Download Force II PPT

1. What is the weight of a 15 kg rock?
147 N
2. A skateboard (mass 12 kg ) accelerates from rest to
3.8 m/s over a distance of 9.4m. What is the
skateboard ’s rate of acceleration?
0.768 m/s2
3. What force was applied to the skateboard in to
achieve this rate of acceleration?
9.22 N
4. An unbalanced force of 25 N is applied to a 12.5kg mass. What should be the acceleration
experienced by the mass?
2 m/s2
11/13 Forces Assignment Parts 2 & 3 due Friday
• NOW: Pick up Force Notes II. ON Force ntoes one, answer
example J.
• Last Friday: A quiz over Fun with Forces & Fun with Weight.
• Monday: finished lab.
• Tuesday: reviewed lab, turned it in and took a quiz.
• If you were absent for any of these you need to make them up today
after school or tomorrow morning.
• PM Test corrections & retakes Mon-Thur this week. You must
show completed PM notes to be eligible for corrections &
retakes. Retakes pm only. I have pm duty until 2:55
• There is an issue with the website. I will fix it today if possible
11/14
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Have your HW out (Force, Friction, & Statics Part 2 and 3A)
You will turn this in on Thursday next week, but may use it on your Quiz today.
Quiz today. It is over anything in Forces that we covered through Wednesday (ie no
friction).
Hints: remember:
– mass must be in kg, distance in m
– Weight is a Force (mag) The quiz is keyed to 9.8 m/s2
– When looking at horizontal motion, use horizontal acceleration
– If velocity is constant, forces are balanced
NOW: On Force Notes I, answer example K.
You will have a quiz Monday over Newton’s Laws. No calculations
Assignment 3B due Tuesday, Assignment 3c due Wednesday
Forces Test Thursday 11/20
Thursday: We began friction and got up to example N. We will continue here today.
There is an issue with the website. I am still trying to resolve it.
11/17
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HW by tomorrow have Force, Friction, & Statics Part 2, 3A&B completed. Part C by
Wednesday.
We had a Quiz Friday. It was over anything in Forces that we covered through Wednesday
(ie no friction). If you were absent you may make it up Today after school or tomorrow am
or pm.
Friday we worked through example K and N.
Have Notes II out
You will have a quiz Wednesday over Newton’s Laws at the end of the period. No
calculations
Forces Test Thursday 11/20
Thursday: We began friction and got up to example N. We will continue here today.
Problem with the website has been resolved.
11/18 Pick up notes III if you did not get them
Monday
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HW by tomorrow have Force, Friction, & Statics Part 2, 3A&B completed. Part C by
Wednesday.
We had a Quiz Friday. It was over anything in Forces that we covered through Wednesday
(ie no friction). If you were absent you may make it up Today after school
Friday we worked through example K and N. Yesterday we worked through Ex P
(some classes we still need to finish it)
Have Notes II out and Notes III
You will have a quiz Wednesday over Newton’s Laws at the end of the period. No
calculations
I will not be available after school Wednesday due to a faculty meeting
Forces Test Thursday 11/20
Ex H The brakes of a 1000-kg car exert 3000 N.
a. How long will it take the car to come to a stop from a velocity of
30 m/s?
m = 1000 kg
F  3000
F = -3000 N
2
a
=
3
m/s

vi = 30 m/s
m 1000
vf = 0 m/s
v f  vi
0  30
= 10 s

t
3
a
b. How far will the car travel during this time?
d = vit+ .5at2
= 30(10)+ .5 (-3)(10)2
= 150 m
Ex I A net horizontal force of 4000 N is applied to a car at rest
whose weight is 10,000 N. What will the car's speed be after 8 s?
FA = 4000 N
Fw = 10,000 N
t = 8s
Fw 10000
m

