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Force. Chapter 4. Isaac Newton (1642 to 1727). Born 1642 (Galileo dies) Invented calculus Three laws of motion Principia Mathematica. Newton’s Three Law’s of Motion. All objects remain at rest or in uniform, straight-line motion unless acted upon by an outside force. ( inertia)

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Force

Force

Chapter 4


Isaac newton 1642 to 1727
Isaac Newton (1642 to 1727)

  • Born 1642 (Galileo dies)

  • Invented calculus

  • Three laws of motion

  • Principia Mathematica


Newton s three law s of motion
Newton’s Three Law’s of Motion

  • All objects remain at rest or in uniform, straight-line motion unless acted upon by an outside force. (inertia)

  • Force = mass X acceleration

  • Every force has an equal and opposite force.


The first law
The First Law

Inertia –tendency of an object to remain at rest or in constant motion.

mass - measure of inertia.

Mass & inertia are directly proportional


The first law1
The First Law

Ball in jar example:


The first law2
The First Law

Direction of jar


The first law3
The First Law

Now stop pushing the jar (same as not wearing a seatbelt)

Jar gets stopped

Marble keeps going


The first law4
The First Law

School bus example

This is the way you want to keep going

The bus has turned, so you feel pulled to the side.


The first law5
The First Law

Bee in a car example

  • Bee is in air when car starts

  • Bee is on the seat when car starts


Does it take less force to push the elephant (ignore friction) on earth or on the moon?



Inertial reference frames
Inertial Reference Frames “weightless” in space?

  • Non-accelerating (constant velocity) reference frame

    • All laws of physics are identical

    • Cannot tell if you are moving in an Inertial Reference Frame

    • Speed of light question


The second law
The Second Law “weightless” in space?

Force = mass X acceleration

SF = ma

Sum of all the forces acting on a body


The second law use the force
The Second Law: Use the Force “weightless” in space?


The second law1
The Second Law “weightless” in space?

Situation One:

Non-moving Object

  • Still has forces

Force of the material of the rock

Force of gravity


The second law2
The Second Law “weightless” in space?

Situation Two: Moving Object

SF = ma

ma = Fpedalling – Fair - Ffriction

Fpedalling

Fair

Ffriction


The second law3
The Second Law “weightless” in space?

  • Unit of Force = the Newton

    SF=ma

    SF = (kg)(m/s2)

    1 N = 1 kg-m/s2

  • A force of 1 Newton can accelerate a 1 kg object from rest to 1 m/s in 1 s.


The second law example 1
The Second Law: Example 1 “weightless” in space?

A 60.0 kg bike and rider accelerates at 0.5 m/s2. How much extra force did the rider’s legs have to provide?

SF = ma = (60.0 kg)(0.5 m/s2)

SF = 30 kg m/s2 = 30 N


The second law example 2
The Second Law: Example 2 “weightless” in space?

A force of 5 Newtons can accelerate a watermelon 2.5 m/s2. What is the mass of the watermelon?

(No actual watermelons were harmed in the production of this example problem)


The second law example 3
The Second Law: Example 3 “weightless” in space?

What Force is needed to accelerate a 5 kg bowling ball from 0 to 20 m/s over a time period of 2 seconds?


The second law example 4
The Second Law: Example 4 “weightless” in space?

Calculate the net force required to stop a 1500 kg car from a speed of 100 km/h within a distance of 55 m.

100 km/h = 28 m/s

v2 = vo2 + 2a(x-xo)

a = (v2 - vo2)/2(x-xo)

a = 02 – (28 m/s)2/2(55m) = -7.1 m/s2


S “weightless” in space?F = ma = (1500 kg)(-7.1 m/s2) = -1.1 X 104 N

(negative sign tells us that the force is the in the opposite direction of motion)

Direction of original motion

Fbrakes


The 3 rd law
The 3 “weightless” in space?rd Law

“For every force, there is an equal and opposite force.”


The 3 rd law1
The 3 “weightless” in space?rd Law

Runner example:

  • Does the runner push on the earth?

  • Why does the runner move more?

  • Does the earth move at all?

Backward force for the earth

Forward force for the runner


The 3 rd law2
The 3 “weightless” in space?rd Law

Fforward

Fgases


The 3 rd law3
The 3 “weightless” in space?rd Law

Sled of bricks on Ice:

  • Would the sled move?

ICE


The 3 rd law4
The 3 “weightless” in space?rd Law

Why would Cyclops be in trouble?


Mass vs weight
Mass vs. Weight “weightless” in space?

