1. Chapter 4: Newton's Laws of Motion
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2. Goals for Chapter 4
To understand force – either directly or as the net force of multiple components.
To study and apply Newton's first law.
To study and apply the concept of mass and acceleration as components of Newton's second law.
To differentiate between mass and weight.
To study and apply Newton's third law & identify two forces and identify action-reaction pairs.
To draw a free-body diagram representing the forces acting on an object.
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3. Dynamics, a New Frontier
Stated previously, the onset of physics separates into two distinct parts:
statics
dynamics
So, if something is going to be dynamic, what causes it to be so?
A force is the cause. It could be either:
pushing
pulling
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4. Mass and Weight Mass is a measure of "how much material do I have?"
Weight is "how hard do I push down on the floor?”
Weight of an object with mass m must have a magnitude w equal to the mass times the magnitude of acceleration due to gravity:
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5. Measurement of Mass
Since gravity is constant, we can compare forces to measure unknown masses.
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6. Types of Force Illustrated I
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7. Types of Force II Single or net Contact force Normal force
Frictional force
Tension
Weight
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8. Dynamics, Newton’s Laws
First Law: Every body stays in its state of motion( rest or uniform speed) unless acted on by a non zero net force.
Second Law: The acceleration is directly proportional to the net force and inversely proportional to the mass:
Third Law: Action is equal to reaction.

9. Newton's First Law
"Every object continues either at rest or at a constant speed in a straight line…."
What this common statement of the first law often leaves out is the final phrase "…unless it is forced to change its motion by forces acting on it."
In one word, we say "inertia."
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10. We Determine Effect with the Net Force –
The top puck responds to a non-zero net force (resultant force) and accelerates.
The bottom puck responds to two forces whose vector sum is zero:
The bottom puck is in equilibrium, and does NOT accelerate.
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11. Forces are Inertial and Non-inertial
The label depends on the position of the object and its observer.
A frame of reference in which Newton's first law is valid is called an inertial frame of reference.
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12. Mass and Newton's Second Law
Object's acceleration is in same direction as the net force acting on it.
We can examine the effects of changes to each component.
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13. Mass and Newton's Second Law II
Let's examine some situations with more than one mass.
In each case we have Newton's second law of motion:
Force in N, mass in kg, and acceleration in m/s2. 1 N = (1 kg)(1 m/s2)
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14. Newton's Third Law
"For every action, there is an equal and opposite reaction."
Rifle recoil is a wonderful example.
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15. Newton’s laws and concept of mass
Famous equation:
𝐸=𝑚 𝑐 2 (Einstein) 𝐹 =𝑚 𝑎 (Newton)
𝑁 = 𝑘𝑔 𝑚 𝑠 2
1N=1kg*1m/s2 (mkgs-systems)
Force=1dyn (cmgs-systems)
1𝑁= 1𝑘𝑔∗ 10 3 𝑔∗1𝑚∗ 10 2 𝑐𝑚 1𝑘𝑔∗ 𝑠 2 ∗1𝑚 = 10 5 𝑔 𝑐𝑚 𝑠 2 = 10 5 𝑑𝑦𝑛
Newton’s Laws
1st law: Without a force the state of motion is the same
2nd law: 𝐹∝𝑎, and 𝑎= 𝐹 𝑚
3rd law: Action = Reaction

16. Force to stop a car
What force is required to bring a 1500kg car to rest to rest from a speed of 100 km/h within 55m?
Conversion of units; 100 𝑘𝑚 ℎ ∗ 𝑚 1𝑘𝑚 ∗ 1ℎ 3.6𝑥10 3 𝑠 =28 𝑚 𝑠
𝐹 =𝑚 𝑎 and 𝑣 2 = 𝑣 𝑜 2 +2𝑎(𝑥− 𝑥 𝑜 )
So, 𝑎= 𝑣 2 − 𝑣 𝑜 2 2(𝑥− 𝑥 𝑜 ) = 0− ∗55 =−7.1 𝑚 𝑠 2
𝐹 =1500 −7.1 = −1.1𝑥10 4 𝑁

