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CHAPTER 4 FORCES prepared by Yew Sze Ling@Fiona, KML
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Chapter 4 Forces
Curriculum Specification Remarks
Before After Revision
4.1 Basic of forces and free body diagram
a) Identify the forces acting on a body in different
situations:
i. Weight ii. Tension
iii. Normal force iv. Friction v. External force (pull or push)
(C1, C2, C3, C4)
b) Sketch free body diagram. (C3, C4)
c) Determine static and kinetic friction.
Nf ss , Nf kk
(C3, C4)
4.2 Newton’s Laws of Motion
a) State Newton’s laws of motion. (C1, C2)
b) Apply Newton’s laws of motion. (C3, C4)
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CHAPTER 4 FORCES prepared by Yew Sze Ling@Fiona, KML
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4.1 Basic of Forces and Free Body Diagram
Force is defined as physical quantity which causes an object to move, stop, change its direction or change its physical form.
A force is a push or pull upon an object resulting from the object’s interaction with other object.
A force requires an agent. That is, a force has specific, identifiable cause (something does the pushing or pulling).
Force is a vector quantity. It has both a magnitude and a direction.
The S.I. unit of force, F is kg m s-2 or newton (N).
Additional Knowledge: Contact force and non-contact force
All forces (interactions) between objects can be placed into two broad categories: contact force or non-contact force.
Contact forces must be touching or connected to have an effect on an object. For example: applied force, tension force, normal force and friction force.
Non-contact forces don’t have to touch to have an effect. They work over a distance or a field of influence. For example: gravitational force/ weight force, electrical force and magnetic force.
Types of Forces
There are many forces we will deal with over and over. In this chapter, we are going to focus on five
of them: weight, tension, normal force, friction and external force (pull or push).
Weight
Explanation Diagram
Weight is defined as the force exerted on a body under gravitational field on or near the surface of the earth.
The agent for the weight force is the entire earth pulling on an object.
Direction: The weight vector always points vertically downward
Equation:
gmW
where g = 9.81 m s-2
.
Symbol:
W
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CHAPTER 4 FORCES prepared by Yew Sze Ling@Fiona, KML
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Tension
Explanation Diagram
Tension is the magnitude of the pulling force exerted by a string, on another
object.
The agent for the tension force is string/rope/wire etc.
Direction: The direction of the tension force is always in the direction of the
string or rope (directed away from the
object).
The tension is the same at all points in the rope (Same string possessed same tension.).
Symbol:
T
Normal Force
Explanation Diagram
Normal force is defined as a reaction force that is exerted by the surface to an
object in contact with the surface.
The agent for the normal force is the surface.
Direction: Normal force is always perpendicular to the surface.
Symbol:
N
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Friction
Friction is defined as a force that opposes any impending relative motions or the relative motion of two surfaces in contact.
The agent for friction is rough surface.
In our syllabus, the surface is always smooth unless stated rough surface.
Friction is directly proportional to the reaction force: Nf
Equation:
Nf
where μ is the coefficient of friction.
Coefficient of friction is defined as the ratio between frictional force to normal force. It depends on the nature of the surfaces.
There are two types friction: static friction ( sf ) and kinetic friction ( kf ).
The value of kinetic friction between two surfaces is less than the value of static friction:
sk ff
Explanation Diagram
Static Friction,
sf
Static friction is the force at the interface between two stationary surfaces which
prevents the surfaces from sliding over each
other (frictional force before the object
start moving, v = 0)
Keywords use in exam:
When the object is about to move
When the object starts to move
Just before the object begins to move
Direction: Static friction points opposite the direction in which the object would move if
there were no friction. (it points in the direction necessary to prevent motion)
Nf ss
Kinetic Friction,
kf
Kinetic friction is the force that exist between 2 objects when there is relative
motion at the interface of the surfaces in
contact
This is a force that “opposes the motion”.
Direction: opposite direction of velocity (“the motion”).
