Date: ________________________
MOTION TERMS AND GRAPHS
TERMS
Kinematics: the study of motion without considering the forces that produce motion.
Dynamics: the study of the causes of motion.
Scalar: a quantity that has magnitude (size) but no direction. Examples include energy, time and
speed.
Vector: a quantity that has both magnitude and direction. Examples include velocity, displacement
and acceleration.
Position and Displacement
Distance: the amount of space between two objects or points. Distance is a scalar.
Symbol:
Unit: m (metre)
∆𝑑
Position: the distance and direction of an object from a reference position. Position is a vector.
Symbol:
Unit: m (metre)
𝑑⃗
Displacement: a vector that represents the change in position from the starting position to ending
position. Displacement tells us how far an object has moved.
Symbol:
∆𝑑⃗
Unit: m (metre)
Formula if positions are given:
∆𝑑⃗ = 𝑑⃗2 − 𝑑⃗1
𝑑⃗1 = initial position
𝑑⃗2 = final position
Formula if displacements are given:
∆𝑑⃗𝑇 = ∆𝑑⃑1 + ∆𝑑⃑2
∆𝑑⃑1 = displacement 1
∆𝑑⃑2 = displacement 2
APPLIED TO A GRAPH: If an object is travelling at a constant velocity, the slope of a position vs
time graph will be equal to the velocity.
Example 1:
The slope of the graph on the left is 3 m/s, which represents the velocity. The velocity is graphed
on the right.
SPH4U Dynamics
Lesson 1 – Motion Terms and Graphs
Page 1 of 4
Date: ________________________
Speed and Velocity
Average Speed: the total distance divided by the total time for a trip, measured in m/s (Recall
that speed and distance are scalars)
Symbol:
𝑣𝑎𝑣
Formula:
𝑣𝑎𝑣 =
Unit: m/s
∆𝑑
∆𝑡
Note: Δ is the Greek letter delta, and means “the change in”.
For example, Δt means “the change in time”.
Velocity: a vector that represents the rate of change of position. The vector has both magnitude
and direction.
Average Velocity:
Symbol:
Formula:
𝑣⃑𝑎𝑣
𝑣⃑𝑎𝑣 =
Unit: m/s
∆𝑑⃗
∆𝑡
or 𝑣⃑𝑎𝑣 =
𝑑⃗1 = initial position in metres
∆𝑡 = elapsed time
𝑡1 = initial time in seconds
𝑑⃑2 −𝑑⃑1
𝑡2 −𝑡1
𝑑⃗2 = final position in metres
𝑡2 = final time in seconds
Example 2: A cyclist takes 15.7 s to cover a total distance of 46.3 m by riding 12.5 m [E] then
33.8 m [W]. Find the cyclist’s displacement, average velocity and average speed.
Known: ∆𝑡 =
Unknown:
∆𝑑⃗1 =
∆𝑑 =
∆𝑑⃗𝑇 =?
∆𝑑⃗2 =
𝑣⃗𝑎𝑣 = ? 𝑣𝑎𝑣 =?
SPH4U Dynamics
Lesson 1 – Motion Terms and Graphs
Page 2 of 4
Date: ________________________
APPLIED TO A GRAPH: If an object is travelling at velocity that is not constant, the average
velocity is the slope of the line segment that connects the positions at the beginning and end of the
time interval. This line segment is called a secant.
Secant: a straight line connecting two separate points on a curve.
Example 3: The average velocity during the time interval t1 = 0.0 s to t2 = 3.0 s is the slope of the
line connecting the two corresponding points on the curve.
Known: t1 =
𝑑⃑1 =
Position m [N]
Position vs. Time
7
6
5
4
3
2
1
0
t2 =
⃑
𝑑2 =
Unknown: 𝑣⃑𝑎𝑣 = ? (Find the slope of the
secant.)
𝑣⃑𝑎𝑣 =
0
2
4
𝑑⃑2 −𝑑⃑1
𝑡2 −𝑡1
6
Time (s)
Instantaneous Velocity: the velocity of an object at a particular instant
Symbol
𝑣⃗
Unit m/s
Instantaneous Speed: the speed of an object at a particular instant
Symbol
𝑣
Unit m/s
APPLIED TO A GRAPH: If an object is travelling at velocity that is not constant, the instantaneous
velocity is the slope of a straight line that intersects the position-time graph only at that particular
time. This line is called a tangent.
Tangent: a straight line that intersects a curve at a point and has the same slope of the curve at
the point of intersection.
SPH4U Dynamics
Lesson 1 – Motion Terms and Graphs
Page 3 of 4
Date: ________________________
Example 4: The instantaneous velocity at t = 2.0 s is the slope of the line that is tangent to the
curve at that time.
Known: t1 =
𝑑⃑1 =
Position vs. Time
𝑑⃑2 =
Unknown: 𝑣⃗ = ? (Find the slope of the
7
6
Position m [N]
t2 =
tangent.)
5
𝑣⃗ =
4
𝑑⃑2 −𝑑⃑1
𝑡2 −𝑡1
3
2
1
0
0
2
4
6
Time (s)
Note: When the textbook refers to velocity, it means instantaneous velocity. It will state
specifically if a velocity is an average.
Acceleration
Acceleration: a vector that represents the rate of change of velocity.
