Transcript Slide 1

Force
10/01/2010
𝑭 = 𝒎𝒂
𝑾 = 𝒎𝒈 (Weight)
𝑻 = 𝒎𝒂 (Tension)
Friction Force
𝒇 = 𝝁𝒔 𝑭𝑵
𝒇 = 𝝁𝒌 𝑭𝑵
𝝁𝒔 ,𝝁𝒌  coefficient of static and
kinetic friction
MIDTERM on 10/06/10 7:15 to 9:15 pm
Bentley 236
 2008 midterm posted for practice.
 Help sessions Mo, Tu 6-9 pm.
 2-side hand written sheet for equation/notes.
 18 MC, 3 long problems
CAPA 4a due this Friday (10/01/10) at
11:59 PM
CAPA 4b due next Tuesday (10/05/10) at
11:59 PM
No quiz on this Friday.
Quiz 4 on 10/05/10, next week Tuesday
Clicker
A net force of 250N is exerted to the right on a large box of mass 50kg. What is the
acceleration of the box?
(1) 0.2 m/s2 to the right
(2) 0.2 m/s2 to the left
(3) 1.25 m/s2 to the right
(4) 1.25 m/s2 to the left
(5) 5.0 m/s2 to the right
(6) 5.0 m/s2 to the left
(7) 12.5 m/s2 to the right
(8) 12.5 m/s2 to the left

