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U1_2 Forces and Newtons Laws

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           1
TYPES OF FORCE

Generally a force is a push, pull, twist or squeeze. Drag, Tension, Friction,
Weight, Gravity, Thrust are applied to particular forces. For example, Thrust is
used for the driving force provided by the engine of a car. However, despite all
the names and situations where forces occur, outside the nucleus of atoms there
are only 3 types of force!

* Gravitational Force is a natural phenomenon by which all objects with mass
attract each other, and is one of the fundamental forces of physics. In everyday
life, gravitation is most commonly thought of as the agency that gives objects
weight. It is responsible for keeping the Earth and the other planets in their
orbits around the Sun; for keeping the Moon in its orbit around the Earth, for
the formation of tides; for convection (by which hot fluids rise); for heating the
interiors of forming stars and planets to very high temperatures; and for various
other phenomena that we observe. Gravitation is also the reason for the very
existence of the Earth, the Sun, and most macroscopic objects in the universe;
without it, matter would not have joined into these large masses and life, as we
know it, would not exist.

Modern physics describes gravitation using the general theory of relativity, but
the much simpler Newton's law of universal gravitation provides an excellent
approximation in most cases.

"Gravitation" is a general term describing the attractive influence that all
objects with mass exert on each other. The gravitational force between an
object and the Earth is weight.




* Electrical Force. Matter is made of atoms. An atom is basically composed of
three different components - electrons, protons, and neutrons. An electron can
be removed easily from an atom. Normally, an atom is electrically neutral, which
means that there are equal numbers of protons and electrons. Positive charge of
protons is balanced by negative charge of electrons. It has no net electrical
charge.
When atoms gain or lose electrons, they are called "ions."



                                        2
The electrical force is responsible for all interactions between objects. When
atoms collide, they exert an electrical force on each other and as a result of
electrical attraction between them they can chemically bond together. When the
atoms in your foot meet those in the ground, when you walk for example, both
sets of atoms exert an electrical force on each other – this force is sensed by
your foot – “you have touched the ground”. All contact forces are electrical
forces and all objects are affected by these contact forces.




 * Magnetic Force is the force between 2 magnetic objects. At an atomic level it
is a force between moving charges. It is only important in examples using
magnetised objects.




                                       3
Newton's laws of motion




Newton developed Galileo's ideas about motion and stated three laws (named
after him). The laws are very nearly correct under all circumstances although
more exact laws are required under certain extreme conditions (in particular, as
the speed of an object approaches the velocity of light).

Newton’s First Law:

A body remains at rest or continues with uniform velocity (constant speed in
a straight line) unless acted upon by a resultant force.

The first law enables a force to be defined as that which changes the motion of
a body (speeds the body up, slows the body down, causes the body to change
direction). Another way of stating this is that a resultant force causes a body to
accelerate.

The first law also implies that matter has a reluctance to change its state of
rest or motion. This property is called the inertia of the body. The mass of a
body is a measure of its inertia.




                                        4
Newton's second law

Newton's second law indicates what happens when a resultant force is applied to
a body.

Experiments show that when the resultant force applied to a body of constant
mass is changed then the acceleration produced is proportional to the resultant
force. Experiments also show that when the resultant force is kept constant
and the mass of the body changed then acceleration is inversely proportional to
the mass.

These experimental results are summarised by the equation which
mathematically expresses the second law:

aαF
a α 1/m
a α F/m or F α ma
F = constant x ma
Resultant force = Mass x Acceleration
F = m x a Where F means resultant force

Definition of the Newton
A force of 1 Newton is the force which when applied to a body of mass 1 kg
produces an acceleration of 1 ms-2.

Combining the definition and the above equation, k = 1

Units
In this equation the following units must be used:
Mass must be expressed in kg, Acceleration will always be in ms-2 and Forces will
always be in N.

It is also important to realise that the net or resultant force is always in the
same direction as the acceleration. For example for a car increasing in speed, the
resultant force and acceleration are both forward. For a car decelerating right
there is an acceleration left and a net force left.

