Transcript m/s 2 - Dallas School District
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
1. All objects remain at rest or in uniform, straight-line motion unless acted upon by an outside force. (
inertia)
2. Force = mass X acceleration 3. Every force has an equal and opposite force.
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 Law
Ball in jar example:
Direction of jar
The First Law
The First Law
Now stop pushing the jar (same as not wearing a seatbelt)
Marble keeps going Jar gets stopped
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 Law
Bee in a car example 1. Bee is in air when car starts 2. 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?
Does it take less force to move the elephant if he is “weightless” in space?
Inertial Reference Frames
• 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
Force = mass X acceleration
S
F = ma
Sum of
all
the forces acting on a body
The Second Law: Use the Force
The Second Law
Situation One: Non-moving Object
• Still has forces Force of the material of the rock Force of gravity
The Second Law
Situation Two: Moving Object S
F = ma ma = F pedalling – F air - F friction F pedalling F friction F air
The Second Law
• Unit of Force = the Newton S F=ma S
F = (kg)(m/s 2 ) 1 N = 1 kg-m/s 2
• 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
A 60.0 kg bike and rider accelerates at 0.5 m/s 2 . How much extra force did the rider’s legs have to provide?
S
F = ma = (60.0 kg)(0.5 m/s 2 )
S
F = 30 kg m/s 2 = 30 N
The Second Law: Example 2
A force of 5 Newtons can accelerate a watermelon 2.5 m/s 2 . What is the mass of the watermelon?
(No actual watermelons were harmed in the production of this example problem)
The Second Law: Example 3
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
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 v 2 = v o 2 + 2a(x-x o ) a = (v 2 a = 0 2 - v o 2 )/2(x-x o ) – (28 m/s) 2 /2(55m) = -7.1 m/s 2
S
F = ma = (1500 kg)(-7.1 m/s 2 ) = -1.1 X 10 4 N
(negative sign tells us that the force is the in the opposite direction of motion) F brakes Direction of original motion
The 3
rd
Law
“For every force, there is an equal and opposite force.”
The 3
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
Law
F forward F gases
The 3
rd
Law
Sled of bricks on Ice: • Would the sled move?
ICE
The 3
rd
Law
Why would Cyclops be in trouble?
Mass vs. Weight
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
• 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
What is the weight of a 60.0 kg man?
Weight Example 2
If a freshman weighs 500 N, what is her mass?
Weight Example 4
A 60.0 kg person weighs 100.0 N on the moon. What is the acceleration of gravity on the moon?
Weight = mg moon g moon = Weight/m g moon = 100.0 N/60.0 kg = 1.67 m/s 2 (Note that this is almost exactly 1/6 th gravity) of earth’s
Calculate the weight of a 56 kg person:
a) On earth (549 N) b) On the moon (g=1.7 m/s 2 ) (95.2 N) c) Suppose that person is taken to a different planet and has a weight of 672 N. Calculate the acceleration of gravity on that planet.
d) 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
Normal Force – The force exerted by a surface
perpendicular
to the contact
F N F N F g = W F g = W
Normal Force: Example 1a
A 10.0 kg present is sitting on a table. Calculate the weight and the normal force.
F N F g = W
Since the box is
not
moving S F = 0 S F = F N 0 = F N – W – 98.0 N F N = +98.0 N
Normal Force: Example 1b
Suppose someone leans on the box, adding an additional 40.0 N of force. Calculate the normal force.
Again, the box is
not
moving so S F = 0 S F = F N 0 = F N – W – 40.0 N – 98.0 N – 40.0 N F N = +138.0 N
Normal Force: Example 1c
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
not
moving so S F = 0 S F = F N +40.0 N – W 0 = F N + 40.0 N – 98.0 N F N = +58.0 N
Normal Force: Example 1d
What happen when the person pulls upward with a force of 100 N?