Mass (kg)

  • Amount of matter in an object

  • Independent of gravity (the # of atoms in an object does not change)

    Weight (N)

  • Force from gravity pulling on an object

  • Weight = mg

  • English unit is a pound


Cereal boxes
Cereal Boxes “weightless” in space?

  • Metric unit of mass = kilogram

  • English unit of mass = slug

    1 lb = 4.45 N

    2.2 lb equivalent to 1 kg


Weight example 1
Weight Example 1 “weightless” in space?

What is the weight of a 60.0 kg man?


Weight example 2
Weight Example 2 “weightless” in space?

If a freshman weighs 500 N, what is her mass?


Weight example 4
Weight Example 4 “weightless” in space?

A 60.0 kg person weighs 100.0 N on the moon. What is the acceleration of gravity on the moon?

Weight = mgmoon

gmoon = Weight/m

gmoon = 100.0 N/60.0 kg = 1.67 m/s2

(Note that this is almost exactly 1/6th of earth’s gravity)


Calculate the weight of a 56 kg person: “weightless” in space?

  • On earth (549 N)

  • On the moon (g=1.7 m/s2) (95.2 N)

  • Suppose that person is taken to a different planet and has a weight of 672 N. Calculate the acceleration of gravity on that planet.

  • If the person dropped a ball from a height of 1.80 m on that planet, calculate the speed that it would hit the ground. (6.57 m/s)


The normal force
The Normal Force “weightless” in space?

Normal Force – The force exerted by a surface perpendicular to the contact

FN

FN

Fg = W

Fg = W


Normal force example 1a
Normal Force: Example 1a “weightless” in space?

A 10.0 kg present is sitting on a table. Calculate the weight and the normal force.

FN

Fg = W


Since the box is “weightless” in space?not moving SF = 0

SF = FN – W

0 = FN – 98.0 N

FN = +98.0 N


Normal force example 1b
Normal Force: Example 1b “weightless” in space?

Suppose someone leans on the box, adding an additional 40.0 N of force. Calculate the normal force.


Again, the box is “weightless” in space?not moving so SF = 0

SF = FN – W – 40.0 N

0 = FN – 98.0 N – 40.0 N

FN = +138.0 N


Normal force example 1c
Normal Force: Example 1c “weightless” in space?

Now your friend lifts up with a string (but does not lift the box off the table). Calculate the normal force.


Again, the box is “weightless” in space?not moving so SF = 0

SF = FN+40.0 N – W

0 = FN + 40.0 N – 98.0 N

FN = +58.0 N


Normal force example 1d
Normal Force: Example 1d “weightless” in space?

What happen when the person pulls upward with a force of 100 N?

SF = FN+ Fp – mg

SF = 0 +100.0N – 98N = 2.0N

ma = 2N

a = 2N/10.0 kg = 0.2 m/s2

Fp = 100.0 N

Fg = mg = 98.0 N


Example
Example “weightless” in space?

A 2.0 kg ball is thrown into the air with a force of 30.0 N.

  • Calculate the acceleration. (5.2 m/s2)

  • Calculate speed at the top of the throw, 2.00 m. (4.56 m/s)

  • Draw a free body diagram while the ball is being tossed.

  • Draw a free body diagram while the ball is above the thrower.


Free body diagrams ex 1
Free Body Diagrams: Ex. 1 “weightless” in space?

Calculate the sum of the two forces acting on the boat shown below. (53.3 N, +11.0o)


A basketball player throws a basketball through 1.2 m from rest to a speed of 1.50 m/s. The basketball has a mass of 0.56 kg and travels vertically.

  • Calculate the acceleration of the basketball. (0.938 m/s2)

  • Calculate the force the player exerted on the ball. (6.01 N)

  • Draw a free body diagram of the ball while being thrown.

  • Draw a free body diagram of the ball in the air.

  • Calculate how long it will take the ball to reach the peak height from when it leaves the player’s hands. (0.153 s)


Free Body Diagrams: Ex. 2 rest to a speed of 1.50 m/s. The basketball has a mass of 0.56 kg and travels vertically.

A hockey puck slides at constant velocity across the ice. Which of the following is the correct free-body diagram?


A person drags the box (10.0 kg) at an angle as shown below.

  • Calculate the acceleration of the box (3.46 m/s2)

  • Calculate the normal force. (78.0 N)

Fp = 40.0 N

30o

FN

mg


Here is the free body diagram

Fpx = (40N)(cos30o)= 34.6N

Fpy = (40N)(sin30o) = 20.0 N

Fp = 40.0 N

30o

FN

mg


Your lawnmower has a mass of 18.0 kg and the handle makes 60.0o angle with the ground. You push with a force of 9.00 N.