 It is important to realise that in many questions, F = ma is used together with
                             the equations of motion!




                                        5
When Einstein came up with his special theory of relativity, which stated,
amongst other things, that at speeds approaching light speed, objects increase in
mass, it became clear that F = ma was invalid for high speeds. The difference
between Newton’s theory and Einstein’s only amounts to a 1% discrepancy for an
object travelling 1/10th speed of light. Nuclear physicists use Einstein’s theory,
but the calculations used to launch NASA rockets were based on Newtons Laws.

Task: Research in more detail, the limitations of F = ma. The use of
theories, models and ideas to develop and modify scientific explanations can
be discussed.

Free body diagrams

A free body diagram is one, which shows all the forces acting on a body in a
particular situation. A free body diagram should show the forces acting on the
body you are interested in and nothing else. The object can be drawn realistically
or more commonly as a single point, with force arrows pointing out wherever
possible.

                            Normal reaction force




Drag/Friction                                   Forward push force




                           Weight

Normal Force: This is the reaction force at 90 degrees to a surface, pushing up
on an object.




                                        6
Draw   free body diagrams for the following:
(a)     a person sitting on a stool (model the person as a particle).
(b)     a block of wood on an inclined plane which has a rough surface.
(c)     a cart being pulled along a level road by a tractor.
(d)     a tractor pulling a cart along a level road.

In each case name the forces and state how they arise.




Teaching Examples
(a)   A runner in a sprint race accelerates uniformly to reach 9 ms-1 in 3 s
      from the start of the race. If her mass is 50 kg, what force must she
      exert in order to do this?




(b)     An aeroplane lands with a velocity of 55 ms-1. Reverse thrust from the
        engines is used to slow it down to a velocity of 25 ms-1, in a time of 3 s.
        If the mass of the aeroplane is 3 x 104 Kg and the plane decelerates
        uniformly, what is the size of the reverse thrust supplied by the
        engines?




                                          7
Year 12 Newton's Second Law Questions

1.    A resultant force of 1.0 N acts on an initially stationary mass of 1.0 kg for
      a time of 1.0 s. How far will it move it?

2.    Forces of 12 N and 16 N act on a body of mass 4.0 kg. If the forces are
      perpendicular to each other calculate the size of the acceleration of the
      body.

3.    The speed of a locust at take off is 3.4 ms-1. If the body is accelerated
      through a distance of 40 mm and the mass of the locust is 3.0 g calculate
      the average force exerted by the ground on the hind legs of the locust.

4.    A lift has a mass of 1.2 x 103kg. Draw a free-body diagram showing all the
      forces acting on the lift.




Calculate the tension in the supporting cable when the lift is:
(a)    descending at uniform velocity
(b)    descending with downward acceleration of 2ms-2
(c)    at rest
(d)    ascending with upward acceleration 2ms-2
(e)    ascending with uniform speed.

5.    A force of 8.0 N gives a mass m1 an acceleration of 12 ms-2, and a mass m2
      an acceleration of 48 ms-2. Calculate the acceleration that the force
      would give the two masses when they are attached together.



                                         8
6.    A lift of mass 490kg is raised by means of a cable and carries a person of
      mass 84kg.

(a)   Draw the free body diagrams for
      (i)   the lift.
      (ii)  the person.
      Show on your diagrams the names and directions of the forces acting.

(b)   If the lift is moving upward with an acceleration of 0.20 ms-2, calculate
the force on the person due to the floor of the lift.

7. A force of 30 N halves a body's velocity in 9.0m. If the mass of the body is
5.0Kg, calculate the original velocity and the time for which the force acts.

8. A spring balance carrying a mass of 20.0Kg in a lift registered 250 N. What
was the acceleration of the lift? Draw free-body diagram from which you can
calculate the balance reading for
(a)    free-fall, and
(b)    movement at constant velocity.




9. A force of 8.0 N gives a mass m1 an acceleration of 12 ms-2 and a mass m2, an
acceleration of 48 m/s-2. Calculate the acceleration that this force would give
the two masses when they are attached.