 FNET 250N
a

 5.0m/s2
m
50kg
Acceleration same direction as the force.
Example 1
You and a friend are sliding a large 100-kg box across the floor. Your friend pulls to
the right with a force of 250N. You push to the right with a force of 300N. The
frictional force of the floor opposes the motion with a force of 500N. What is the
acceleration of the box?
300N
100 kg
250N
500N
Connections - Gravity
Abell 1060 / Hydra
Newton’s Law of Gravitation
m1
m2
r
G = 6.67 x 10-11 Nm2/kg2 (Universal gravitational constant).
On Earth,
where
𝑭=𝑮
𝒈=
𝑴𝑬. 𝒎
𝒓𝑬𝟐
𝑴𝑬
𝑮 𝟐
𝒓𝑬
Gravitational constant (G), mass of the Earth (ME), and radius of the Earth (rE) are
constants. So we can determine the gravitational acceleration due to gravity (g) as
9.81 m/s2.
Example 2
A force of 300 N pointing towards East is applied to a block of 100 kg. Another
force of 300 N is applied towards North. If the friction is negligible, find the
acceleration of the block.
300N
100 kg
Net Force
300N
Example 3
A group is pushing a box of 100 kg. A pushes to the East with 100 N, B to the
North-East with 200 N, C to the South with 150 N, and D to the North-West with
100 N. What will be the net acceleration of the box?
B
200N
D
100N
100 kg 100N
150N
C
A
Example 4
A FedEx employee pushes a 50-kg box initially at rest on the floor to the right. 10 s
later the box is moving with a speed of 1.5 m/s. Find the net force used by the
employee.
50 kg
(1)
V = V0 + at
(2)
Dx = ½ (V0 + V) t
(3)
Dx = V0t + ½ at2
(4)
V2 = V02 + 2 a Dx
Example 5
A FedEx employee pushes a 50-kg box initially at rest on the floor to the right. 10 s
later the box is displaced 20 m from the initial position. Find the net force used by
the employee.
50 kg
(1)
V = V0 + at
(2)
Dx = ½ (V0 + V) t
(3)
Dx = V0t + ½ at2
(4)
V2 = V02 + 2 a Dx
Example 6
CLICKER!
The Moon’s gravitational acceleration is six times less than that of the Earth.
What is the weight of an astronaut on Moon whose mass is 80 kg? Gravitational
acceleration on Earth is 9.81 m/s2.
a) 262 N
b) 131 N
W=mg
c) 49 N
d) 0.02 N
= 80 x (9.81)/6 = 130.8 N
Example 7
The mass of the Moon is 7.36 E22 kg and the mass of the Earth is 5.9742E24 kg.
If the distance between the Earth and Moon is 3.84E8 m, find the gravitational force
between them.
mMoon
mEarth
r
Lab
Acceleration down an incline:
If θ = 0 (Horizontal), no acceleration (a=0)
If θ = 90 (Vertical), a=g
If in between, a = g sinθ
x = v0t + ½at2 which reduces to x = ½at2 if v0 = 0 m/s
Lab – Data inexact
Various slopes – max and min
from steepest and shallowest
reasonable lines
Draw Straight lines, not
“Connect the Dots”
Clicker
The graph shows the speed of a car as a function of time. Selected the correct
statement.
1). FB = FC
2). FA > FE = FD
3). FE > FB > FC
Larger the slope  higher acceleration (a = Dv/t)  larger force.
Normal Force
The force exerted by the surface. It is always perpendicular to the surface. (The
object must be located on a surface.)
Normal force has an equal magnitude and opposite indirection with the weight or
weight component.
Free Body Diagram
The drawing of forces acting on a body.
- weight (W) downward.
- on a surface; normal force (FN) perpendicular to the surface.
- ropes; add tensions (T).
FN
T2
T1
W
W
Equilibrium
The net force acting on a body is zero.
(The forces cancel out. Therefore, an object at rest remains at rest, and an object in
motion continues to travel with a constant velocity).
Clicker
A large crate weighs 100N. You push straight down on the top of the crate with a force of
75N. What is the force with which the floor pushes up on the crate?
(1) 25N
(2) 75N
(3) 100 N
(4) 125N
(5) 175N
FFLOOR
ΣFY = 0
FFLOOR – 100 N – 75 N = 0
FFLOOR = 100N + 75N
What if aY wasn't zero? Just plug in value.
FG
FYOU
Clicker
What is the magnitude of the upward force (normal force) of the table on the box?
(1) 100N
(2) 100N + F cosθ
(3) 100N - F cosθ
(4) 100N + F sinθ
(5) 100N - F sinθ
(6) F cosθ
(7) F sinθ
FY
+y
+x
Fx
ΣFY = 0
FN – 100 N – Fsinθ = 0
FN = 100 N + Fsinθ
Clicker
You are on an elevator which is accelerating upward.
How does the normal force (FN) compare to your weight (W)?
1.
2.
3.
normal force > weight
normal force = weight
normal force < weight
Apparent Weight
We can increase or decrease the weight by adding or subtracting additional vertical
force on the object.
(lift is stationary)
W = FN
(lift moves upward)
Wapp > FN
(lift moves downward)
Wapp < FN
Example 8
You are on an elevator which is accelerating upward at a rate of 2.00m/s2. Your mass is 80.0kg,
and the elevator has a mass of 500.kg. A cable is used to pull the elevator upward. What is the
force of the cable on the elevator?
(lift is stationary)
W = FN
(lift moves upward)
Wapp > FN
(lift moves downward)
Wapp < FN
Frictional Forces
𝒇𝒔
𝒎𝒂𝒙
= 𝝁𝒔 𝑭 𝑵
𝝁𝒔 = 𝒄𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 𝒔𝒕𝒂𝒕𝒊𝒄 𝒇𝒓𝒊𝒄𝒕𝒊𝒐𝒏
Example 9
You push a crate on the floor. The crate starts moving at 100N force. If the coefficient of fraction
is 0.15, find the mass of the crate.
Incline
y
x
FN
f
q
Fx
q
W
Multiple Masses Tied by Ropes
T2 > T1
T1= m1 a
T2= (m1 + m2) a
Clicker
You are standing still on ice (consider this to be frictionless). Your friend (mass=60kg)
pushes you (mass=80kg). Your acceleration is +1.0m/s2. What is your friend's acceleration?
(1) 0 m/s2
(4) +1.33 m/s2
(7) -1.33 m/s2
(2) +0.75 m/s2
(5) -0.75 m/s2
(3) +1.0 m/s2
(6) -1.0 m/s2
Σ Horizontal Forces on You = maYOU
FFRIEND ON YOU = (80kg)*(+1m/s) = +80 N
You exert same magnitude force in opposite direction on them.
ΣHorizontal Forces on Friend = maFRIEND
-80N = (60kg)aFRIEND
aFRIEND = - 1.33 m/s2 (opposite direction)