F p = 100.0 N
S F = F N + F p – mg S F = 0 +100.0N – 98N = 2.0N
ma = 2N a = 2N/10.0 kg = 0.2 m/s 2
F g = mg = 98.0 N
Example
A 2.0 kg ball is thrown into the air with a force of 30.0 N.
a) Calculate the acceleration. (5.2 m/s 2 ) b) Calculate speed at the top of the throw, 2.00 m. (4.56 m/s) c) Draw a free body diagram while the ball is being tossed.
d) Draw a free body diagram while the ball is above the thrower.
Free Body Diagrams: Ex. 1
Calculate the sum of the two forces acting on the boat shown below. (53.3 N, +11.0
o )
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.
a) Calculate the acceleration of the basketball. (0.938 m/s 2 ) b) Calculate the force the player exerted on the ball. (6.01 N) c) Draw a free body diagram of the ball while being thrown.
d) Draw a free body diagram of the ball in the air.
e) 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
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. a. Calculate the acceleration of the box (3.46 m/s 2 ) b. Calculate the normal force. (78.0 N)
F N F p = 40.0 N 30 o mg
Here is the free body diagram F px = (40N)(cos30 o )= 34.6N
F py = (40N)(sin30 o ) = 20.0 N
F p = 40.0 N 30 o F N mg
Your lawnmower has a mass of 18.0 kg and the handle makes 60.0
o angle with the ground. You push with a force of 9.00 N.
a) Calculate acceleration. (0.25 m/s 2 ) b) Calculate the normal force. (184 N) c) 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 35.0
o with the horizontal. The child and sled experience an acceleration of 1.64 m/s 2 . in the horizontal direction.
a) Calculate the horizontal force on the sled. (82 N) b) Calculate the Force of the pull (the hypotenuse) (100N) c) Calculate the vertical force of the pull on the sled. (57.4 N) d) 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 65.0
o , with a horizontal acceleration of 0.50 m/s 2.
a) Calculate the x component of the push. (10.0N) b) Calculate the actual force of the push (22.8 N) c) Calculate the y-component of the force (20.5 N) d) Calculate the normal force on the stroller. (217 N) e) Calculate the speed of the stroller (from rest) if the stroller accelerates for 3.0 m (1.73 m/s)
Example 3
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.
a) Calculate the acceleration of the boxes (1.33 m/s 2 ) b) Calculate the contact force between the two boxes. (6.67 N) 20.0 N 10.0 kg 5.00 kg
Tension
• Flexible cord (can only pull) • F T • Neglect mass of the cord
Tension: Example 1
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 F p = 40.0 N
First let’s find the acceleration (consider the boxes as one mass) S F = F p (this is the only horizontal force) ma = F p a= F p /m = 40.0 N/(12.0 kg + 10.0 kg)
a=1.82 m/s 2
F T
To find the tension, let’s deal with each box one at a time S F = F p – F T
10.0 kg
m 1 a = F p – F T
F N F p = 40.0 N
F T = m 1 a – F p F T = [(10 kg)(1.82 m/s 2 ) – 40 N
m 1 g
F T = -21.8 N
The second box only has a horizontal pull from the tension.
S F = F T
12.0 kg
m 2 a 2 = F T
F N F T
F T = (12.0 kg)(1.82 m/s 2 ) F T = 21.8 N
m 2 g
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.
a) Calculate the acceleration of both boxes. (3.5 m/s 2 ) b) Calculate the force of tension in the cord. (17.6 N)
a) Calculate the acceleration. (3.68 m/s 2 ) b) 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
Calculate the acceleration of the elevator and the tension in the cable. (Atwood’s Machine)
Draw free-body diagrams for both the elevator and counterweight
Ma = Mg -F T ma = F T - mg M = 1150 kg m = 1000 kg (a = 0.68 m/s 2 , F T = 10,500N)
A 10.00 kg and a 5.00 kg box are connected by a cord. The boxes are not moving.
a. Calculate the tension the cords (49N, 147 N) b. 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/s 2 .
a. Calculate the tension in the bottom, connecting cord. (21.1 N) b. 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.
a. Calculate the tension in the bottom, connecting cord. (28.6 N) b. Calculate the acceleration. (4.5 m/s 2 )
Static Friction
• Friction that opposes motion
before
moves it • coefficient of static friction = m s • Generally greater than kinetic friction ( m k ) F fr = m s F N
Kinetic Friction
• Friction that opposes motion
while
moves it • coefficient of kinetic friction (unit-less) = m s • Generally less than static friction F fr = m k F N
Friction: Example 1
A 10.0 kg box is on a table the coefficients of friction are m s =0.40, m k = 0.30. What is the force of friction if the force pulling on the box is: a) 0 N b) 10N c) 20N d) 38N e) 40N
F fr F N mg F p
First let’s calculate the maximum force from the static friction.