ag
9.8 = 1020.4 kg
F
4000
a
2

=
3.92
m/s
m 1020.4
vf = vi + at
= 0 +3.92(8)
= 31.36 m/s
Forces Part II
Friction
When surfaces are pressed
together, we can identify
four forces
• Friction Force: FF
• Force opposing motion
• Measured in Newtons (N)
• Applied Force: FA
• The push or pull applied to the object
• Measured in Newtons (N)
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Fw Force of Weight or Gravity
(Mass in kg) (Acceleration due to Gravity)
Kg x 9.8 m/s2
Measured in Newtons (N)
• Normal Force: FN
• Usually a 3rd Law reaction to gravity, that is
equal and opposite of Force of Weight (Fw)
• Perpendicular to the surface.
• Measured in Newtons (N)
• FN is NOT FNET
• Don’t confuse them just because they begin
with N!
• A 50kg object is moving
horizontally at a constant velocity.
Is there acceleration? Is there a net
force?
FN
FA
FF
FW
Ex J No Friction
• Fnet = ma
A 50 kg object experiences an applied force of 400 N,
what is the Fnet? (Disregard friction) What is the
acceleration?:
• Fnet = FA
• Fnet = Net Force, results in acceleration
• a = Fnet /m = (400N )/ (50 kg) = 8 m/s2
Example J
FN
FA
FW
Friction
• Is a force that opposes motion.
Ex K Friction and
Constant velocity. Is there a net
force?
• Fnet = 0
• FA+ -FF = ma or FA + - FF = Fnet
A 50 kg object moves at a constant velocity when acted
upon by an applied force of 400 N. What is the FF?
What is the Fnet? What is the acceleration?:
• FF = FA
• Fnet = FA + - FF = 0 N
• If Fnet = 0 then acceleration = ?
• 0
Example K
Friction and constant velocity
FN
FF
FA
FW
How does friction affect net force?
• Fnet = ma
• ma = FA+ FF
Where:
• Fnet = Net Force
• FF = Friction Force
• FA = Applied Force
• You are actually subtracting FF from FA, since Ff is
in the opposite direction
• Friction will reduce the net force
• Do all surfaces provide the same
amount of friction? How is this
described?
FN
FF
FA
FW
Friction
What does it depend on?
• Depends on the surface of the
materials.
• Depends on how tightly the surfaces
are pressed together.
• FF = Force of Friction
Coefficient of Friction
• The coefficient of friction is a measure of how
difficult it is to slide a material of one kind over
another; the coefficient of friction applies to a pair
of materials, and not simply to one object by itself
• Coefficient of Friction Reference Table Engineer's Handbook
When Surfaces are Pressed
Together
• Coefficent of Friction µ can be calculated
• It is a ratio of FF and FN
FF