17. A Force May Be Resolved Into Components
The x- and y-coordinate axes don't have to be vertical and horizontal.
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18. Adding force vectors

19. pulling a sled, Michelangelo’s assistant
The two forces in an action-reaction pair always act on different objects
For forward motion: FAG> FAS FSA > FSG

20. weight=𝑚𝑔
When drawing force diagrams consider only forces acting on the same object
For forward motion:
𝐹 𝑆𝐴 > 𝐹 𝑆𝐺
𝐹 𝐴𝐺 > 𝐹 𝐴𝑆

21. Weight, normal force and a box

22. Normal force = 𝐹 𝑁 and mystery box
𝐹 =𝑚𝑔 and (mass) 𝑚=10 𝑘𝑔
a) 𝐹 = 𝐹 𝑁 −𝑚𝑔=𝑚𝑎 and 𝑎=0
𝐹 𝑁 −𝑚𝑔=0 So, 𝐹 𝑁 =𝑚𝑔 (normal force)
b) 𝐹 𝑁 −𝑚𝑔−40𝑁=0; 𝐹 𝑁 =98𝑁+40𝑁=138𝑁
c) 𝐹 𝑁 −𝑚𝑔+40𝑁=0; 𝐹 𝑁 =98𝑁−40𝑁=58𝑁
d) 𝐹 𝑦 = 𝐹 𝑃 −𝑚𝑔=100𝑁−98𝑁=2𝑁= 𝑚𝑎 𝑦
𝑎 𝑦 = 𝐹 𝑦 𝑚 = 2𝑁 10𝑘𝑔 =0.2 𝑚 𝑠 2

23. Clicker question
You use a cord and pulley to raise boxes up to a loft, moving each box at a constant speed. You raise the first box slowly. If you raise the second box more quickly, what is true about the force exerted by the cord on the box while the box is moving upward?
The cord exerts more force on the faster-moving box.
The cord exerts the same force on both boxes.
The cord exerts less force on the faster-moving box.
The cord exerts less force on the slower-moving box.
Answer: b) The cord exerts the same force on both boxes.
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24. Use Free Body Diagrams In Any Situation
Find the object of the focus of your study, and collect all forces acting upon it.
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25. Unwanted weight loss(descending)

26. Weight loss in a descending elevator
𝐹 =𝑚𝑎 and 𝑚=65 𝑘𝑔 with 𝑎=0.2∗𝑔 (elevator)
𝐹 𝑁 =normal force indicated by scale
𝐹 𝑁 −𝑚𝑔=𝑚(−𝑎)  𝐹 𝑁 =𝑚𝑔−0.2∗𝑔∗𝑚=𝑚0.8𝑔=509.6𝑁 (apparent weight)
𝐹 𝑁 =𝑚𝑔=65 𝑘𝑔∗9.8 𝑚 𝑠 2 =637 𝑁 (real weight)
𝑚= 0.8∗ =52 𝑘𝑔 (apparent mass)

27. Stepping on a scale in an elevator and push “up”
Stepping on a scale in an elevator and push “up”. Your normal weight is 625N.
Your weight going up
Tension in the string
a) Work out your mass
𝑚= 𝑊 𝑔 =638 𝑘𝑔
𝑎=+2.5 𝑚 𝑠 2
What does the scale read? 𝐹 𝑦 =𝑚 𝑎 𝑦 =𝑁−𝑊=𝑚𝑎
Scale reading;
N=𝑚𝑎+𝑊=63.8∗ ∗9.8=784𝑁
Apparent mass = =80 𝑘𝑔
b) What is the tension in a string holding a 3.85kg package?
𝐹 𝑦 =𝑚 𝑎 𝑦 =𝑇−𝑊
T=𝑚𝑎+𝑊=3.85∗ ∗9.8=47.4𝑁