F max = m s F N F max = F max = m s mg (0.40)(10.0 kg)(9.8 m/s 2 ) F max = 39 N (the box will not move until at least 39 N are applied)
S F = F p – F fr a) Since the force of the pull is zero, the box will not move: S F = F p – F fr 0 = 0 N – F fr (F fr = 0 N) b) We are still under 39 N S F = F p – F fr 0 = 10 N – F fr (F fr = 10 N)
c) Still under 39 N S F = F p – F fr 0 = 20 N – F fr (F fr = 20 N) d) Still under 39 N S F = F p – F fr 0 = 38 N – F fr (F fr = 38 N)
e) Now we are over 39 N, so kinetic friction takes over: F F fr fr = m k F N = (0.30)(10.0 kg)(9.8 m/s S F = F p – F fr 2 ) = 29 N ma = 40 N – 29 N = 11N (10.0 kg)(a) = 11 N a= 1.1 m/s 2
A 450 g wooden box is pulled at a constant speed with a force of 1.10 N. Calculate the value of m
k
.
(0.25)
A 2.50 kg sled is pulled at a constant speed. What force is required if the m
k
is 0.11 on ice?
Friction: Example 3
The coefficient of kinetic friction is 0.20. a) Calculate acceleration (1.4 m/s 2 ) b) Calculate the force of tension in the cord. (16.8 N) m 1 =5.0 kg m 2 =2.0 kg
Your little sister wants a ride on her sled. Should you push or pull her?
q
Inclines
F N mg
q
Inclines
F N mg q mgcos q mgsin q
q
Inclines
F N = mgcos q F fr mgsin q
Example 1
A 75.0 kg sled (and rider) slides down a hill that has a 23.0
o degree incline.
a) Calculate the force of acceleration. (287 N) b) Calculate the value of acceleration. (3.83m/s 2 ) c) Calculate the speed of the sled after it has gone 5.00 m. (6.2 m/s) d) Calculate the normal force on the sled. (677 N)
Example 2
A 20.0 kg child slides down a slide with a 45.0
o incline.
a) Calculate the force of acceleration. (139 N) b) Calculate the value of acceleration. (6.93 m/s 2 ) c) Calculate the speed of the child after 1.50s (10.4 m/s) d) Calculate the normal force on the child. (139 N)
The coefficient of kinetic friction is 0.25.
a) Calculate the acceleration and F T .
b) Calculate how far the box will move in 0.50 s.
25.0 kg 15.0 kg (2.14 m/s2, 115 N, 26.8 cm)
Two zombies (50.0 kg and 100.0 kg) are pushed across a floor with a m k of 0.40. The 100.0 kg zombie is pushed with a force of 750.0 N.
a) Calculate the acceleration of the zombies. (1.08 m/s 2 ) b) Calculate the contact force between the zombies. (250 N) 750.0 N
A duck slides down a 20.0
o bank. The bank is 2.5 m long and very smooth.
a) Calculate the acceleration of the duck. (3.35 m/s 2 ) b) Calculate the duck’s speed at the bottom of the bank. (4.1 m/s) c) The normal force of the duck is 9.2 N. Calculate the mass of the duck. (1.00 kg)
Inclines: Example 3
a) What is the acceleration of the skier if snow has a m k of 0.10
b) What is her speed after 4.0 s?