Symbol:
Formula:
Unit: m/s2
𝑎⃑
𝑎⃑ =
∆𝑣⃗⃗
∆𝑡
=
𝑣⃗⃗2 −𝑣⃗⃗1
𝑡2 −𝑡1
∆𝑣⃗ = change in velocity, m/s
∆𝑡 = time interval, s
Note: a positive acceleration means the velocity is increasing, a negative acceleration means
that the velocity is decreasing
APPLIED TO A GRAPH: If an object is travelling at velocity that is not constant, the acceleration
can be calculated by finding the slope of the velocity-time graph for the required time interval. If
the velocity-time graph is a straight line, the acceleration is uniform (constant). If the velocitytime graph is a curve, the slope of the secant will give the average acceleration and the slope of the
tangent will give the instantaneous acceleration.
HOMEWORK – Nelson Physics 12 – See Outline
SPH4U Dynamics
Lesson 1 – Motion Terms and Graphs
Page 4 of 4
Date: ________________________
MOTION EQUATIONS
For uniformly accelerated motion:
⃗⃗ = (
∆𝒅
⃗⃗𝒇 + 𝒗
⃗⃗𝒊
𝒗
) ∆𝒕
𝟐
⃗⃗𝒇 = 𝒗
⃗⃗𝒊 + 𝒂
⃗⃗𝒂𝒗 ∆𝒕
𝒗
⃗⃗ = 𝒗
⃗⃗𝒊 ∆𝒕 +
∆𝒅
𝟏
⃗⃗ ∆𝒕𝟐
𝒂
𝟐 𝒂𝒗
⃗⃗
⃗⃗𝟐𝒇 = 𝒗
⃗⃗𝟐𝒊 + 𝟐𝒂
⃗⃗𝒂𝒗 ∆𝒅
𝒗
𝟏
⃗⃗ = 𝒗
⃗⃗𝒇 ∆𝒕 − ⃗𝒂⃗𝒂𝒗 ∆𝒕𝟐
∆𝒅
𝟐
⃗⃗⃗⃗ - displacement in m
∆𝒅
∆𝒕 - elaspsed time in s
⃗⃗𝒇 - final velocity in m/s
𝒗
⃗⃗𝒂𝒗 – average acceleration in m/s
𝒂
⃗⃗𝒊 - initial velocity in m/s
𝒗
2
Freely falling objects:
An object is in free fall when it is dropped down or thrown up. It experiences the force of gravity,
resulting in an acceleration of 9.8 m/s2.
Example 1:
Two cyclists are at rest on a path, Cyclist A is 53.5 m ahead of Cyclist B. The cyclists begin riding.
Cyclist A travels at a constant velocity of 4.20 m/s [N]. Cyclist B travels at a constant acceleration
of 0.750 m/s2 [N]. a) How long will it take for Cyclist B to catch up to Cyclist A? b) What will the
final velocity of Cyclist B be? c) What will the displacement of Cyclist B be when the bikes meet?
SPH4U Dynamics
Lesson 2 – Motion Equations
Page 1 of 3
Date: ________________________
SPH4U Dynamics
Lesson 2 – Motion Equations
Page 2 of 3
Date: ________________________
Example 2:
A car travelling with a velocity of 25.4 m/s [S] starts to slow down at 4.95 m/s 2 [N], eventually
stopping. a) Calculate the braking time. B) Determine the braking displacement.
SPH4U Dynamics
Lesson 2 – Motion Equations
Page 3 of 3
Date: ________________________
VECTOR ADDITION and SUBTRACTION
Displacement Vectors and their Properties:
A vector can be represented by an arrow. The length of the line represents the vector’s magnitude,
and the direction the arrowhead is pointing represents the vector’s direction.
Examples:
Scale: 1 cm = 10 m
∆𝑑⃗ = 25 𝑚 [𝐸]
∆𝑑⃗ = 35 𝑚 [𝑁 25°𝑊]
To determine total displacement if given two vectors, one of three methods can be used, the scale
diagram method, the sine and cosine laws method and the perpendicular components of a vector
method.
The Scale Diagram Method (Graphical vector addition)
To add or subtract non-collinear vectors, the tip-to-tail method must be used. This method is not
very accurate.
Tools: pencil, ruler, protractor
1.
Choose a scale and draw origin lines and note directions.
2. Choose a vector to start with and draw it, using proper scale and direction.
3. Choose the next vector and draw it starting at the tip of the previous vector.
4. Continue drawing vectors as in 3 as required.
5. Draw a resultant vector from the tail of the first vector to the tip of the final vector.
6. Measure the length of the resultant vector and the angle from the origin. Use the scale
factor on the length to determine the magnitude of the vector sum.
Example 1: Add the vectors ∆𝑑⃗1 = 25 𝑚 [𝐸]
and ∆𝑑⃗2 = 35 𝑚 [𝑁 25°𝑊] together using the scale
diagram method.
SPH4U Dynamics
Lesson 3 – Vector Addition and Subtraction
Page 1 of 4
Date: ________________________
The Sine and Cosine Laws Method
This method is accurate, however it will only work when adding two vectors.
1.
Follow steps 1 to 5 in the Scale Diagram Method.