28. Forces and Free Body Diagrams I
The vertical forces are in equilibrium so there is no vertical motion.
But there is a net force along the horizontal direction, and thus acceleration.
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29. Forces and Free Body Diagrams II
Like the previous example, we account for the forces and draw a free body diagram.
Again, in this case, the net horizontal force is unbalanced.
In this case, the net horizontal force opposes the motion and the bottle slows down (decelerates) until it stops.
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30. Superposition of Forces: Resultant and Components
An example of superposition of forces:
In general, the resultant, or vector sum of forces, is:
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31. Pulling a box
Note: FN is less than mg

32. Pulling the box 𝐹 =𝑚𝑎 and 𝑚=10 𝑘𝑔 𝐹 𝑝𝑥 =40∗𝑐𝑜𝑠30=34.6𝑁
𝐹 𝑝𝑦 =40∗𝑠𝑖𝑛30=20.0𝑁
Horizontal
𝐹 𝑝𝑥 =𝑚 𝑎 𝑥  𝑎 𝑥 = 𝐹 𝑝𝑥 𝑚 = =3.46 𝑚 𝑠 2
Vertical
𝐹 𝑦 =𝑚 𝑎 𝑦
 𝑎 𝑦 =0 box does not accelerate vertically
and 𝐹 𝑁 −𝑚𝑔+ 𝐹 𝑝𝑦 =𝑚 𝑎 𝑦
𝐹 𝑁 −98𝑁+20𝑁=0=78𝑁
box accelerates horizontally

33. Box sliding down

34. Clicker question In which case does the rope break first
a) He climbs with a constant rate
b)He just hangs in the rope
c)He accelerates upward with constant a
d) He decelerates downward with the same
constant a

35. A gymnast climbs a rope
In which case does the rope break first? Need to see when rope tension is largest.
Consider the following cases;
a) He climbs with a constant rate
𝐹 =𝑚𝑎 and 𝑎=0 So, 𝑇=𝑚𝑔
b) He just hangs on the rope
𝐹 =𝑇−𝑚𝑔=0; So, 𝑇=𝑚𝑔
c) He climbs up with constant acceleration
𝐹 =𝑇−𝑚𝑔=𝑚𝑎; So, 𝑇=𝑚(𝑔+𝑎)
d) He slides downward with constant acceleration
𝐹 =𝑇−𝑚𝑔=−𝑚𝑎; So, 𝑇=𝑚(𝑔−𝑎) +a -a

36. Getting the car out of the mud

37. Getting the car out of the sand
Getting the car out of the sand. All you need is a rope and Newton’s second law;
𝐹 =𝑚𝑎 and 𝐹 𝑥 =− 𝐹 𝑇1 cos 𝜃 + 𝐹 𝑇2 cos 𝜃 and 𝐹 𝑇1 = 𝐹 𝑇2
𝐹 𝑦 = 𝐹 𝑃 −2 𝐹 𝑇 sin 𝜃 =𝑚𝑎=0 (At the break or loose point)
𝐹 𝑇 = 𝐹 𝑃 2 sin 𝜃 and 𝐹 𝑃 =300𝑁, 𝜃= 5 𝑜
𝐹 𝑇 = 300 2𝑠𝑖𝑛5 =1700𝑁 (almost 6 times large)

38. Why does the force needed to pull the backpack up increase?
Precautions in Yosemite Park
Why does the force needed to pull the backpack up increase?

39. Atwood’s Machine FT=? a=?
Draw a free body diagram for each mass and choose up positive, then a2= a , a1= -a
FT - m1g = m1a1= -m1a
FT – m2g = m2a2= m2a
(m1–m2)g=(m1+m2)a

40. a) When and at what acceleration does the elevator cable break
a) When and at what acceleration does the elevator cable break? Mass of elevator 𝑚=1800 𝑘𝑔 with load.
Note: The cable of the elevator can stand 28000N
𝐹 =𝑚𝑎 and 𝑎= 𝐹 𝑇 −𝑚𝑔 𝑚 = 𝐹 𝑇 𝑚 −𝑔= −9.8=5.75 𝑚 𝑠 2
If we go to the moon with this system; the cable breaks at larger acceleration
𝑎= 𝐹 𝑇 −𝑚𝑔 𝑚 = −1.62=13.9 𝑚 𝑠 2
b) Work out the acceleration of the system shown if it is pulled up with a force of 200N.
𝐹 = 𝑚𝑎 =200𝑁− ∗9.8=53𝑁
So, 𝑎= 53𝑁 15𝑘𝑔 =3.53 𝑚 𝑠 2