Nf kk
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CHAPTER 4 FORCES prepared by Yew Sze Ling@Fiona, KML
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External Force
Explanation Diagram
External force is a force exerted on a system by an agency outside the system.
External force can either be push or pull force exerted on the object.
The agent for external force is human or other objects.
Direction: Always directed outwards from the body. Usually will be given by
the question.
Symbol:
F
Even though the person is pushing the object
from the left, but the direction of force is
coming out from the right (outward from body).
Summary: Identifying Forces
Always ask yourselves 5 questions:
1. Is there mass of object?
Got mass → Weight
2. Is the object in contact with a surface?
In contact → Normal force
3. Is the coefficient of friction given (μ) or is it mentioned that the surface is rough?
Got μ or rough surface → Friction
4. Is there string/ rope/ chain connected to the object?
Got string/ rope/ chain → Tension
5. Is there any applied forces (pushing or pulling) acting on the object?
Got pushing or pulling → External force
Free Body Diagram
Free body diagram is a diagram that shows all the forces acting on just one object. The object is
represented by a dot. The force arrow is drawn from the centre of the point outward in the
direction in which the force is acting. For example:
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Resultant Forces
Resultant force is also known as net force. It is defined as the vector sum of all forces acting on the object.
If both the forces are acting in a straight line (same component), we can add or subtract the forces to show the overall effect on the object called the resultant force.
If both the forces are not acting in a straight line, then we have to solve it by using the resolving method.
Resolving method: Refer to Chapter 1 notes.
4.2 Newton’s Laws of Motion
Newton’s First Law
An object continues in a state of rest or in a state of motion at a constant velocity (constant speed in a constant direction), unless compelled to change that state by a net force.
In other words, the vector sum of all forces (net force) acting on it is zero.
0F
The first law gives the idea of inertia.
Additional Knowledge: Inertia and Mass
Inertia is defined as the tendency of an object to resist any change in its state of rest or motion.
Inertia depends on mass. A bigger mass needs a bigger force to overcome its inertia and change its motion.
inertiamass
Easy to start moving Hard to start moving
Mass is defined as a measure of a body’s inertia.
Mass is a scalar quantity.
The terms mass and weight are often confused with one another. Mass is the property of object, it is the amount of matter an object contains. The value of mass is independent of location.
Weight on the other hand is a force, the pull of gravity acting on the object. To see the
difference, suppose we take an object to the moon. The object will weigh only one-sixth as much as it did on earth, since gravity is weaker. But its mass will be the same.
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Example of Newton’s First Law: Equilibrium of a Particle
A particle is said to be in equilibrium when the vector sum of all forces acting on it is zero, in other words, when all forces which act upon a particle are balanced.
0F
In term of components:
0 xF ; 0 yF
According to Newton’s first law there are two types of equilibrium:
Static equilibrium – Object is at rest (v = 0 m s-1)
Dynamic equilibrium – Object is moving in a straight line with constant velocity
Problem solving strategy:
1. Sketch a simple diagram of the system to help conceptualize the problem.
2. Identify all forces acting on the object.
3. Draw free body diagram.
4. Choose a convenient coordinate axes for each body and construct a table to resolve the forces into x- and y-components.
5. Apply the condition for equilibrium of a particle in component form:
0 xF ; 0 yF
6. Solve the component equations for the unknown.
Newton’s Second Law
When a net external force ∑ �⃗� acts on an object of mass, the acceleration that result is directly proportional to the net force and has a magnitude that is inversely proportional to the mass.
m
Fa
amF
The direction of the acceleration is the same as the direction of the net force.
Equation above is a vector expression and hence is equivalent into two components:
xxmaF ; yy maF
Example of Newton’s Second Law
Velocity not constant, change in acceleration
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Problem solving strategy:
1. Sketch a simple diagram of the system to help conceptualize the problem.
2. Identify all forces acting on the object.
3. Draw free body diagram.
4. Choose a convenient coordinate axes for each body and construct a table to resolve the forces into x- and y-components.