2. Use the cosine line to determine the magnitude of the resultant vector.
3. Use the sine law to determine the angle of the resultant vector.
Example 2: Add the vectors ∆𝑑⃗1 = 25 𝑚 [𝐸]
and ∆𝑑⃗2 = 35 𝑚 [𝑁 25°𝑊] together using the sine and
cosine laws method.
SPH4U Dynamics
Lesson 3 – Vector Addition and Subtraction
Page 2 of 4
Date: ________________________
The Perpendicular Components of a Vector Method
This method is accurate.
1.
To add vectors, each vector must first be resolved into perpendicular components. The
following steps detail displacement vectors, but the method can be applied to all vectors.
To resolve a vector:
i.
On an x-y coordinate system grid, draw the vector originating at 0,0.
ii.
Label the angle between the x-axis and the vector θ.
iii.
Draw a perpendicular vector line from the x-axis to the tip of the vector. This line
is the y-component of the vector.
iv.
Draw a horizontal line from the origin (0,0) to the tail of the y-component vector.
This line is the x-component of the vector.
v.
Use sin θ and cos θ trig ratios to solve for the x and y component vectors. Add
negative signs to any components pointing in the negative x or negative y directions.
2. Add all the x- component vectors together, to get a total x-vector ∆𝑑⃗𝑥 .
3. Add all the y-component vectors together, to get a total y-vector ∆𝑑⃗𝑦 .
4. The total x and y vectors represent the horizontal and vertical components of the total
vector ∆𝑑⃗𝑇 . Calculate the magnitude of ∆𝑑⃗𝑇 by using the Pythagorean Theorem.
2
|∆𝑑⃗𝑇 | = √(∆𝑑𝑥 )2 + (∆𝑑𝑦 )
5.
Calculate the angle between the total vector and the positive horizontal axis by using
𝜃 = tan−1 (
|∆𝑑𝑦 |
)
|∆𝑑𝑥 |
6. The total resultant vector is the magnitude found in step 4 and the angle found in step 5.
SPH4U Dynamics
Lesson 3 – Vector Addition and Subtraction
Page 3 of 4
Date: ________________________
Example 3: Add the vectors ∆𝑑⃗1 = 25 𝑚 [𝐸]
and ∆𝑑⃗2 = 35 𝑚 [𝑁 25°𝑊] together using the
perpendicular components method.
SPH4U Dynamics
Lesson 3 – Vector Addition and Subtraction
Page 4 of 4
Date: ________________________
VELOCITY AND ACCELERATION IN TWO DIMENSIONS
Change in Velocity:
A velocity is changed when:
•
The magnitude has changed
•
The direction has changed
•
Both the magnitude and direction have changed
Acceleration:
Whenever an object’s velocity is changing, the object is accelerating.
Example 1: A car drives along a straight highway for 129.7 km [N 25.30˚ E]. The car then turns
onto another road and travels 25.73 km [N 14.70˚W]. The trip takes a total of 2.250 h.
a) Calculate the average velocity of the car in km/h. b) Calculate the average speed of the car in
km/h.
SPH4U Dynamics
Lesson 4 – Velocity and Acceleration in Two Dimensions
Page 1 of 4
Date: ________________________
SPH4U Dynamics
Lesson 4 – Velocity and Acceleration in Two Dimensions
Page 2 of 4
Date: ________________________
Multiplying Vectors by Scalars:
When a vector is multiplied by a scalar value (assuming that the scalar does not have a magnitude of
1), the magnitude of the vector changes.
When a vector is multiplied by a negative scalar value then the resultant vector’s direction changes.
The direction will be opposite to the direction of the original vector (ie. 180˚ different)
Acceleration and Subtracting Vectors in Two Dimensions:
⃗⃗⃗ + (-𝐵
⃗⃗ can be written 𝑨
⃗⃗). We use this for vector subtraction. The negative of a vector is the
𝐴⃗ – 𝐵
same in magnitude and opposite in direction.
Vector subtraction must be applied to acceleration calculations, since 𝑎⃗𝑎𝑣 =
𝑣⃗⃗𝑓 −𝑣⃗⃗𝑖
∆𝑡
.
The velocity vectors must be broken down into horizontal (x) and vertical (y) components. Then the
components must be subtracted to find the net horizontal and vertical components. Once net
components are found, the magnitude and angle of ∆𝑣⃗ can be found using Pythagorean theorem and
the tan ratio.
Example 2: A car initially travelling at 3.7 m/s [S 22˚ W] turns and then travels at
3.7 m/s [N37˚W]. The turn takes 3.5 s. Calculate the car’s average acceleration.
SPH4U Dynamics
Lesson 4 – Velocity and Acceleration in Two Dimensions
Page 3 of 4
Date: ________________________
SPH4U Dynamics
Lesson 4 – Velocity and Acceleration in Two Dimensions
Page 4 of 4
Date: ________________________
PROJECTILES
Projectile: An object that is launched through the air along a parabolic trajectory and accelerates
due to gravity.
Properties of Projectile Motion:
•
The horizontal motion of a projectile is constant.
•
The horizontal component of acceleration of a projectile is zero.
•
The vertical acceleration of a projectile is constant because of gravity.
•
The horizontal and vertical motions of a projectile are independent, but they share the
same time.