Free Body Diagram
First we need to resolve the force of gravity into x and y components:
F Gy = mgcos30 o F Gx = mgsin30 o The pull down the hill is:
F Gx = mgsin30 o
The pull up the hill is: F fr = m k F N
F fr = (0.10)(mgcos30 o )
S F = pull down – pull up S F = mgsin30 o – (0.10)(mgcos30 o ) ma = mgsin30 o – (0.10)(mgcos30 o ) ma = mgsin30 o – (0.10)(mgcos30 o ) a = gsin30 o – (0.10)(gcos30 o ) a = 4.0 m/s 2 (note that this is independent of the skier’s mass)
To find the speed after 4 seconds: v = v o + at v = 0 + (4.0 m/s 2 )(4.0 s) = 16 m/s
Inclines: Example 4
Suppose the snow is slushy and the skier moves at a constant speed. Calculate m k S F = pull down – pull up ma = mgsin30 o – ( m k )(mgcos30 o ) ma = mgsin30 o – ( m k )(mgcos30 o ) a = gsin30 o – ( m k )(gcos30 o )
Since the speed is constant, acceleration =0 0 = gsin30 o – ( m k )(gcos30 o ) ( m k )(gcos30 o ) = gsin30 o m k = gsin30 o gcos30 o = sin30 cos30 o o = tan30 o =0.577
Example 3
A 1.00 kg block is placed on a 37.0
o incline and does not slide down. a) Draw a free body diagram b) Calculate the normal force (7.83 N) c) Calculate the force of acceleration (5.90 N) d) Calculate the minimum value of m
s
. (0.753) e) If the maximum value of m
s
is 0.800, calculate the angle at which it will slide? (38.7
o )
Example 4
A 100.0 kg filing cabinet slides down a ramp that has a coefficient of sliding friction ( m
k
) of 0.38. The ramp has an angle of 25.0
o .
a) Calculate the normal force. (888 N) b) Calculate the acceleration of the cabinet. (0.766 m/s 2 ) c) If the coefficient of static friction is (0.60), calculate the angle of the ramp at which it will start slipping. (31.0
o )
The acceleration of the following system is 1.00 m/s 2 . The table is not smooth and has friction.
a) Calculate the force of tension in the cord. (44.0 N) b) Calculate m
k
(0.35) c) 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.
a. Calculate the tension in the cord (199 N) b. Calculate the acceleration of the blocks (0.157 m/s 2 ) 50.0 kg 20.0 kg 25.0
o
For the following diagram, assume the m
k
and the ball falls toward the ground. = 0.18, a) Calculate the acceleration of the system (4.86 m/s 2 ) b) Calculate the tension in the cord. (494 N) 50.0 kg 100.0 kg 20 o
Formula Wrap-Up
S
F=ma Weight = mg F max F fr =
m
s F N =
m
k F N
(before it moves)
2. 116 kg 4. 1260 N 6. a) 196 N(both) b) 294 N c) 98.0 N 8. 153 N 10. 1.29 X 10 3 N in direction of ball’s motion 12. 1.3 X 10 4 N 14. 5.04 X 10 4 N(max) 4.47 X 10 4 N (min) 16. -2.5 m/s 2 (down) 18. a) 7.4 m/s 2 (down) b) 1176 N
24. Southwest 26. a) F bat 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/s 2 34. q = 22 o 38. 100 N, if m k =0 no force is required 40. a) static b) kinetic c) kinetic 42. 2.3
44.4.1 m 66. -21,000 N and 1.1 m
44. 4.1 m 46. a) x min = v 2 / m s g b) 47 m c) 280 m 48. Bonnie’s (slips if angle is > 8.5
o ) 50. 0.32
52. a) 3.67 m/s 2 b) 8.17 m/s 54. a) 1.85 m/s 2 56. 101 N, m k b) 0.82 m up, 1.49s
= 0.719
58. a = 0.098 m/s 2 time = 4.3 s 70. a) 16 m/s 74.a) 75.0 kg d) 98.0 kg b) 12 m/s b) 75.0 kg e) 52.0 kg c) 75.0 kg
Force Table Lab Example
Calculate the resultant vector (size and direction) of these three forces.