                                        9
Forces questions page 31 – OCR Physics




                                         10
Link Between Weight and Mass

Suppose an object has a mass of m kg.

If the object is released it falls to the ground with an acceleration due to the
force of gravity of g ms-2 (acceleration of free fall).

For this object:   mass                 =        m kg,
                   acceleration         =        g ms-2
                   resultant force      =        WN (the weight of the object)

Remember, weight is the force of gravity acting on an object!

Applying           resultant force      =        mass x acceleration
                         W              =              m x g

Therefore a body of mass m has a weight of mg
The acceleration due to gravity g has a value of 9.81 ms-2 on Earth
So, a mass of 1 kg has a weight 9.81N, a mass of 5kg has a weight of (5 x 9.81) N.

To change Kg  N x by 9.81
To change N  Kg x by 9.81




For a mass of 1kg on the Earth its weight W will be given by
W = mg
W = 1 x 9.81 = 9.81N

On the Moon g has a value of 1.6ms-2
A mass of 1kg has a force on it:
W = 1 x 1.6 = 1.6N (1/6th of the weight on Earth).




                                            11
Self Study Exercise: FRICTION

Friction always occurs when there is relative motion e.g. an object falling
through air, a fish swimming in the sea, an object being dragged along a
surface.

Prepare answers to the following questions:

(a)  What is friction?
(b)  How does it arise? (the theory)
(c)  How does it affect the motion of a body?
(d)  Give some examples of where friction is useful.
(c)  How may friction be overcome?
(f)  How may the friction force be measured?
(g)  What two types of friction are there?
(h)  What is meant by the coefficient of friction and how may it be
measured?


Friction in fluids: Viscosity

Water resists the motion of boats, and air resists the motion of aircraft. This is
fluid friction, which is called viscosity. Viscosity tends to prevent motion within
the fluid – it is caused by internal friction between moving layers of fluid.




The viscosity of fluids varies greatly. Water is much easier to pour than treacle.
It is less viscous. Gases have a much lower viscosity than liquids.

The viscosity of most fluids gets less as they get hotter. Warm treacle for
example is quite “runny”.


To demonstrate viscous force we can drop an object in air, water and in
oil/washing up liquid.

Viscosity changes with the shape of the object being dropped. The greater the
Surface Area, the more the viscosity. Streamlining reduces viscosity.

Air resistance is an example of viscosity.


                                        12
Drag
Air resistance or aerodynamic drag acts when a body moves through air or any
fluid. It is caused by the fact that an object has to push air/fluid out of the
way in order to move through it, and this requires a force.
The force that is exerted by two surfaces rubbing together (friction) does not
depend on the speed at which the two surfaces move over each other. However,
the aerodynamic drag caused by an object moving through air does depend on
speed - the faster the object moves, the greater the aerodynamic drag. In fact,
the drag increases as the square of the velocity.

D α v2

Drag depends on several things:

*Velocity
*Roughness of the surface passing through the fluid
*Cross-sectional Area
*Shape (streamlined or not)

The physics of drag can be complicated. It is easy to understand that that a
slow, thin, streamlined object will have less drag than a fast, fat, flat object.
However, quantifying to what extent these issues affect drag, is difficult.




A model of a plane undergoes tests in a wind tunnel to measure the planes drag
coefficient Cd, in the following equation:
F = ½ρCdAv2
F is the drag force, ρ is the density of air, A is the frontal area of plane, v is the
velocity of the plane.
We don’t need to remember this equation, only that Drag α v2 or D = K v2

Air resistance becomes more important the faster you want to go. Careful design
can reduce aerodynamic drag, by producing shapes that can cut through the
air/water and cause as little disturbance to it as possible.


                                          13
Terminal Velocity

Ignoring air resistance and resistive forces, if the force causing an object to
accelerate increases, acceleration will increase.

However, resultant force often decreases because resistive forces increase
when the velocity increases, so acceleration will decrease.

Because aerodynamic drag increases greatly as an object's velocity increases,
objects tend to reach a terminal velocity as their velocity increases - whether
they are a parachutist falling through air or a car travelling along a race track.

