1. Forces In One Dimension
Physics:
Chapter 4

2. 4-1: Force In physics, a force is a push or a pull.
Unbalanced forces  Acceleration
Acceleration is in same direction as unbalanced force
F is a vector; represents size & direction of a force
F represents only the magnitude.
Force and Motion
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3. 4-1: Force Identify object(s) of interest = system
Everything around the object that exerts forces on it is called the external world.
Each force has a cause called the agent.
A contact force exists when an object from the external world touches a system and thereby exerts a force on it.
Field forces are exerted without contact.
Force and Motion
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4. 4-1: Force
A physical representation that shows the forces acting on a system is called a free-body diagram.
Force and Motion
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5. 4-1: Force Force vectors in the same direction: combine them
Combining Forces
Force vectors in the same direction: combine them
Forces in opposite directions: subtract
resulting vector in the direction of the greater force.
Vector sum of all the forces on an object is net force.
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6. 4-1: Force Acceleration and Force
As the applied force increases, so does the object’s acceleration.
As the top graph shows, acceleration is proportional to the force.
If the mass is increased, but the force is kept constant, the acceleration decreases.
As the bottom graph shows, acceleration is inversely proportional to the mass.
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Force and Motion

7. 4-1: Force Newton’s Second Law Newton’s Second Law
Newton’s second law states that the acceleration of an object is proportional to the net force acting on the object and inversely proportional to the mass of the object.
To use Newton’s second law to solve a problem, follow these steps.
Draw a free-body diagram showing the direction and relative strength of each force acting on the system.
Then add the force vectors to find the net force.
Next, use Newton’s second law to calculate the acceleration.
Finally, if necessary, use what you know about accelerated motion to find the velocity or position of the object.
Newton’s Second Law
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Force and Motion

8. 4-1: Force Newton’s Second Law ● 20.0 N 15.0 N Problem Response
Use with Example Problem 1.
Problem
Kwaku pushes one end of a table with a force of 15.0 N. Salali pushes on the other end of the table with a force of N. What is the net force on the table?
SOLVE FOR THE UNKNOWN
Determine the net force.
Response
SKETCH AND ANALYZE THE PROBLEM
Draw the situation and a free-body diagram.
Identify the knowns and unknowns.
That is, Fnet = 5.0 in the direction Salali is pushing.
KNOWN
UNKNOWN
FKwaku = 15.0 N
Fnet = ?
FSalali = −20.0 N
EVALUATE THE ANSWER
The net force is in the direction of the force with the large magnitude.
Force and Motion
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9. 4-1: Force
Newton’s First Law
An object that is at rest will remain at rest, and an object that is moving will continue to move in a straight line with constant speed, if and only if the net force acting on that object is zero
Law of Inertia
Inertia—tendency of an object to resist changes in velocity.
If net force on object is zero, then the object is in equilibrium.
If there is no net force acting on the object, then the object does not experience a change in speed or direction and is in equilibrium.
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Force and Motion

10. 4-2: Weight and Drag Force
Weight is the gravitational force experienced by an object. (a field force)
Measured in Newtons (N).
Fg = mg or w = mg
Near Earth’s surface, g is 9.8 N/kg toward Earth’s center. (= 9.8 m/s2)
Weight and Drag Force
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11. 4-2: Weight and Drag Force
35.0 kg ●
502 N
Fg = mg
Weight
Use with Example Problem 2.
Problem
Arnold needs to lift a 35.0-kg rock. If he exerts an upward force of 502 N on the rock, what is the rock’s acceleration?
SOLVE FOR THE UNKNOWN
Determine the net force.
Use Newton’s Second Law to find the acceleration.
Response
SKETCH AND ANALYZE THE PROBLEM
Draw the situation & a free-body diagram.
List the knowns and unknowns.
KNOWN
UNKNOWN
mrock = 35.0 kg
a = ?
g = −9.8 m/s2
EVALUATE THE ANSWER
The sign makes sense because the net force is upward.
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Weight and Drag Force