  • Calculate acceleration. (0.25 m/s2)

  • Calculate the normal force. (184 N)

  • How far will the lawnmower have travelled in 2.50 s. (78.1 cm)


You pull a child and sled (50.0 kg) with a rope at an angle of 35.0o with the horizontal. The child and sled experience an acceleration of 1.64 m/s2. in the horizontal direction.

  • Calculate the horizontal force on the sled. (82 N)

  • Calculate the Force of the pull (the hypotenuse) (100N)

  • Calculate the vertical force of the pull on the sled. (57.4 N)

  • Calculate the normal force on the sled. (433 N)


A man pushes a stroller (20.0 kg) across the floor. He pushes at an angle of 65.0o, with a horizontal acceleration of 0.50 m/s2.

  • Calculate the x component of the push. (10.0N)

  • Calculate the actual force of the push (22.8 N)

  • Calculate the y-component of the force (20.5 N)

  • Calculate the normal force on the stroller. (217 N)

  • Calculate the speed of the stroller (from rest) if the stroller accelerates for 3.0 m (1.73 m/s)


Example 3
Example 3 pushes at an angle of 65.0

A 10.0 kg box is placed next to a 5.00 kg box. The 10.0 kg box is pushed with a force of 20.0 N.

  • Calculate the acceleration of the boxes (1.33 m/s2)

  • Calculate the contact force between the two boxes. (6.67 N)

20.0 N

10.0 kg

5.00 kg


Tension
Tension pushes at an angle of 65.0

  • Flexible cord (can only pull)

  • FT

  • Neglect mass of the cord


Tension example 1
Tension: Example 1 pushes at an angle of 65.0

Two boxes are connected by a cord as shown. They are then pulled by another short cord. Find the acceleration of each box and the tension in the cord between the boxes.

12.0 kg

10.0 kg

Fp= 40.0 N


First let’s find the acceleration (consider the boxes as one mass)

SF = Fp (this is the only horizontal force)

ma = Fp

a= Fp/m = 40.0 N/(12.0 kg + 10.0 kg)

a=1.82 m/s2


To find the tension, let’s deal with each box one at a time

SF = Fp – FT

m1a= Fp – FT

FT = m1a – Fp

FT= [(10 kg)(1.82 m/s2) – 40 N

FT= -21.8 N

10.0 kg

FT

FN

Fp= 40.0 N

m1g


The second box only has a horizontal pull from the tension. time

SF = FT

m2a2 = FT

FT = (12.0 kg)(1.82 m/s2)

FT = 21.8 N

12.0 kg

FN

FT

m2g

Note that the sign of the tension varies depending on which box you consider.


A 5.00 kg box and a 12.0 kg box are connected by a cord. The 12.0 kg box is pulled horizontally with a force of 60.0 N.

  • Calculate the acceleration of both boxes. (3.5 m/s2)

  • Calculate the force of tension in the cord. (17.6 N)


  • Calculate the acceleration. (3.68 m/s The 12.0 kg box is pulled horizontally with a force of 60.0 N.2)

  • Calculate the tension in the cord (91.9 N)

    b) Calculate how far the box will move in 1.50 s. (4.13 m)

25.0 kg

15.0 kg


Atwood s machine
Atwood’s machine The 12.0 kg box is pulled horizontally with a force of 60.0 N.

Calculate the acceleration of the elevator and the tension in the cable. (Atwood’s Machine)



Ma = Mg -F counterweightT M = 1150 kg

ma = FT - mg m = 1000 kg

(a = 0.68 m/s2, FT = 10,500N)


A 10.00 kg and a 5.00 kg box are connected by a cord. The boxes are not moving.

  • Calculate the tension the cords (49N, 147 N)

  • Suppose the boxes accelerate at 0.80 m/s2 upward. Now calculate the tensions. (53 N, 159 N)


A 5.00 kg and a 2.00 kg box are connected by a cord. The top box (5.00 kg) is lifted by a second cord with an acceleration of 0.750 m/s2.

  • Calculate the tension in the bottom, connecting cord. (21.1 N)

  • Calculate the tension in the top cord. (73.85 N)


A 5.00 kg and a 2.00 kg box are connected by a cord. The top box (5.00 kg) is lifted by a second cord with a Force of 100 N.

  • Calculate the tension in the bottom, connecting cord. (28.6 N)

  • Calculate the acceleration. (4.5 m/s2)


Static friction
Static Friction top box (5.00 kg) is lifted by a second cord with a Force of 100 N.