Analyzing Projectile Motion:
Horizontal motion:
Vertical motion:
Initial x velocity
𝑣𝑖𝑥 = 𝑣𝑖 cos 𝜃 (this value is constant)
Range (x displacement) ∆𝑑⃗𝑥 = 𝑣𝑖𝑥 ∆𝑡 or ∆𝑑⃗𝑥 = (𝑣𝑖 cos 𝜃)∆𝑡
Initial y velocity
𝑣𝑖𝑦 = 𝑣𝑖 sin 𝜃
(y velocity changes)
Final y velocity
𝑣𝑓𝑦 = 𝑣𝑖 sin 𝜃 − 𝑔∆𝑡
1
y displacement
∆𝑑𝑦 = (𝑣𝑖 sin 𝜃)∆𝑡 − 𝑔∆𝑡 2
Final y velocity
2
𝑣𝑓𝑦
= (𝑣𝑖 sin 𝜃)2 − 2𝑔∆𝑑𝑦
2
𝜃 is the projection angle, the angle 𝑣⃗𝑖 makes with the horizontal.
Example 1: A diver jumps horizontally off a 8.75 m high cliff with an initial speed of 0.764 m/s.
a) How long will it take for the diver to hit the water below? b) Determine the range of the diver.
SPH4U Dynamics
Lesson 5 – Projectiles
Page 1 of 3
Date: ________________________
The Range Equation
We know:
∆𝑑⃗𝑥 = (𝑣𝑖 𝑐𝑜𝑠𝜃)∆𝑡 (1)
To determine ∆𝑡:
1
∆𝑑𝑦 = (𝑣𝑖 sin 𝜃)∆𝑡 − 𝑔∆𝑡 2 (2)
2
If the projectile lands at the same height from which it was launched, ∆𝑑𝑦 = 0, so:
1
0 = (𝑣𝑖 sin 𝜃)∆𝑡 − 𝑔∆𝑡 2 (3)
2
Factor ∆𝑡 out from the right side of (3)
1
0 = ∆𝑡 (𝑣𝑖 sin 𝜃 − 𝑔∆𝑡)
2
From (4), either:
∆𝑡 = 0
(4)
(𝑡𝑎𝑘𝑒 − 𝑜𝑓𝑓)
OR
1
𝑣𝑖 sin 𝜃 − 𝑔∆𝑡 = 0 (6) (landing)
2
Solving (6) for ∆𝑡:
∆𝑡 =
2𝑣𝑖 sin 𝜃
𝑔
(7)
Substitute (7) into (1)
∆𝑑𝑥 = 𝑣𝑖 cos 𝜃 (
∆𝑑𝑥 =
2𝑣𝑖 2
𝑔
2𝑣𝑖 sin 𝜃
𝑔
) and rewrite as
cos 𝜃 sin 𝜃 (8)
Using the trig identity 2 sin 𝜃 cos 𝜃 = sin 2𝜃 to substitute in (8):
∆𝑑𝑥 =
SPH4U Dynamics
Lesson 5 – Projectiles
𝑣𝑖 2
𝑔
sin 2𝜃 (9) “The Range Equation”
Page 2 of 3
Date: ________________________
Equation 9 only applies when the projectile lands at the same height from which it was
launched. The largest range occurs when 𝜃 = 45°.
Example 2: A golf ball is hit on a flat fairway at a launch angle of 33˚ with a speed of 12.1 m/s.
a) How long does the golf ball stay in the air? b) How far does the golf ball travel?
SPH4U Dynamics
Lesson 5 – Projectiles
Page 3 of 3
Date: ________________________
RELATIVE MOTION
Frame of Reference: A coordinate system relative to which motion is described or observed.
Relative Velocity: the velocity of an object relative to a specific frame of reference.
Relative Velocity Notation:
⃗⃗𝑨𝑩
𝒗
The first subscript (A) represents the moving object, the second subscript (B)
represents the frame of reference
For example if an airplane (P) is travelling at a velocity of 251 km/h [N] relative to a frame
⃗⃗𝑷𝑬 = 𝟐𝟓𝟏
of reference from Earth [E], it would be written as 𝒗
𝒌𝒎
𝒉
[𝑵].
If we consider that the airplane above is flying in windy air (A) we must also consider the
⃗⃗𝑷𝑨 and the velocity of the air relative to the Earth
velocity of the plane relative to the air 𝒗
⃗⃗𝑨𝑬 . It relationship for the velocity of the plane relative to the earth will be:
𝒗
𝑣⃗𝑃𝐸 = 𝑣⃗𝑃𝐴 + 𝑣⃗𝐴𝐸
Example 1: If a plane is flying at a velocity of 253 km/h [S] relative to the air and the air velocity
is 24 km/h [N], what is the velocity of the plane relative to the Earth?
In general, the equation for relative motion takes the form:
𝑣⃗𝐴𝐶 = 𝑣⃗𝐴𝐵 + 𝑣⃗𝐵𝐶
Note that the outside subscripts on the right side of the equation (A and C) are in the same order
as the subscripts on the left side of the equation, and the inside subscripts on the right side are
the same (B).