5. Apply suitable Newton’s Law of Motion:
Axis PARALLEL to motion:
0F for constant velocity OR maF for constant acceleration
Axis PERPENDICULAR to motion 0F
6. Solve the component equations for the unknown.
Newton’s Third Law
Whenever one object exerts a force on a second object, the second object exerts an oppositely directed force of equal magnitude on the first object.
In other word, for every action (force), there is a reaction (opposing force) of equal magnitude but acts in opposite direction.
BAAB FF
Example of Newton’s Third Law
Action–reaction forces that
are equal and opposite
Action–reaction forces that
are equal and opposite Action–reaction forces that are equal and opposite
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Exercise
Revision
1. A particle is being acted upon by three coplanar forces
with magnitudes and directions as indicated in the
figure. Calculate the magnitude and direction of
resultant force.
Basic of Forces and Free Body Diagram
Draw the free-body diagram of the box.
Draw the free-body diagram of the box.
Draw the free-body diagram of the boxes.
Draw the free-body diagram of the boxes.
Draw the free-body diagram of the boxes.
Draw the free-body diagram of the box.
F2 = 2.0 N
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Newton’s Laws of Motion
1. An object of weight W = 49 N is suspended by two strings
which are at 30° and 70° to the vertical as shown below. The object is in equilibrium. Calculate the tension in each string.
2. Calculate the magnitude and direction of a force that balance
the three forces acted at point A as shown in Figure above.
3. You are driving a BMW 525i on a straight testing track at a constant speed of 150 km/h. You
pass a Perodua Kancil doing a constant 75 km/h. Which cars’ net force is greater and which
Newton’s Law of Motion is used?
4. According to a simplified model of a mammalian heart, at each pulse approximately 20 g of
blood is accelerated from 0.25 m s-1
to 0.35 m s-1
during a period of 0.10 s. What is the magnitude of the force exerted by the heart muscle?
5. A force of 15 N is applied at an angle of 30° to the horizontal on a 0.750 kg block at rest on a
frictionless surface as shown in figure.
Calculate
a) the acceleration of the block. b) the magnitude of normal force exerted by the table on the
box.
6. Two objects of masses m1 = 10 kg and m2 = 15 kg are connected by a
light string which passes over a smooth pulley as shown in Figure.
Calculate
a) the acceleration of the object of mass 10 kg.
b) the tension in each string.
7. Two masses mA = 2.0 kg and mB = 5.0 kg
are on inclines and are connected together
by a string as shown in Figure. The
coefficient of kinetic friction between each
mass and its incline is μk = 0.30. If mA
moves up, and mB moves down, determine
their acceleration.
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8. A 72 kg water skier is being accelerated by a ski boat on a flat (“glassy”) lake. The
coefficient of kinetic friction between the skier’s skis and the water surface is μk = 0.25
(Figure).
a. What is the skier’s acceleration if the rope pulling the skier behind the boat applies a horizontal tension force of magnitude T = 240 N to the skier (θ = 0°)?
b. What is the skier’s horizontal acceleration if the rope pulling the skier exerts a force of T = 240 N on the skier at an upward angle θ = 12°?
c. Explain why the skier’s acceleration in part (b) is greater than that in part (a).
9. A 65 kg woman descends in an elevator that accelerates at 2.0 m s-2
downward. She stands on a scale that reads in kg.
a. During this acceleration, what is her weight and what does the scale read?
b. What does the scale read when the elevator descends at a constant speed of 2.0 m s
-1?
10. A 20.0 kg box rests on a table.
a. What is the weight of the box and the normal force acting on it?
b. A 10.0 kg box is placed on top of the 20.0 kg box, as shown in Figure. Determine the normal
force that the table exerts on the 20.0 kg box and
the normal force that the 20.0 kg box exerts on the 10.0 kg box
11. Two blocks, A of mass 10 kg and B of mass 30 kg, are side by side and in contact with each
another. They are pushed along a smooth floor under the action of a constant force F of
magnitude 200 N applied to A as shown in figure below.
Calculate
a) the acceleration of the block.
b) the force exerted by A on B.