Falling bodies

Steady speed means forces are balanced - Viscosity up and Weight down for
example, for a falling object. Steady speed of a falling object is called terminal
velocity. In thick liquids Terminal Velocity reached very quickly but in less
viscous liquids E.g. Air, this takes time to happen.

As an object falls through a fluid, the force of gravity tries to increase its
velocity. The viscous force or fluid friction, which increases as the object
speeds up, acts so as to slow it down. At a certain velocity, which depends on
both the viscosity of the fluid and the shape of the object, the two forces
balance each other. The object now moves with a constant velocity, which is
called the terminal velocity. (Remember, no resultant force means constant
velocity).




                                         14
                         Demonstration of viscous force


                                              Oil




The tube contains engine oil. We will investigate the motion of a ball bearing
falling through the oil. To do this, we will mark the position of the ball bearing
every second after it is released at the surface of the oil. We will measure the
distance fallen by the ball bearing after the times 1s. 2s. 3s. … As shown in the
table.

Time/seconds                     Distance/cm
     0
     1
     2
     3
     4
     5

Now plot a graph of distance (y-axis) against time (x-axis). What does the graph
indicate?




                                         15
Free fall and Terminal Velocity

A man jumping out of a plane will accelerate towards the ground under the
influence of the force of gravity from the Earth. The acceleration with which he
falls is called the acceleration due to gravity or the acceleration of free fall. At
the surface of the earth the approximate value of this acceleration is 10 ms-2.
The acceleration is not uniform however…

At the top of the jump, the man is instantaneously at rest so his air resistance is
zero. The force acting on him is greatest at this his acceleration at this point
has its maximum value.

A little later, the man is moving more rapidly and his air resistance is now
significant. The magnitude of his weight is still greater than his air resistance,
so he is still accelerating downwards, but not as quickly as at first.
Later still his velocity has reached a point where his air resistance is equal to his
weight. Now the resultant force acting on him is zero - and he is no longer
accelerating. The velocity at which this happens is called the terminal velocity.
For a human being without a parachute, terminal velocity is about 56ms-1.

On opening the parachute, the air resistance increases dramatically due to the
parachute's large surface area. Now the air resistance is greater than the
weight so the resultant force on the man is upwards. The man accelerates
upwards and his velocity decreases.

Eventually the man's velocity decreases to a new terminal velocity. This terminal
velocity is much lower than the previous terminal velocity about 10ms-1. Hitting
the ground at this speed still requires some care - it is like jumping off a wall 5
m high!




                                         16
Air resistance of a falling body:

   Just out of      10 seconds      Later on          Parachute deployed
   Plane            later
                                                                AR
                                        AR

                        AR




            mg          mg              mg                    mg

Only force is       Speed up =      AR = mg.                  Bigger force up.
Down.               AR up           Balanced force            Speed drops–deceleration.
Unbalanced =        (AR opposite    = steady speed            TV reached again, at a
Accelerate down.    to motion).     Called Terminal           lower speed than before
                    Bigger force    Velocity.                 And man safely hits
                    Still down.                               Ground.
                    Still                                     Bigger SA = Bigger AR.
                    accelerates
                    Down.
                    AR still
                    Increases


Gravity is always constant, but Air resistance increases as speed increases until
Terminal Velocity reached, where it stays constant. Taking down as positive, mg –
AR = ma. As velocity increases, AR increases therefore, a decreases. When mg =
AR, a = 0 as there is no resultant force. It can be noted here that, without the
effect of air resistance, a falling body would continually accelerate and never
reach a terminal velocity.




                                               17
 Graphs of Force, Acceleration and Velocity for a falling body
 (Take down as positive)

                            Weight
Force
                                                    Acc’n


                             Resultant                           Time
                                             Time



                                     A.R.