FN
FF    FN
Coefficient of friction
• What is it equal to?
• What is the unit for Coefficient of friction?
• If Coefficient of friction is small, what does
that mean about the FF?
• If Coefficient of friction is large, what does
that mean about the FF?
• The higher the coefficient of friction, the
more difficult to slide
What forces cause the swimmer to
move forward?
• Her push on the wall?
Or
• The wall’s push on her?
I call these FAWN problems
Example L
• A 400 N force is applied
horizontally to a 50kg object.
Calculate the acceleration of the
object if  = 0.3.
FN
FF
FA
Fg
Example Cont’d.
• FA =
• Fg =
Fg =
• FN =
• FF =
400 N
m·g (50kg)(9.8m/s2)
490 N
490 N also
(0.3)(490N) = 147 N
Example cont’d.
FN = 490N
FF = 147N
FA = 400N
Fg = 490N
To solve for Acceleration must
calculate Net Force
FNET = ma = FA+ -FF
• FNET = ma= 400N – 147N
• ma = 253N
Now use Net Force and mass of
object in F=ma formula
a = F/m
a = 253N/50kg
a =5.06 m/s2
Ex M
Putting it all together
A 50 kg object accelerates horizontally at 0.2 m/s2 from
rest for 5 seconds. If the coefficient of friction is 0.01,
what is the Fnet? What is the FF ? What is the FA ?How
far does it travel in 5 seconds?:
• Fnet = (50 kg) (0.2 m/s2) = 10 N
• F F = μ FN
• FN = (50kg)(9.8 m/s2) = 490 N
• FF = (0.01) (490 N) =4.9N
• FA = Fnet – -FF
• FA = 10 N + 4.9 N = 14.9 N
How does friction affect net force?
Putting it all together
A 50 kg object accelerates horizontally at 0.2 m/s2 from
rest for 5 seconds. If the coefficient of friction is 0.01,
what is the Fnet? What is the FF ? What is the FA ?How
far does it travel in 5 seconds?:
• Fnet = (50 kg) (0.2 m/s2) = 10 N
• Set up DVVAT
• d=?
• vi = 0 m/s
• vf = ?
• a =0.2 m/s2
• d = (.5)(.2)(5)2
• d = 2.5m
As the coefficient of friction
decreases and the object and the
applied force remain the same,
What happens to:
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FA
FW
FN
FNET
TENSION aka FT
• is the magnitude of the pulling force exerted by a string,
cable, chain, or similar object on another object.
• It is the opposite of compression. It is a “response force”
• That is to say, if one pulls on the rope, the rope fights back
by resisting being stretched
• Ropes, strings, and cables can only pull. They cannot push
because they bend.
• is measured in newtons
• is always measured parallel to the string on which it
applies.
• What does the rope provide?
• A lift (vertical force) and a pull (horizontal force)
• If there was no angle, would there be any vertical
force?
• No
• If the angle was at 90°, how would that affect the
force components?
• Force would only be in the vertical plane
• How would you calculate the horizontal and vertical
force components if the angle of the rope with the
floor was 57° and the Force of tension (FA) in the
rope was 400 N?
Ex N. This crate is be pulled with a rope across a friction
based horizontal surface at a constant velocity. The rope
exerts a tension of 400 N at an angle of 57°. What is the
coefficient of friction?
50 kg
57°
A box is pulled into motion with a rope across a horizontal surface.
The rope makes an angle of 57° to the floor. The Force of tension
(FA) in the rope is 400N
FAY
400 N
50 kg
A. Determine FAX or FHoriz
= cos (57) (400N)
= 217.86 N
57º
FAX
B. Determine FAY or FVert
= sin (57) (400N)
= 335.47 N
I work in a circle
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Determine FW
Determine FAX
Determine FAY
Determine FN
The FAY supplies part of the upward force. The total
upward force is FAY + FN and together these are equal but
opposite the FW. FN = FW – FAY
• Determine FF
• Determine coefficient of friction
FAY 335.47 N
400 N
50 kg
57°
FAX 217.86 N
Determine Fw
= (50 kg)(9.8 m/s2)
= 490 N
Determine FN
= FW- FAY
= 490 N – 335.47 N
= 154.53 N
FAY 335.47 N
400 N
50 kg
57°
FAX 217.86 N
Determine FF
FF= Fax (Constant Velocity)
FF= 217.86N
Determine µ= FF/ FFN
= 217.86N/ 154.53 N
= 1.41
1. Cameron has to clean up his room. He places all of his
dirty clothes on a large sheet of cardboard and drags them
to the laundry room His arms make an angle of 37° with
the floor and he pulls with a force of 52.5N. If he pulls at
a constant velocity, what is the force of friction?
2. Mehtab has a messy room too! She piles all her stuffed
animals into a box. The box has a mass of 128 kg. She
pulls into motion by a rope across the carpet. (μ = 0.264)
with a force of 512 N at 35.9° above the horizontal. How
far does the crate travel in 8 seconds?
Practice
A sled with a weight of 80 N is dragged with
a rope across the ground at a constant
velocity. The Force Tension on the rope is
100N at 40º with the horizontal.
What are the components?
What is the normal force?
What is the force of friction?
What is the coefficient of friction between the
crate and the surface?
Practice
A sled with a weight of 80 N is dragged with
a rope across the ground at a constant
velocity. The Force Tension on the rope is
100N at 40º with the horizontal.
What are the components?
What is the normal force?
What is the force of friction?
What is the coefficient of friction between the
crate and the surface?
The End