If we add another frame of reference, the equation would be:
𝑣⃗𝐴𝐷 = 𝑣⃗𝐴𝐵 + 𝑣⃗𝐵𝐶 + 𝑣⃗𝐶𝐷
SPH4U Dynamics
Lesson 6 – Relative Motion
Page 1 of 3
Date: ________________________
Example 2: A bowler on a cruise ship throws a ball. The ball travels 4.7 m/s [forward] relative to
the ship. The ship travels in the same direction as the ball at 21.0 m/s relative to the Earth.
a) What is the ball’s velocity relative to the Earth? b) If the ball moved in a direction opposite to
that of the ship, what would the ball’s relative velocity to the Earth be? c) If the water is moving
forward at 1.1 m/s , and both the ship and the ball are travelling forward, what is the ball’s relative
velocity to the Earth?
SPH4U Dynamics
Lesson 6 – Relative Motion
Page 2 of 3
Date: ________________________
Example 3: A sailboat is motoring into shore. The boat’s velocity relative to the air is 21.0 km/h
[S 25˚ E] and the wind is blowing at 6.5 km/h [S 15˚ W]. Determine the boat’s velocity relative to
the shore.
SPH4U Dynamics
Lesson 6 – Relative Motion
Page 3 of 3
Date: ________________________
FORCES AND FREE BODY DIAGRAMS
Force ( ⃗𝑭⃗): A push or a pull.
Newton: the SI unit of force, symbol N
Contact force: a force that acts between two objects when they touch each other. i.e. a person
pushing on a box
Non-contact force: a force that acts between two objects without the objects touching; also
called action-at-a-distance force. i.e. the force of gravity between a person who is in the air and
Earth
Force of gravity ( ⃗𝑭⃗𝒈 ): the force of attraction between all objects due to mass
Normal force ( ⃗𝑭⃗𝑵 ): a force perpendicular to the surface between objects in contact
Tension ( ⃗𝑭⃗𝑻 ): a force exerted by objects that can be stretched. i.e. the force in a rope that is
being pulled
Friction ( ⃗𝑭⃗𝒇 ): a force that opposes the sliding of two surfaces across one another; always acts in
the direction opposite to motion or attempted motion.
Static Friction ( ⃗𝑭⃗𝑺 ): a force that resists attempted motion between two surfaces in contact.
Kinetic Friction ( ⃗𝑭⃗𝑲 ): a force exerted on a moving object by a surface in the direction of motion
opposite to the motion of the object.
Air resistance ( ⃗𝑭⃗𝒂𝒊𝒓 ): the friction between objects and the air around them.
Applied force ( ⃗𝑭⃗𝑨 ): a force due to one object pushing or pulling on another
FREE BODY DIAGRAMS (FBD)
A free body diagram is a vector diagram showing all forces acting simultaneously on an object.
Steps
1.
Indicate direction.
2. Draw a point indicating the object on which the forces are acting.
3. From the point center, draw all forces acting on the object starting from the point. The
arrow indicates the direction of the force.
SPH4U Dynamics
Lesson 7 – Forces and Free Body Diagrams
Page 1 of 3
Date: ________________________
Example 1:
A car is being pulled out of a ditch by a tow truck. Draw a system diagram. List all the forces
acting on the car, and then draw a free body diagram.
NET FORCE
The net force is the vector sum of all forces acting on an object. The symbol is ∑ 𝐹⃗ , and the unit is
Newton.
Example 2: A football is thrown with a force of 12.0 N at an angle of 17˚ above the horizontal.
The force of gravity on the ball is 3.7 N. Calculate the net force on the ball as it is thrown.
SPH4U Dynamics
Lesson 7 – Forces and Free Body Diagrams
Page 2 of 3
Date: ________________________
Example 3: Two students are pulling on ropes attached to a wagon. The tension in both ropes is
6.5 N and the angle between the ropes is 25˚. A force of friction of 0.45 N acts on the wagon.
The wagon is moving east. What is the net force acting on the wagon?
SPH4U Dynamics
Lesson 7 – Forces and Free Body Diagrams
Page 3 of 3
Date: ________________________
NEWTONS THREE LAWS OF MOTION
Newton’s First Law of Motion – If the external net force on an object is zero, the velocity of the
object will not change.
This implies:
•
An object at rest will stay at rest if the net force on it is zero.
•
An object that is moving a constant velocity will continue to do so if the net force on it is
zero. A net force is not required to maintain that constant velocity.
•
A net force is required to change the velocity of an object in magnitude, direction or both.
•
External forces are required to change the motion of an object. Internal forces have no
effect on an object’s motion.
Recall:
Inertia is a measure of an object’s resistance to a change in velocity. A heavier object has greater
inertia than a lighter object.
Newton’s Second Law of Motion – If the net external force on an object is not zero, the object
will accelerate in the direction of the net force. The magnitude of the acceleration is directly
proportional to the magnitude of the net force and inversely proportional to the object’s mass.
∑ 𝐹⃗ = 𝑚𝑎⃗
∑ 𝐹⃗ – net force on the object in Newtons
𝑚 – mass of the object in kilograms
𝑎⃗ – acceleration of the object in m/s2
Recall 1 𝑁 = 1 𝑘𝑔 ∙ 𝑚/𝑠 2
SPH4U Dynamics
Lesson 8 – Newton’s Laws
Page 1 of 4
Date: ________________________
Example 1: Two people are pulling some friends on a toboggan. The total mass being pulled is 160.0
kg. Bob pulls with a force 55.0 N [E 25.0˚ N] and Kathy pulls with a force of 75.0 N [E 20.0˚ S].
