Transcript Force and Motion

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
Ex J 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
1000
m
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
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)
•
•
•
•
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 L No Friction
• Fnet = ma
A 50 kg object experiences an applied force of 400 N,
what is the Fnet? What is the acceleration?:
• Fnet = FA
• Fnet = Net Force, results in acceleration
• a = Fnet /m = (400N )/ (50 kg) = 8 m/s2
Example L
FN
FA
FW
Friction
• Is a force that opposes motion.
Ex M 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 M
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
I call these FAWN problems
Example N
• A 400 N force is applied 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 O
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:
•
•
•
•
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 P. 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 of57°. 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
•
•
•
•
•
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
The End