41. Two buckets on a string Buckets are at rest; 𝑚 1 = 𝑚 2 =3.5 𝑘𝑔
Upper bucket
𝐹 𝑇1 = 𝐹 𝑇2 + 𝑚 1 𝑔=34+3.5∗9.8=68𝑁
Lower bucket
𝐹 𝑇2 − 𝑚 2 𝑔= 𝑚 2 𝑎=0; 𝐹 𝑇2 =𝑚 2 𝑔=3.5∗9.8=34𝑁
b) Buckets are pulled with constant acceleration 𝑎=1.6 𝑚 𝑠 2
𝐹 𝑇2 − 𝑚 2 𝑔= 𝑚 2 𝑎; 𝐹 𝑇2 =𝑚 2 (𝑔+𝑎)=40𝑁
𝐹 𝑇1 = 𝐹 𝑇2 + 𝑚 1 (𝑔+𝑎)=80𝑁

42. Where does the thread break?
Where does the tread break?
𝐹 =𝑚𝑎
1) 𝑇 1 − 𝑇 2 −𝑚𝑔=𝑚(−𝑎)  𝑇 1 = 𝑇 2 −𝑚𝑔+𝑚𝑎
𝑎>𝑔 and 𝑇 1 > 𝑇 2
Threads break below
2) 𝑇 1 − 𝑇 2 −𝑚𝑔=𝑚(−𝑎)  𝑇 2 = 𝑇 1 +𝑚𝑔−𝑚𝑎
𝑎<𝑔 and 𝑇 2 > 𝑇 1
Threads break above

43. Sliding and hanging blocks
How large are acceleration and tension?
𝐹 =𝑚𝑎
x-component ( 𝑚 1 ) 𝐹 𝑇 = 𝑚 1 𝑎
y-component ( 𝑚 1 ) 𝐹 𝑁 − 𝑚 1 𝑔=0
y-component ( 𝑚 2 ) 𝑚 2 𝑔− 𝐹 𝑇 = 𝑚 2 𝑎
Adding first and third components;
𝑚 2 𝑔= 𝑚 1 + 𝑚 2 𝑎
 𝑎= 𝑚 2 𝑔 𝑚 1 + 𝑚 2
Putting 𝑎 into first equation;
 𝐹 𝑇 = 𝑚 1 𝑚 2 𝑔 𝑚 1 + 𝑚 2

44. A parachutist relies on air
A parachutist relies on air. Let upward direction be positive and Fair =620N is the force of air resistance.
𝐹 =𝑚𝑎
What is the weight of the parachutist?
𝑊=𝑚𝑔=55𝑘𝑔∗9.8 𝑚 𝑠 2 =539𝑁
What is the net force on the parachutist?
𝐹 𝑦 = 𝐹 𝑎𝑖𝑟 −𝑊=620𝑁−539𝑁=81𝑁
What is the acceleration of the parachutist?
𝑎 𝑦 = 𝐹 𝑦 𝑚 1 + 𝑚 2 = 81𝑁 55𝑘𝑔 =1.5 𝑚 𝑠 2 (upward)

45. Clicker question
You are in a spacecraft moving at a constant velocity. The front thruster rocket fires incorrectly, causing the craft to slow down. You try to shut it off but fail. Instead, you fire the rear thruster, which exerts a force equal in magnitude but opposite in direction to the front thruster. How does the craft respond?
It stops moving.
It speeds up.
It moves at a constant speed, slower than before the front thruster fired.
It continues at a constant speed, faster than before the front thruster fired.
Answer: c) It moves at a constant speed, slower than before the front thruster fired.
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