallel to the x -axis  4 – 2t = 0 when t = 2  the velocity is parallel to the x -axis when t = 2.

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

s = 0, a = – 9 8, u = 14 and t =?

Using s = ut + at^2 we have 0 = 14 t –  9  8 t^2

 t (14 – 4  9 t ) = 0

 t = 0 (at start) or t = 2 seconds.

Answer: The ball takes 2 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-1^ downwards.

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 or deceleration.

Example: A particle is initially travelling at a speed of 2 m s -1^ and immediately accelerates at 3 m s -2^ for 10 seconds; it then travels at a constant speed before decelerating at a 2 m s - until it stops. ( a ) Find the maximum speed and the time spent decelerating. ( b ) sketch a speed-time graph. ( c ) 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 -1^ is maximum speed. For deceleration from 32 m s -1^ at 2 m s -2^ the time taken is 32  2 = 16 seconds. In the graph, T is the total time taken.

Distance travelled in first 10 secs is area of trapezium = (2 + 32)  10 = 170 metres,

distance travelled in last 16 secs is area of triangle =  16  32 = 256 metres,

 distance travelled at constant speed = 1130  (170 + 256) = 704 metres  time taken at speed of 32 m s -1^ 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  and F sin .

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 The resultant force  is 4 cos 60 + 9 – 7 cos 25 = 465585 N and the resultant force  is (7sin 25 + 8) – 4 sin 60 = 749423 N

giving this picture

 R =^4 ^655852 ^7 ^494232 ^8 ^82 N

and tan  =

  = 58 1 o

 Answer: resultant is 882 N at an angle of 58 1 o^ below the 9 N force.

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

x

y

F sin ^ F

 F cos 

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  = 36.9o, and now combine R 1 with the 2 N force to find the final resultant R 2 using the cosine and or sine rule. Probably easier to resolve each force in two perpendicular directions as in the previous example.

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 = 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

 Answer: a = – 4 and b = – 2

3 N

2 N

4 N

48 o

48 o

o

4 N

3 N

R 2 N

R 1 N

2 N

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: DRAW A DIAGRAM SHOWING ALL 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.

2 g cos 25

2 g sin 25

R

F

R

2 g

F

25 o^25 o

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 =  R, where  is the coefficient of friction and R is the normal reaction.

Example: A particle of mass 3 kg lies in equilibrium on a slope of angle 25o. 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  R = 3 g cos 25 = 26 645 Res  F = 3 g sin 25 = 12 4

But the maximum friction force is Fmax =  R = 0 6  26 645 = 16 0 N

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.

R (^) F

3 g sin 25 (^3) g cos 25

25 o

R (^) F

3 g