NCEA Level 2
Physics
MECHANICS - FORCES
90522
MASS & WEIGHT
What is the difference between mass & weight?
MASS is the amount of substance we have and is measured in kg. This is a
constant value wherever you go.
WEIGHT is the force of gravity acting on the mass. This is measured in
Newtons (N) and varies dependent on the planet you happen to be standing
upon.
The weight force can be calculated by:
F = mg
Where F = weight force (N)
m = mass (kg)
g = acceleration due to gravity. On Earth 9.81 ms-2
EXERCISE
PAGE 52
RUTTER
FORCE
What is a force?
A force is a vector that has both magnitude and direction.
It can be made up of two or more forces that produce a resultant force.
5N
15N to the right
10N
These force can act at angles and thus produce a resultant force which can
be calculated using Pythagoras if a right angle triangle is formed.
10N
?
5N
5N
10N
Calculate the force
vector?
EXERCISE
PAGE 55 – 58
& PAGE 63 - 65
RUTTER
NEWTONS 3 LAWS OF MOTION
Sir Isaac Newton set up three laws of motion for macro particles:
NEWTONS FIRST LAW
“Every body continues in a state of rest or
uniform motion in a straight line unless
impressed forces
act upon it.”
AAAAAAAH
STATE OF REST
UNLESS ……. !
NEWTONS SECOND LAW
“In order to change the motion of an object it
must be accelerated. To accelerate an object
force must be applied”
This uses the following equation:
F = ma
Where F = weight force (N)
m = mass (kg)
a = acceleration (ms-2)
Example 1:
A 1.0kg mass is connected by a string to
a trolley which has a mass of 3.0kg.
What will the trolley’s acceleration be?
a
3.0kg
a
(Take g = 10ms-2 and neglect friction in
the axles of the trolley and the pulley.)
1.0kg
Fw
SOLUTION:
The objects which accelerate are the 3.0kg
trolley and the 1.0kg mass.
Acceleration of trolley =
acceleration of mass + trolley:
a = F/m
Total mass of objects = 4.0kg
= 10/4.0
Force causing acceleration is Fw:
= 2.5ms-2
Fw = mg = 1.0 x 10
Fw = 10N
NEWTONS THIRD LAW
“Forces occur in pairs. So for every action there
is an equal and opposite reaction”
THE CLUB EXERTS A FORCE F ON THE BALL
THE BALL EXERTS AN EQUAL AND OPPOSITE FORCE F
ON THE CLUB
F
Newton's Third Law.
The rocket's action is to push down on the ground with the force of its
powerful engines, and the reaction is that the ground pushes the
rocket upwards with an equal force.
EXERCISE
PAGE 53 - 55
RUTTER
FRICTION & TENSION
FRICTION
 Produced when two surfaces come into contact and ‘grip’ together without slipping
past each other.
 One surface slide over another.
Friction always opposes the relative motion of an object. Must be allowed for when
involved in the situation. Usually friction is subtracted from the applied force to give a
resultant.
Fres = F - Ffriction
N.B: In vector notation the negative shows acting in the opposite direction.
Example 2:
The thrust from the outboard motor of a powerboat is 1000N. If the boat has a mass
of 500kg and the friction force opposing the motion of the boat through the water is
200N, what is the power boat’s acceleration?
1000N
SOLUTION:
The resultant force on the powerboat is:
Fres = 1000 – 200
= 800N (forward)
Using Fres = ma, the acceleration is:
a = Fres/m
a = 800/500
a = 1.6ms-1
200N
Fres
TENSION
This is a force found in connections and strings. It pulls in both
directions along the string or rope.
A stationary 0.5kg mass hangs on the end
of a string from a ceiling. Tension is caused
in the string by the weight force (5N) of the
hanging mass. Tension acts downward on
the ceiling due to the weight force of the
mass.
The tension also acts upward on the
hanging mass to balance the weight force
acting down.
The effect on the string is that tension acts
to pull the string apart. If the string is not
strong enough then it will snap!