Determine the acceleration of the toboggan. (Assume friction is negligible)
SPH4U Dynamics
Lesson 8 – Newton’s Laws
Page 2 of 4
Date: ________________________
Newton’s Third Law of Motion – For every action force, there exists a simultaneous reaction force
that is equal in magnitude but opposite in direction.
This is also known as the action-reaction principle.
Example 2: A figure skater with a mass of 62.0 kg pushes horizontally against the rink board
toward the north for 0.65 s with a constant force. The skater glides to a maximum speed of 58
cm/s. Assume that friction is negligible and that the skater is initially at rest. Calculate a) the
magnitude of the acceleration b) the force exerted by the board on the skater c) the force
exerted by the skater on the board d) the displacement of the skater from the wall after 1.50 s.
SPH4U Dynamics
Lesson 8 – Newton’s Laws
Page 3 of 4
Date: ________________________
WEIGHT - The force of gravity that a planet or moon exerts on an object.
•
We consider the force of gravity to be acting at a single point on an object called the
center of gravity (c.o.g).
•
In humans, the c.o.g., is located in the trunk, where most of the body mass occurs.
⃗⃗⃗⃑
𝐹𝑔 = 𝑚𝑔⃑
where: ⃗⃗⃗⃑
𝐹𝑔 – weight (force of gravity) in N (Newtons)
m – mass in kg
𝑔⃑ – acceleration due to gravity in m/s2
On Earth, we will use 𝑔⃑ = 9.81 m/s2 [down]. Your textbook uses 9.8 m/s2
Notes:
1. The weight of an object varies throughout the universe, because the acceleration
due to gravity varies on different planets. Even on earth the distance from an object and the
center of the earth can vary.
2. The mass of an object is constant throughout the universe.
NORMAL FORCE
Normal force: The force a surface exerts on the object it contacts.
•
Force is perpendicular to the surface .
•
When an object is on a horizontal surface with no additional applied force, the magnitude of
the normal force is always equal to the magnitude of the weight of an object. Force always
acts in the opposite direction to the weight if the object is on a flat surface. |𝐹⃑𝑁 | = |𝐹⃑𝑔 |
•
The symbol for normal force is 𝐹⃑ N.
SPH4U Dynamics
Lesson 8 – Newton’s Laws
Page 4 of 4
Date: ________________________
APPLYING NEWTON’S LAWS
Equilibrium – A state in which an object has zero net force acting on it. ∑F x = 0 and ∑Fy = 0.
Tip for objects in equilibrium – When solving problems involving objects in equilibrium, draw the
FBD first. Then set the positive x-axis in the direction that will give you the fewest component
vectors to work with.
Example 1: A 1.50 kg dynamics cart is tied with a string to a pin at the top of a ramp that is
inclined at an angle of 15.0˚ to the horizontal. Assume the string is parallel to the ramp, and that
friction is negligible. a) Calculate the magnitude of the tension in the string. b) Calculate the
normal force acting on the cart.
SPH4U Dynamics
Lesson 9 – Applying Newton’s Laws
Page 1 of 2
Date: ________________________
Example 2: A 75.0 kg rock climber is attached to a rope that is allowing her to hang horizontally
with her feet on the wall. The tension in the rope is 825 N and the rope makes an angle of 35.0˚
with the vertical. Determine the magnitude of the force exerted on the wall by the climber’s feet.
SPH4U Dynamics
Lesson 9 – Applying Newton’s Laws
Page 2 of 2
Date: ________________________
MORE ABOUT APPLYING NEWTON’S LAWS
Tip for Accelerating Objects – Draw the FBD and set the positive x-axis in the direction of the
net force, which is also the direction of acceleration.
Example 1: A 75 kg trunk is released from the top of a very slippery ramp that slopes at an angle
of 35˚ from the ground. The height of the ramp is 1.75 m. Calculate the speed of the trunk when
it reaches the ground. Assume friction is negligible.
SPH4U Dynamics
Lesson 10 – More Applying Newton’s Laws
Page 1 of 2
Date: ________________________
Example 2: Kim attached a kite with a mass of 325 g to a string and flew it in a strong easterly
wind. The string made an angle of 60.0˚ with the horizontal and exerted a force of 2.75 N on the
kite. The kite accelerated upward at 1.20 m/s2. Determine the force exerted by the wind on the
kite.
SPH4U Dynamics
Lesson 10 – More Applying Newton’s Laws
Page 2 of 2
Date: ________________________
FRICTION
There are two types of friction, static and kinetic.
•
Both always act 180˚ away from the direction of motion.
•
The magnitude of the friction is related to the roughness of the two surfaces and the
normal force
•
The coefficient of friction is the ratio of friction to the normal force. The value depends
of the characteristics of the two surfaces.