  • Friction that opposes motion before it moves

  • coefficient of static friction = ms

  • Generally greater than kinetic friction (mk)

    Ffr = msFN


Kinetic friction
Kinetic Friction top box (5.00 kg) is lifted by a second cord with a Force of 100 N.

  • Friction that opposes motion while it moves

  • coefficient of kinetic friction (unit-less) = ms

  • Generally less than static friction

    Ffr = mkFN


Friction example 1
Friction: Example 1 top box (5.00 kg) is lifted by a second cord with a Force of 100 N.

A 10.0 kg box is on a table the coefficients of friction are ms =0.40, mk = 0.30. What is the force of friction if the force pulling on the box is:

  • 0 N

  • 10N

  • 20N

  • 38N

  • 40N

Ffr

Fp

FN

mg


First let’s calculate the maximum force from the static friction.

Fmax = msFN

Fmax = msmg

Fmax = (0.40)(10.0 kg)(9.8 m/s2)

Fmax = 39 N

(the box will not move until at least 39 N are applied)


S friction.F = Fp – Ffr

  • Since the force of the pull is zero, the box will not move:

    SF = Fp – Ffr

    0 = 0 N – Ffr (Ffr = 0 N)

    b) We are still under 39 N

    SF = Fp – Ffr

    0 = 10 N – Ffr (Ffr = 10 N)


  • Still under 39 N friction.

    SF = Fp – Ffr

    0 = 20 N – Ffr (Ffr = 20 N)

    d) Still under 39 N

    SF = Fp – Ffr

    0 = 38 N – Ffr (Ffr = 38 N)


e) Now we are over 39 N, so kinetic friction takes over: friction.

Ffr = mkFN

Ffr = (0.30)(10.0 kg)(9.8 m/s2) = 29 N

SF = Fp – Ffr

ma = 40 N – 29 N = 11N

(10.0 kg)(a) = 11 N

a= 1.1 m/s2


A 450 g wooden box is pulled at a constant speed with a force of 1.10 N. Calculate the value of mk.

(0.25)



Friction example 3
Friction: Example 3 required if the

The coefficient of kinetic friction is 0.20.

  • Calculate acceleration (1.4 m/s2)

  • Calculate the force of tension in the cord. (16.8 N)

m1=5.0 kg

m2=2.0 kg



Inclines
Inclines push or pull her?

FN

mg

q


Inclines1
Inclines push or pull her?

FN

q

mgcosq

mg

q

mgsinq


Inclines2
Inclines push or pull her?

FN

= mgcosq

Ffr

mgsinq

q


Example 1
Example 1 push or pull her?

A 75.0 kg sled (and rider) slides down a hill that has a 23.0o degree incline.

  • Calculate the force of acceleration. (287 N)

  • Calculate the value of acceleration. (3.83m/s2)

  • Calculate the speed of the sled after it has gone 5.00 m. (6.2 m/s)

  • Calculate the normal force on the sled. (677 N)


Example 2
Example 2 push or pull her?

A 20.0 kg child slides down a slide with a 45.0o incline.

  • Calculate the force of acceleration. (139 N)

  • Calculate the value of acceleration. (6.93 m/s2)

  • Calculate the speed of the child after 1.50s (10.4 m/s)

  • Calculate the normal force on the child. (139 N)


The coefficient of kinetic friction is 0.25. push or pull her?

a) Calculate the acceleration and FT.

b) Calculate how far the box will move in 0.50 s.

(2.14 m/s2, 115 N, 26.8 cm)

25.0 kg

15.0 kg


Two zombies (50.0 kg and 100.0 kg) are pushed across a floor with a mk of 0.40. The 100.0 kg zombie is pushed with a force of 750.0 N.

  • Calculate the acceleration of the zombies. (1.08 m/s2)

  • Calculate the contact force between the zombies. (250 N)

750.0 N


A duck slides down a 20.0 with a o bank. The bank is 2.5 m long and very smooth.

  • Calculate the acceleration of the duck. (3.35 m/s2)

  • Calculate the duck’s speed at the bottom of the bank. (4.1 m/s)

  • The normal force of the duck is 9.2 N. Calculate the mass of the duck. (1.00 kg)


Inclines example 3
Inclines: Example 3 with a

  • What is the acceleration of the skier if snow has a mk of 0.10

  • What is her speed after 4.0 s?