Ceiling reaction F
String
Tension 5.0N
Tension 5.0N
0.5kg
Weight 5.0N
Example 3:
Two masses of 1.0kg and 2.0kg are suspended by strings. Calculate the size of the
tensions in the strings shown in the diagram alongside.
SOLUTION:
Ceiling
Free body force diagrams can be drawn to show
forces acting on each mass.
Tension 2
The forces acting on the 1.0kg mass are shown:
2.0kg
Tension 1
Fw = mg
Fw = 1.0 x 10
1.0kg
Fw = 10N
Fw = 10N
Thus tension 1 = 10N
Tension 1
1.0kg
Weight Fw
The forces acting on the
2.0kg mass are shown
alongside – the weight force
of the 2.0kg mass is:
Tension 2
Fw = 2.0 x 10
Fw = 20N
and so the total force acting
downwards is:
2.0kg
Tension 1
Fw
20 + 10 = 30N.
Thus tension 2 must be 30N
to balance the forces
upwards.
Fw = 2.0 x 10
= 20N
EXERCISE
PAGE 55 - 58
RUTTER
TORQUE
Often the force required to get an object to change its velocity either
a. Can not applied straight on an object.
b. Is too great a force that a person is unable to exert it.
In this case a tool has to be required. E.g. a spanner on a bolt.
In this case the force is applied further away from
the object and produces a turning force or torque
(). The turning force is directly related to the
force applied and the distance it is applied from
the pivot point. Hence the equation:
 = Fd
Where  = torque, turning force(Nm)
F = Force (N)
d = distance from pivot (m)
Example 4:
A force of 10N which acts at 8.0cm from the pivot is needed to lift the cap off a bottle
of soft drink.
Calculate the torque:
SOLUTION:
=Fxd
10N
= 10 x 0.080
8.0cm
= 0.80 Nm anticlockwise
This torque is the same as a force of 40 N acting
2.0 cm from the pivot.
EXERCISE
PAGE 58
RUTTER
TORQUE & EQUILIBRIUM
COUPLES:
This where two equal and oppositely directed forces act at a distance apart. A couple
causes rotation.
The diagram shows the
situation of a couple applied to
a jam jar lid.
The torque produced by the
couple is caused by two equally
sized forces.
The size of the torque is:
SOLUTION:
 = (F x d/2) + (F x d/2)
 = Fd
F
d
F
PRINCIPLES BEHIND TORQUE:
Pivot point
1
2
1
1 = 2
2
1 = 2
FOR A SYSTEM TO BE IN EQUILIBRIUM, THE SUM OF THE CLOCKWISE
TORQUE ABOUT ANY POINT MUST EQUAL THE SUM OF THE
ANTICLOCKWISE TORQUE ABOUT THAT POINT
EQUILIBRIUM is when either:
1. The resultant force or vector sum of the forces equals zero.
2. The sum of all torques acting on the object is zero.
Often torque can be replaced with the word moments
Anticlockwise
torque
Clockwise
torque
Anticlockwise
Torque
= 8.0N
x 0.03m
=0.24 Nm
Calculate the
anticlockwise
and clockwise
torque
Clockwise Torque = 6.0N x 0.04m =0.24 Nm
THE SYSTEM IS IN EQUILIBRIUM
Example 5:
A metre ruler of negligible mass is pivoted at P, at the 20cm mark. A mass weighing
15N is suspended from the 80cm mark.
What weight, Fw, must be suspended at the 10cm mark so that the metre ruler is
balanced?
0.20m
0.10m
0.80m
P
Fw
0.0m
15N
1.0m
SOLUTION:
For the ruler to be balanced, the sum of clockwise moments must equal the sum of
anticlockwise moments. Taking moments about pivot point, P:
Clockwise moments
= Fd
= 15 x (0.80 - 0.20)
= 15 x 0.60
= 9.0Nm
Anticlockwise moments
= Fd
= Fw(0.20 – 0.10)
= 0.10 x Fw Nm
Since sum of clockwise moments = sum of anticlockwise moments:
9.0 = 0.10Fw
Fw = 90N
The two weights add to give a total force of 105N (90 + 15) acting downward on the
pivot. There must also be a reaction force of 105N upward on the metre ruler at the
pivot. Overall, the forces acting on the metre ruler are 105N upward and 105N
downward so that the metre ruler is in euilibrium.