Kinetic Friction
•
The friction between two surfaces when the surfaces are moving with respect to each
other
𝐹𝐾 = 𝜇𝐾 𝐹𝑁
𝐹𝐾 – force of kinetic friction in Newtons
𝜇𝐾 – coefficient of kinetic friction, unit-less (Table on p. 85)
𝐹𝑁 – Normal force of the moving object in Newtons
Static Friction
•
The friction between two surfaces when the surfaces are not moving with respect to each
other
𝐹𝑆 = 𝜇𝑆 𝐹𝑁
𝐹𝑆 – force of static friction in Newtons
𝜇𝑆 – coefficient of static friction, unit-less (Table on p. 85)
𝐹𝑁 – Normal force of the moving object in Newtons
SPH4U Dynamics
Lesson 11 – Friction
Page 1 of 3
Date: ________________________
Example 1: A person can exert a force of 2.50 x 102 N. A rectangular box has a mass of 42.5 kg.
The person’s arms make an angle of 30.0˚ with the top of the box. If the coefficient of kinetic
friction between the floor and the box is 0.25, would it be more effective for the person to push or
pull the box?
SPH4U Dynamics
Lesson 11 – Friction
Page 2 of 3
Date: ________________________
Example 2: A mass, m1 sits on a table connected with a massless string to mass m 2. Mass m2 is
hanging over the table ledge, with the sting over a frictionless pulley. The coefficient of friction
between m1 and the table is 0.228. If m1 has a mass of 0.250 kg, what is the maximum mass of m2
before m1 starts sliding across the table?
SPH4U Dynamics
Lesson 11 – Friction
Page 3 of 3
Date: ________________________
FRAMES OF REFERENCE
A frame of reference is the observer’s coordinate system for describing a motion. A person on a
sidewalk watching a moving bus is in a stationary frame of reference, with the ground as reference.
A passenger in the bus who sees the person on the sidewalk is in a moving frame of reference, with
the car as reference.
Inertial frame of reference: a frame of reference that moves at a zero or constant velocity; the
law of inertia holds. A stationary observer on the ground has an inertial frame of reference
relative to the ground.
Non-inertial frame of reference: a frame of reference that accelerates with respect to an
inertial frame; the law of inertia does not hold. A passenger in an accelerating bus has a noninertial reference frame.
Fictitious forces: apparent but non-existent forces that explain the motion in accelerating (noninertial) frames of reference. For illustration, city buses do not have seat belts. The force applied
to a bus to accelerate it is not directly applied to the passenger. When a bus accelerates, the
passenger feels like they are being pushed backwards – really it’s because the seat is accelerating
forward, but initially they aren’t. When the bus turns, a passenger feels like they are being pushed
the opposite way – it’s really because their body is initially continuing in a constant forward velocity
while the seat is turning (therefore accelerating), so it feels like their body is being pushed away
from the turn. Physicists use the concept of fictitious forces to explain these motions.
Apparent weight: the magnitude of the normal force acting on an object in an accelerating (noninertial) frame of reference. Some examples:
•
If a person is standing on a scale in an accelerating elevator, the apparent weight goes down
when the elevator accelerates downward, and goes up when the elevator accelerates upward.
•
In a free-fall amusement park ride, acceleration is 9.8 m/s2 downward, and the normal
force is zero.
•
An astronaut on the ISS is in constant free-fall, so the normal force acting on the
astronaut is zero.
SPH4U Dynamics
Lesson 12 – Frames of Reference
Page 1 of 3
Date: ________________________
Example 1: A small cork is suspended with a thread from the ceiling of a bus. When the bus
accelerates at a constant rate backward, the string makes an angle of 8.0˚ with the vertical.
Calculate the magnitude of the acceleration of the bus.
SPH4U Dynamics
Lesson 12 – Frames of Reference
Page 2 of 3
Date: ________________________
Example 2: An elevator accelerates upward at a rate of 1.2 m/s2 until it reaches a safe constant
velocity. As it approaches the stopping position, it undergoes a downward acceleration of magnitude
0.85 m/s2. If a passenger has a mass of
65 kg, determine their apparent weight when a) the elevator is accelerating upward b) the
elevator is moving at a constant velocity c) the elevator is accelerating downward.
SPH4U Dynamics
Lesson 12 – Frames of Reference
Page 3 of 3
Date: ________________________
CENTRIPETAL ACCELERATION
Uniform circular motion: the motion of an object with a constant speed along a circular path of
constant radius. Because the direction of the object is continually changing, the velocity is also
changing.
Centripetal acceleration: the instantaneous acceleration that is directed toward the centre of the
circular path.
𝑎𝑐 =
𝑣2
𝑟
𝑎𝑐 - the magnitude of the centripetal acceleration in m/s2
𝑣 - speed of the object moving on the circular path, in m/s
𝑟 - the radius of the circular path in m
Where did this formula come from?
a)
b)
In part a) of the above diagram, an object is travelling at constant speed in a circular path from A’
to A over a short time interval Δt. The distance the object travels is s.
1) Recall 𝑎⃗𝑎𝑣 =
∆𝑣⃗⃗
∆𝑡
In part b) of the diagram, ∆𝑣⃗ = 𝑣⃗2 − 𝑣⃗1 is shown graphically. We are assuming the time interval is
very small. Notice ∆𝑣⃗ is pointed inward towards the center of the circle, so the acceleration will be
inward towards the center of the circle.
2) We know |𝑣⃗1 | = |𝑣⃗2 | so we can assume that the magnitude of the velocity at any point in the
circular path can be stated as 𝑣.