   Velocity




                                             Time




                                            18
Falling Objects, Accelerating Objects

1. A free-fall sky-diver jumps from a high altitude balloon. The diagram shows
   the forces acting on her five seconds later.
                                      X




                                        Y
a)   Name the forces X and Y.




b)   What happens to the size of X as she continues to fall?




c)   Explain, in terms of X and Y, why she eventually reaches a steady terminal
     velocity.




d)   Explain how her motion changes when her parachute opens.




e)   As she nears the ground, she remembers the instruction to bend her knees
     on landing. Explain why this makes the landing “softer”.




                                        19
2. A raindrop falling through air eventually travels at its terminal velocity.




Use one of the statements below to complete the sentences that follow:

Zero; downward and increasing; downward and decreasing; upward and
constant; downward and constant; upward and decreasing.

At its terminal velocity the

   (i). Velocity of the raindrop is
        _______________________________________

   (ii). acceleration of the raindrop is
         ____________________________________

  (iii). force of the air on the raindrop is
         __________________________________

  (iv). weight of the raindrop is ________________________




                                         20
3.                                       4000 N          Mass = 250 Kg




                                     2500 N (weight)

Use the information given in the diagram to calculate:
a. the resultant upward force of the rocket

b. the acceleration of the rocket




Freefall and Drag questions page 33 – OCR Physics

Project Work

Write a short report on the following:
How do fast moving jet aircraft use parachutes to decelerate?




                                        21
Equilibrium

Revision

Adding and Subtracting Vectors
Forces are vectors. Vectors can be represented using arrows. The length = the
vector size. The sum can be found by joining the vector arrows together in any
order tip to tail.
The resultant is the single vector that could replace all the others and points in
the direction in which they are going. This is a vector triangle:

                                       a

                                                        b
                                      R



                              R points start – finish



The resultant (or sum) of two vectors is calculated using these vector triangles
and Pythagoras theorem/trigonometry or scale drawing. If the vectors are at
right angles use Pythagoras Theorem and Trigonometry to find the vector
magnitude and vector angle.

We can have more than two vectors being added together. We use the same
method of scale drawing to find the resultant vector, and add them tip to tail as
before:


                                  B


              A




                                           C




                  Resultant




                                               22
Sometimes the resultant is almost zero because the forces can cancel out:

                         A
 Resultant
                                 B



Equilibrium
One condition for equilibrium is as follows:
An object is in equilibrium if the forces acting are balanced (zero resultant
force.
Static solid objects are always in equilibrium.

If you add a number of vectors tail to nose, the resultant vector needed to keep
the body in equilibrium can be found by joining the vectors start to the finish.
The resultant vector closes the loop and points in the opposite direction from
the other vectors (it balances the overall vector that replaces the vectors!).



                     R                         a

                                                       b
                                          R
                                     a
                 c           b
                                                   c




If the vectors are already balanced you end up with a closed polygon. The arrows
will lead back to the start because the resultant is zero.

Look at the diagram below. Forces A, B and C are acting at a point on an object.
They add together to give a zero resultant force. This triangle is called the
triangle of forces


                                     A
                                                                                B
                                                            A

                                          B


        C                                                                C




                                         23
    Force C cancels out the resultant of A and B. Zero resultant means zero
    acceleration. An object that has zero resultant force is said to be in equilibrium.
    An object can be in equilibrium if it is at rest or moving with constant velocity.

    Examples

        1. A climber on a rock face is in equilibrium under the action of 3 forces – his
           weight, the force the rock face exerts on the climber (50 degrees west of
           the vertical) and the force the rope exerts on the climber (40 degrees
           east of the vertical). The diagram is as follows:


                              Force of rope
Force of wall on              on climber Y
climber X




             Weight of
             climber



    Calculate force X and Y, given that the mass of the climber is 80 kg.

    a work out the weight of the climber:
    W = m x g = 80 x 9.81 = 785 N

    b next draw a sketch of the forces, with angles and then draw a triangle of
    forces:                Y



         X
                                   Y

                                              785 N
                     785 N



                                   X

    c we have a right angled triangle which makes calculations easier:

    X = 785 cos 50 = 505 N
    Y = 785 cos 40 = 601 N




                                              24
Questions

  1. A 20 N weight is hanging by a string from the roof. There is tension T in
      the string. A horizontal force H is applied to hold the string at an angle of
      30 degrees to the vertical. By drawing a diagram, obtain the value of the
      force H and Tension T.
  2. Three strings are attached to a small metal ring. Two of the strings make
      an angle of 70 degrees to each other and each is pulled with a force of 7
      N. What force must be applied to the third string to keep the ring
      stationary?
  3. A body is in equilibrium under the action of three forces. One force is 6.0
      N acting due East and one is 3.0 N, in a direction 60 degrees, North of
      East. What is the magnitude and direction of the third force?
  4. An acrobat stands on a tightrope. The rope makes an angle of 7degrees
      with the horizontal on each side. If his weight is 800 N what is the
      tension in each half of the rope?
  5. A bob of mass 0.2Kg is pulled aside by a horizontal force so that the
      string makes an angle of 25 degrees with the vertical. What is the new
      tension in the string?
  6. A sphere of mass 150 g is suspended by a thread which will break under a
      tension of 2 N. If the sphere is slowly pulled aside by a force which is
      always acting horizontally, at what angle to the vertical will the string
      break? What is the value of the horizontal force when this happens?
  7. A mass of 20 kg is pulled up a smooth plane, which makes an angle of 30
      degrees with the horizontal, by a force F acting on it. F is of value 200 N
      and always acts horizontally. Calculate the reaction of the plane on the
      mass and the acceleration of the body.
  8. A mass of weight 5 N is attached to a point B of a string ABC. AB is at 30
      degrees to the vertical, BC is at 45 degrees to the vertical and AC is
      horizontal. Determine the pull in each string.
  9. A boy and girl, raising a bucket of water, pull at an angle to each other.
      Each exerts a force of 150N and their effective pull is 260N. What is
      the angle between their arms? (Scale drawing)
  10. A 100N lamp is suspended over a road by two straight wires, each making
      an angle of 20 degrees with the horizontal and attached to posts on
      opposite sides of the road. Find the tension in each wire.
  11. A kite weighing 50N flies at the end of a string, which makes an angle of
      30 degrees with the horizontal. The force of the wind on the kite makes
      an angle of 45 degrees with the horizontal. Determine the pull on the
      string and also the force of the wind.




                                       25
Equilibrium questions page 35 – OCR Physics



                                      26
   Centre of Gravity




When we considered the climber in the last section of work, climbing up a rock
face, we took him to be a Point Mass. This makes life a lot easier! In problems
involving solid objects, whenever we use the idea of a point mass we often draw
the weight of an object as acting through a single point. This point is called the
centre of gravity. It is the point at which the whole weight of the body may be
considered to act. For a regular object the centre of gravity will be at the
geometrical centre of the body. Finding the C og G for a cylinder, sphere and
uniform beam can be done easily This will not be so if for an irregular object.
Indeed, centre of gravity can be outside the body (boomerang)!
If a body is supported on a pivot directly below the centre of gravity the body
will balance. It will be in equilibrium. The weight of the body is taken to act at
the centre of gravity and is therefore marked using an arrow starting at the c
of g.



                                             C of G




                                          Weight




Stability is important in everyday objects – otherwise, things topple over.
Toppling occurs when the centre of gravity of an object falls outside its base. A
low C of G and a wide base helps make an object more stable, by making as large
as possible the angle through which the object must tilt in order to cause the
centre of gravity to fall outside the base.




                                        27
Experiment to find the C of G of an irregular object

1. Cut the cardboard into a strange shape.
2. Punch a hole near the edge of the cardboard and hang it suspended from a
nail.
3. Make a plumb line by tying a weight to a thread and attaching the thread to
the nail.
4. Mark a line on the cardboard where the thread touches it.
5. Repeat steps 2, 3, and 4, in two more spots on the side of the cardboard.
6. Remove the thread from the cardboard.
7. Balance the cardboard on the end of a pencil. How do you know where to put
the pencil?
8. Describe the centre of gravity for the piece of cardboard.
9. Explain how this principle might help a tightrope walker in a circus.