The weight of an object acts down through the object’s centre of mass. This is the point
at which, if an object is suspended or pivoted, the object will balance.
Centre of mass
Centre of mass of the ruler is the
point in the middle of the ruler.
weight
Example 6:
A house painter uses a uniform wooden plank 2.00m long and of weight 300N. It is
supported at both ends by trestles A and B. The painter (of weight 800N) stands
0.50m from trestle A.
Calculate the size of the forces with which trestles A and B hold up the plank and
painter.
1.00m
FA
FB
0.50m
0.0m
2.0m
A
B
300N
800N
SOLUTION:
The diagram is drawn with the weight force of the plank (acting through the centre of
mass), the reaction forces on the trestles (FA and FB), and the painter’s weight force.
To find the size of the force on pivot B (F B), moments are taken about pivot A. At this
point, FA produces no torque, since it acts through A.
Clockwise moments
= (800 x 0.50) + (300 x 1.00)
= 400 + 300
= 700 Nm
Anticlockwise moments
= FB x 2.00
Clockwise moments
700
= Anticlockwise moments
= 2.00 FB
FB = 350N
FA can be calculated by considering the other condition for equilibrium; ie the vector
sum of forces = 0.
Sum of forces upward = sum of forces downward
FA + FB = 800 + 300
FA + 350 = 1100
FA = 750 N
The calculation could be checked by taking the moments about pivot B.
It would not help to take moments about the centre of the plank or the painter, since
both FA and FB would be present together in the one equation of moments.
EXERCISE
PAGE 59 - 60
RUTTER
BRAIN TEASER
Determine the mass of a retort stand using a 500g mass, a ruler, and a laboratory stool
as a pivot.
STEP ONE
Turn the laboratory stool upside down and slide the retort stand shaft along one of the
metal braces until the retort stand balances. The point along the retort stand shaft
above the pivot is the centre of mass for the retort stand. Mark its position carefully
STEP TWO
Hang the 500g at the top of the retort stand shaft and rebalance the retort stand.
Measure the distance of the centre of mass to the new pivot position and the distance
from the 500g mass to the new pivot position.
STEP THREE
Calculate the weight of the retort stand from the measurements taken using equilibrium
of moments.
A
B
Retort stand
Weight of
retort stand
5.0N
Anticlockwise moment = clockwise moment
Weight of retort stand x A = 5.0 x B
Mass of the retort stand can be determined from W = mg
CIRCULAR MOTION
In order to keep an object in a circular motion a force has to applied to an
object. This force is called centripetal force and act to the centre of the
circle being produced. The effect of this force is that the direction changes
and thus the object accelerates.
F = ma
We know from previous work the acceleration is called centripetal
acceleration and also acts to the centre of the circle being produced.
We know that:
ac = v2/r
Substituting
into:
F = ma
v
Fc & ac
We can calculate
Fc by:
Fc = mv2/r
EXERCISE
PAGE 65 66
RUTTER
Practical: Investigating ‘F’ and Time Period in circular motion (Page 68 Rutter)
SPRINGS & HOOKES LAW
Forces can also be used to extend strings. Study was done into this by Robert Hooke
who was a 17th Century British Physicist. He stated that:
“the amount by which a material body is deformed (the strain) is linearly related to
the force causing the deformation (the stress)”
From this he stated in 1678 the following:
“as the extension increases so does the force”
From this he stated mathematically:
Where F = restoring force (N)
x = extension (m)
k = spring constant (Nm-1)
Fx
This is converted to:
F = -k x
k is written as a negative shows the direction.
On a graph the rules change in that the independent variable ‘F’ goes up the y-axis
and the dependent variable ‘x’ goes a long the x-axis.
F (N)
k = gradient
k is a constant and really is the degree
of “stretch-ability” a spring has when a
force is applied
x (m)
Practical: To Investigate Hookes law
(Page 67 Rutter)
EXERCISE
PAGE 65 - 66
RUTTER