3) Triangle BAC has the same interior angles as triangle AOA’, so the two triangles are similar
triangle.
SPH4U Dynamics
Lesson 13 – Centripetal Acceleration
Page 1 of 4
Date: ________________________
In b), the angle between A and the radius is 90˚, so
𝜃
2
+ 𝛼 = 90°
𝜃 + 2𝛼 = 180°
𝜃 = 180° − 2𝛼
Or, when you add the two equal angles, 𝛼, within triangle BAC to the third angle 𝜃, they must
equal 180˚, so
180 − 2𝛼 = 𝜃
4) In triangle AOA’ there are two side lengths of r, and one side length s. The distance travelled is
s and 𝑠 = 𝑣∆𝑡. Using the ratios of similar sides for the similar triangles:
|∆𝑣⃗⃗|
𝑣
|∆𝑣⃗⃗|
𝑣
=
=
𝑠
𝑟
𝑣∆𝑡
𝑟
|∆𝑣⃗| =
5)
𝑎𝑎𝑣 =
𝑎𝑎𝑣 =
𝑎𝑎𝑣 =
𝑣 2 ∆𝑡
𝑟
|∆𝑣⃗⃗|
∆𝑡
(Substitute the above equation into this one)
𝑣2 ∆𝑡
𝑟
∆𝑡
𝑣2
𝑟
If Δt is very small, 𝑎𝑎𝑣 = 𝑎𝑎𝑐 so 𝑎𝑐 =
𝑣2
𝑟
Period (T) : the time in seconds required for a rotating, revolving, or vibrating object to complete
one cycle.
Frequency (f): the number of rotations, revolutions, or vibrations of an object per unis of time, the
SI unit is Hz.
Additional equations
𝒗=
∆𝒅
∆𝒕
and for a complete circular path ∆𝑑 = 2𝜋𝑟 𝑎𝑛𝑑 ∆𝑡 = 𝑇, so 𝑣 =
Substituting into 𝑎𝑐 =
𝑣2
𝑟
2𝜋𝑟
𝑇
.
will result in
𝑎𝑐 =
SPH4U Dynamics
Lesson 13 – Centripetal Acceleration
4𝜋2 𝑟
𝑇2
Page 2 of 4
Date: ________________________
𝒇=
𝟏
𝑻
, substituting into the above equation will result in
𝑎𝑐 = 4𝜋 2 𝑟𝑓 2
Example 1: A car travels at a constant speed of 10.0 m/s around a circular road of radius 50.0 m.
Find the magnitude of car’s centripetal acceleration.
SPH4U Dynamics
Lesson 13 – Centripetal Acceleration
Page 3 of 4
Date: ________________________
Example 2: A sock spins on the tub of a washing machine at 30.0 rpm (revolutions per minute). If
the radius of the tub is 0.275 m, calculate the period and the centripetal acceleration of the sock.
SPH4U Dynamics
Lesson 13 – Centripetal Acceleration
Page 4 of 4
Date: ________________________
CENTRIPETAL FORCE
Centripetal acceleration is caused by centripetal force.
Centripetal force: the net force that causes centripetal acceleration.
Applying Newton’s Second Law (F = ma) we get:
𝐹𝑐 =
𝑚𝑣 2
𝑟
In the case of a rotating object on a
string, the string tension is:
𝐹𝑇 =
𝑚𝑣 2
𝑟
𝐹𝑐 - the magnitude of the centripetal force in m/s2
𝑣 - speed of the object moving on the circular path, in m/s
𝑟 - the radius of the circular path in m
Example 1: A car with a mass of 2135 kg is travelling around a level curve with a radius of
curvature of 52 m. If the centripetal force on the car is 1.45 x 10 4 N, what is the car’s speed?
SPH4U Dynamics
Lesson 14 – Centripetal Force
Page 1 of 3
Date: ________________________
Example 2: A car with is travelling around a banked curve with a radius of
1.5 x 102 m and a banking angle of 12˚. If the coefficient of static friction between the tires and
the road is 0.60, determine the maximum safe speed of the car.
SPH4U Dynamics
Lesson 14 – Centripetal Force
Page 2 of 3
Date: ________________________
Example 3: A roller coaster travelling at 21 m/s is on a curved track at the lowest point. A
passenger is experiencing an apparent weight that is 3 times her normal weight. Determine the
radius of curvature of the track.
SPH4U Dynamics
Lesson 14 – Centripetal Force
Page 3 of 3
Date: ________________________
SPH4U – CENTRIPETAL FORCE AND ROTATING FRAMES OF REFERENCE
WORKSHEET
Read pages 120 to 124, Nelson Physics 12
1. Define centripetal force and provide some examples.
Read pages 125 to 130, Nelson Physics 12
1. Define the following terms:
Centrifuge –
Centrifugal force –
Coriolis force –
Artificial gravity –
SPH4U Dynamics
Lesson 14 – Centripetal Force Worksheet
Page 1 of 2
Date: ________________________
2. Using the figures below, explain the rider’s sensation on the merry-go-round.
3. Explain how a centrifuge is able to separate blood into its components.
SPH4U Dynamics
Lesson 14 – Centripetal Force Worksheet
Page 2 of 2