                                                     Step 3: Suspend the
                                                     shape from another point
                          Step 2: Suspend the        not too close to the first.
                          shape from a point near    Drop a plumb line again
Step 1: Cut any 2D shape.
                          an edge. Drop a plumb line and mark. The
                          and mark on the object.    intersection of the two
                                                     lines is the centre of
                                                     gravity.




                                       28
Moments

The turning effect of a force is called its moment. The moment of a force
about a pivot is the turning effect or rotation effect the force produces about
that pivot (fulcrum).

Moment of a force (Nm) = Force (N) x perpendicular distance (m) from the
                                     line of action of the force to the
                                     pivot point

For a body in equilibrium, sum of clockwise and anticlockwise moments must
be equal about each and every point. This is called the PRINCIPLE OF
MOMENTS




                                        Weight of arm


The forearm is an example of a pivot system.
   The Elbow is the pivot point
   The weight of forearm x distance from elbow plus the carried weight x
      distance from elbow supplies the clockwise moment
   The muscle x distance from the elbow the anticlockwise moment




                                       29
Moment Experiment
(Ruler and weights)

                                   Distance 2
                   Ruler



                  Distance 1



                                          Weight 2

       Weight 1




                                        Clamp and Stand



Get a metre ruler with a hole drilled in the centre and pivot it on a clamp and
stand. Hang the correct weight on the left, in the correct position, and balance
with the named weight on the right. Note down the position on the right for
balancing to occur. Calculate force x distance on both sides (Remember, 100g =
0.981N).

        Left Hand Side                         Right Hand Side
 W1 (N)     D1 (m)     W1 x D1          W2 (N)     D2 (m)      W2 x D2
  1wt         0.4                        4wt
  3wt         0.4                        4wt
  6wt         0.2                        3wt
  2wt        0.45                        3wt
  2wt         0.3                        3wt

When the lever balances, what is the connection between W1 x D1 and W2 x D2?




                                        30
Couples

A single force can cause a body to turn but it will also cause linear motion. A
couple is a pair of equal and opposite forces along different lines of action (not in
the same straight line). Couples cause rotation but no linear motion/acceleration
because the upward force balances the downward force. The resultant is zero,
but they do tend to cause rotation.

Two equal, but oppositely directed forces acting simultaneously on opposite sides
of an axis of rotation. Since the translatory forces (forces that produce linear
motion) cancel out each other, a force couple produces torque (rotatory forces)
only. The magnitude of the force couple is the sum of the products of each force
and its moment arm.




                                    Force couple


The turning effect of a couple, called its Torque is calculated by multiplying one
of the forces by the perpendicular distance between the forces. Torque is
measured in Newton metres. Electric Torque Wrenches can be used to give the
correct torque or turning effect on a nut.




                                         31
Stability, Moments and Centre of Gravity




                                 Pivot point




If the line of action of the centre of gravity is to the left of the pivot point,
there will be an anticlockwise moment, which will return the bus to its upright
position.

If the line of action of the centre of gravity is to the right of the pivot point,
there will be a clockwise moment, which will cause the bus to topple over.

A low centre of gravity and a wide object will be more stable because the angle
through which you need to tilt before the line of action of the C of G falls to the
right of the pivot point will be greater!




                                         32
YEAR 12 MOMENTS

1. The bar in the diagram is in equilibrium.
                       X

     1m      2m                   3m

                       O

        6N                                  F

What is the moment of the 6N force about O?
What is the moment of the force F about 0?
Calculate F and the reaction force X.
Now calculate the moments of F, X and the 6N force about P.
What is the total clockwise moment about P?
What is the total anticlockwise moment about P?

2.    In the diagram a light plank on two trestles is supporting a man
      and a block of concrete.

             X                              Y
                 1m          1m        1m

             A                              B

                                  400N
                      700N



      Write down the moment of each force about A.
      Use the principle of moments to calculate the reaction force Y.
      What is the total upward force X + Y?
      What is the size of the reaction force X?

      In the diagram the man walks past A towards the end of the plank.
      What is the reaction force at Y at the instant the plank starts to tip?
      How far is the man from A as the plank starts to tip?