Free body diagram
Free Body Diagram with a

First we need to resolve the force of gravity into x and y components:


F with a Gy = mgcos30o

FGx = mgsin30o

The pull down the hill is:

FGx = mgsin30o

The pull up the hill is:

Ffr= mkFN

Ffr= (0.10)(mgcos30o)


S with a F = pull down – pull up

SF = mgsin30o– (0.10)(mgcos30o)

ma = mgsin30o– (0.10)(mgcos30o)

ma = mgsin30o– (0.10)(mgcos30o)

a = gsin30o– (0.10)(gcos30o)

a = 4.0 m/s2

(note that this is independent of the skier’s mass)


To find the speed after 4 seconds: with a

v = vo + at

v = 0 + (4.0 m/s2)(4.0 s) = 16 m/s


Inclines example 4
Inclines: Example 4 with a

Suppose the snow is slushy and the skier moves at a constant speed. Calculate mk

SF = pull down – pull up

ma = mgsin30o– (mk)(mgcos30o)

ma = mgsin30o– (mk)(mgcos30o)

a = gsin30o– (mk)(gcos30o)


Since the speed is constant, acceleration =0 with a

0 = gsin30o– (mk)(gcos30o)

(mk)(gcos30o) = gsin30o

mk= gsin30o= sin30o = tan30o =0.577

gcos30o cos30o


Example 31
Example 3 with a

A 1.00 kg block is placed on a 37.0o incline and does not slide down.

  • Draw a free body diagram

  • Calculate the normal force (7.83 N)

  • Calculate the force of acceleration (5.90 N)

  • Calculate the minimum value of ms. (0.753)

  • If the maximum value of ms is 0.800, calculate the angle at which it will slide? (38.7o)

    (


Example 4
Example 4 with a

A 100.0 kg filing cabinet slides down a ramp that has a coefficient of sliding friction (mk) of 0.38. The ramp has an angle of 25.0o.

  • Calculate the normal force. (888 N)

  • Calculate the acceleration of the cabinet. (0.766 m/s2)

  • If the coefficient of static friction is (0.60), calculate the angle of the ramp at which it will start slipping. (31.0o)


The acceleration of the following system is 1.00 m/s with a 2. The table is not smooth and has friction.

  • Calculate the force of tension in the cord. (44.0 N)

  • Calculate mk(0.35)

  • How long will it take the boxes to move 1.50 m? (1.73s)

10.0 kg

5.0 kg


Neglect friction for the following system. Assume the 50.0 kg block will slide down the incline.

  • Calculate the tension in the cord (199 N)

  • Calculate the acceleration of the blocks (0.157 m/s2)

50.0 kg

20.0 kg

25.0o


  • For the following diagram, assume the kg block will slide down the incline.mk = 0.18, and the ball falls toward the ground.

  • Calculate the acceleration of the system (4.86 m/s2)

  • Calculate the tension in the cord. (494 N)

50.0 kg

100.0 kg

20 o


Formula wrap up
Formula Wrap-Up kg block will slide down the incline.

SF=ma

Weight = mg

Fmax = msFN (before it moves)

Ffr = mkFN


2. 116 kg kg block will slide down the incline.

4. 1260 N

6. a) 196 N(both) b) 294 N c) 98.0 N

8. 153 N

10. 1.29 X 103 N in direction of ball’s motion

12. 1.3 X 104 N

14. 5.04 X 104 N(max) 4.47 X 104 N (min)

16. -2.5 m/s2 (down)

18. a) 7.4 m/s2 (down) b) 1176 N


24. Southwest kg block will slide down the incline.

26. a) Fbat NE, mg S b) mg S

28. b) 62.2 N c) 204 N c) 100N

30. a) 29 N b) 34 N and 68 N

32. a) 320 N b) 0.98 m/s2

34. q = 22o

38. 100 N, if mk=0 no force is required

40. a) static b) kinetic c) kinetic

42. 2.3

  • 4.1 m

    66. -21,000 N and 1.1 m


44. 4.1 m kg block will slide down the incline.

46. a) xmin = v2/msg b) 47 m c) 280 m

48. Bonnie’s (slips if angle is > 8.5o)

50. 0.32

52. a) 3.67 m/s2 b) 8.17 m/s

54. a) 1.85 m/s2 b) 0.82 m up, 1.49s

  • 101 N, mk = 0.719

    58. a = 0.098 m/s2 time = 4.3 s

    70. a) 16 m/s b) 12 m/s

  • a) 75.0 kg b) 75.0 kg c) 75.0 kg

    d) 98.0 kg e) 52.0 kg


Force table lab example
Force Table Lab Example kg block will slide down the incline.

Calculate the resultant vector (size and direction) of these three forces.


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