12. 4-2: Weight and Drag Force
When you step on a scale, the scale exerts an upward support force on you.
Apparent weight is the support force exerted on an object.
Apparent weight depends on how you are accelerating.
Weightlessness is the condition where there are no contact forces acting to support the object and the object’s apparent weight is zero.
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Weight and Drag Force

13. 4-2: Weight and Drag Force
Use with Example Problem 3.
Problem
You are in an elevator, standing on a bathroom scale. You notice that the scale reads a weight that is less than your actual weight. (Assume that the scale is correctly calibrated.)
a. Is the elevator moving at constant velocity, or is it accelerating?
b. If it is accelerating, what is the direction of the acceleration?
Response
a. Since the support force does not equal your weight, you must be accelerating.
b. The support force on you is less than your weight, so the elevator must be accelerating downward.
Weight and Drag Force
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14. 4-2: Weight and Drag Force
A drag force is the force exerted by a fluid on an object opposing motion through the fluid.
This force is dependent on:
the motion of the object,
the properties of the object,
the properties of the fluid (viscosity and temperature) that the object is moving through.
The constant velocity that is reached when the drag force equals the force of gravity is called the terminal velocity.
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Weight and Drag Force

15. 4-3: Newton’s Third Law Interaction Pairs
When you exert a force on your friend to push him forward (FA on B), he exerts an equal and opposite force on you (FB on A), which causes you to move backward.
The forces FA on B and FB on A are an interaction pair.
An interaction pair is two forces that are in opposite directions, have equal magnitudes, and act on different objects.
Newton’s third law states that all forces come in pairs and that the two forces are equal in strength, opposite in direction, and act on different objects.
Newton’s Third Law
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Newton’s Third Law

16. 4-3: Newton’s Third Law Interaction Pairs
When identifying an interaction pair, remember that they always occur in two different free-body diagrams and they always have the symmetry in subscripts.
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Newton’s Third Law

17. 4-3: Newton’s Third Law Interaction Pairs ● Problem ●
FEarth on you ●
Use with Example Problem 4.
Problem
You are walking along when you slip on some ice and fall. For a moment you are in free fall. During this time, what force do you exert on Earth if your mass is 55 kg?
Fyou on Earth ●
SOLVE FOR THE UNKNOWN
Use Newton’s third law and the definition of weight to determine the force you exert on Earth.
Response
ANALYZE AND SKETCH THE PROBLEM
Draw the situation and a free-body diagram.
Identify the knowns and unknowns.
KNOWN
UNKNOWN
myou = 55 kg
Fyou on Earth = ?
g = −9.8 N/kg
EVALUATE THE ANSWER
The force you exert on Earth is equal in magnitude and opposite in direction to the force Earth exerts on you.
Newton’s Third Law
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18. 4-3: Newton’s Third Law Tension
The force exerted by a string or rope is called tension.
At any point in a rope, the tension forces are pulling equally in both directions.
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Newton’s Third Law

19. 4-3: Newton’s Third Law Tension Problem ● Response 100 N Fg = mg
Use with Example Problem 5.
Problem
You are fishing and catch a fish with a mass of 6 kg. If the fishing line can withstand a maximum tension of 100 N, what is the maximum acceleration you can give the fish as you reel it in?
SOLVE FOR THE UNKNOWN
Use Newton’s Second Law to find the maximum acceleration.
Response
ANALYZE AND SKETCH THE PROBLEM
Draw the situation and a free-body diagram.
Identify the knowns and unknowns.
KNOWN
UNKNOWN
mfish = 6 kg
amax = ?
FT, max = 100 N
g = −9.8 N/kg
EVALUATE THE ANSWER
The magnitude of the acceleration is similar to the acceleration due to gravity, so it is reasonable.
Newton’s Third Law
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20. 4-3: Newton’s Third Law Normal Force
The normal force is the perpendicular contact force exerted by a surface on another object.
The normal force is not always equal to the object’s weight, as shown below.
Finding the normal force is important when calculating the effects of friction.
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Newton’s Third Law