3.    The diagram shows a wheel of mass 15 kg and radius 1 m being pulled by a
      horizontal force F against a step 0.4 m high.




                                                            F

                           1m                   0.4m




                                                       33
What initial force is just sufficient to turn the wheel so that it will rise over the
step?
What happens to the size of this horizontal force as the wheel rises?

4. A car bonnet weighs 200 N and is 1.8 m long. The bonnet's weight acts at
its midpoint. If the bonnet is freely hinged at one end and is supported by a
vertical force at the other end, what size must this force be in order to
support the bonnet at 30 degrees to the horizontal?

5. A car maintenance manual specifies that the cylinder head bolts on an
engine should be tightened to a torque of 69 Nm. What force must be applied to
the end of the spanner 30 cm long in order to achieve this?
What might be the result of using a higher torque than this to tighten the nut?

6, A railway engine weighing 7.5 x 105 N is 40 m across the centre span of a
bridge consisting of an iron girder 120 m long, weighing 1.0 x 107N. Calculate the
load on each support at each end of the bridge.




                                80m           40m




                                         34
Moments questions etc. pages 40 and 41 – OCR Physics




                                     35
YEAR 12 EQUILIBRIUM

1. An object is held in the position shown by two strings. If the force of friction
at the pulley is negligible find the values of the forces T1 and T2.



                     T1     T2

                                  20

               50



                    250 N


2. A beam of length l and weight w is supported horizontally from two spring
balances A and B attached to its ends as shown. The reading of A is twice that
of B.



           A
                                  B




(i) Find the position of the centre of gravity of the
beam.
(ii) Explain whether you consider the beam to be uniform.

3. The diagram shows a body of weight 120 N supported by two wires from fixed
points A and B. The angles made by the wires with the horizontal are shown in
the diagram. Calculate the tensions in the wires.

      A
                             B


          60                      30




                                        36
5. An aerial attached to the top of a radio mast 20 m high exerts a horizontal
force on it of 3 X 102 N. A stay-wire from the mid-point of the mast to the
ground is inclined at 60 degrees to the horizontal. Assuming the action of the
ground on the mast can be regarded as a single force, find:
(a)    the force exerted on the mast by the stay-wire,
(b)    the magnitude and direction of the action of the ground.

6. A uniform ladder 5.0 m long and having a mass 40 kg rests with its upper end
against a smooth vertical wall and with its lower end 3.0 m from the wall on rough
ground. Find the magnitude and direction of the force exerted at the bottom of
the ladder.

7. A uniform heavy rod OA of mass 12 kg is pivoted freely at O and held in the
position shown by a force F acting at right angles to the rod and applied at A.
Calculate the magnitude of F.
                       A




   O          60




8. Draw a diagram of a cube of side 1m and clearly indicate the position of its
centre of gravity.
The cube is then placed on a rough surface, which is gradually tilted. To what
angle can the surface be tilted before the cube topples? Assume that the
surface is rough so that no slipping can occur.

9. A rectangular box 2 m tall and with base 1m x 1m is filled with matter of
uniform density. It stands on the back of a flat bed truck. What is the maximum
acceleration that the truck can have if the box is not to topple over (Assume
that the friction between box and truck is such that the box does not slide)?
Start by drawing a free body diagram for the box showing the situation when it
begins to topple and so only makes contact with the truck along its rear edge.

10.a. Figure i shows a uniform cylinder whose parallel end faces are not
perpendicular to its axis. Indicate clearly on the diagram the precise position of
its centre of gravity, G.
b. A student places a 2p coin on a table as shown in Fig ii. A second 2p coin is
places on top of the first one, but slightly displaced to one side. Further 2p coins
are added, each one being given the same displacement relative to the previous
coin.




                                        37
The line through the centres of the coin makes an angle of 10 degrees with the
vertical.
If each coin has a diameter of 25 mm and a thickness of 2.0 mm, show that the
student is able to add no more than 70 coins before the pile topples. Explain your
reasoning.




                                             10




                                       38
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