Transcript CHAPTER 4 The Laws of Motion

CHAPTER 4
The Laws of Motion
Newton’s First Law: An object at rest remains at rest and an
object in motion continues in motion with
constant velocity (constant speed in straight
line) unless acted on by a net external force.
“in motion”
or
“at rest” – with respect to the chosen frame of reference
“net force” – vector sum of all the external forces acting on the
object
– FNet,x and FNet,y calculated separately
Forces:
Contact Forces
*Applied Forces (push or pull)
*Normal Force (supporting force)
*Frictional Force (opposes motion)
Field Forces
*Gravitational
·Magnetic
·Electrostatic
*The typical four forces analyzed in our study of classical mechanics
Newton’s Second Law: The acceleration of an object is
directly proportional to the net
force acting on it
FNet = ma
Mass – The measurement of inertia (“inertial mass”)
Inertia – The tendency of an object to resist any attempt
to change its motion
Book Example:
1. Strike golf ball w/golf club
2. Strike bowling ball w/golf club
Which has greatest inertia?
Which has greatest mass?
Dimensional Analysis
F = ma = kg x m/s2
= newton
=N
1 newton = 1 kg · m/s2
Weight and the Gravitational Force
Mass – an amount of matter (“gravitational mass”)
“Your mass on the Moon equals your mass on Earth.”
Weight – the magnitude of the force of gravity acting on an
amount of matter
F = ma
Fg = mg
w = mg
NOTE: Your text treats weight (w) as a
scalar rather than as a vector.
Example
Your mass is 80kg. What is your weight?
w = 80kg · 9.8m/s2
w = 780 kg·m/s2
w = 780 N
Newton’s Third Law:
If two objects interact, the force
exerted on object 1 by object 2 is
equal in magnitude but opposite
in direction to the force exerted
on object 2 by object 1
Example: (Contact Force)
Book
Table
Book pushes down on table with force of 9.8.N
Table pushes up on book with force of 9.8.N
Net Force on book =9.8N – 9.8N = 0N
Hence, book does not accelerate up or down.
Example: (Field Force)
Earth
F
Moon
FEarth
Moon
Earth pulls on Moon equal to the force the
Moon pulls on Earth.
Problem Solving Strategy
Remember: We are working now with only 4 forces.
• Applied Force
Fa
• Normal Force
FN
• Frictional Force
Ff
• Gravitational Force
Fg
Draw a Sketch
FN
Fa
Ff
Fg
Determine the Magnitude of Forces in “x” and in “y” Direction
FN often equals Fg (object does not accelerate up off
surface or accelerate downward through surface)
FNet,y = FN – Fg = 0 N
FNet,x = Fa – Ff = ma
Label forces on Sketch
Ff < Fa
Solve Problem
Example 1: Sliding “Box” Problem (Horizontal Fa)
“Box” = hockey puck
= shopping cart
= tire
= dead cat
= etc.
A 55 kg shopping cart is pulled horizontally with a force of
25N. The frictional force opposing the motion is 15N. How
fast does the cart accelerate?
FN=540N
Ff=15N
Fa=25N
Fg=540N
Fa = 25N
Ff = 15N
FNet,x = 25N – 15N = ma
= 10.N = 55kg·a
a = .18m/s2
Fg = mg = 55kg · 9.8m/s2
= 540N
FN = Fg = 540N
Example 2: Sliding “Box” Problem (Pulled at an Angle)
A dead cat with a mass of 7.5kg is pulled off the road by a
passing motorist. The motorist pulls the cat by its tail which
is at an angle of 37° to the horizontal. A force of 25N is
applied. The force of friction opposing motion is 18N. How
fast does the cat accelerate?
59N= FN
18N= Ff
Fa,y =15N
Fa,x =20N
Fg =74N
m = 7.5kg
Fa = 25N Fa,x = Fa cos37 = 20.N
Fa,y = Fa sin 37 = 15N
FN + Fa,y = Fg
Ff = 18N
(up forces equal down forces)
FNet,x = Fa,x – Ff = ma
Fg = mg = 74N
20.N – 18N = 7.5kg · a
FN = 74N – 15N
a = .27m/s2
FN = 59N
Friction
Friction opposes motion.
Kinetic Friction opposes motion of a moving object.
Static Friction opposes motion of a stationary object.
Ff = FN
static = coefficient of static friction
kinetic = coefficient of kinetic friction
s > k
Why?
Static condition: peaks and
valleys of the two surfaces
overlap each other.
Kinetic Condition: surfaces
slide over each other
touching only at their peaks
s > k Ff,s > Ff,k
Applied Physics Example:
Anti-lock Brakes
Example 3: Sliding “Box” (Pulled at Angle: advanced)
A box is pulled at a 37° angle with increasingly applied
force. The box which has a mass of 15kg begins to move
when the applied force reaches 50.N. What is the coefficient
of static friction between the box and the surface?
FN
37 °
Fa,y
Fa,x
Ff
Fg
Fa = 50.N
Fa,x = Fa cos 37 = 40.N
Fa,y = Fa sin 37 = 30.N
Fg = mg = 150N
FN + Fa,y = Fg
FN = 120N
Ff,s = Fa,x
At the point where
box started to move
Ff,s = s FN = Fa,x
= s· 120N = 40N
s = .33
Forces on an Inclined Plane
y
Fgx
x

Fgy
 = 30°
Fg
Fg is always directed straight down.
We then choose a Frame of Reference where the x-axis is
parallel to the incline and the y-axis perpendicular to the
incline.
Fg,x = Fg sin
Fg,y = Fg cos
FN = Fg,y (in opposite direction)
Fa and Ff will be along our new x-axis
Example Problem (Inclined Plane)
A 25.0kg box is being pulled up a 30° incline with a force of
245N. The coefficient of kinetic friction between the box and
the surface is .567. Calculate the acceleration of the box.
y
x
Draw a Sketch
Fgx

Fgy
 = 30°
Fg
Determine the Magnitude of the forces in x and y directions
m = 25.0 kg
Fa = 245N (to right along x-axis)
2
Fg = mg = 25.0kg · 9.80m/s
FN = Fg cos = 212N
= 245N (down)
(up along y-axis)
Fg,x = Fg · sin = 245N · sin30
Ff = kFN = .567 · 212N
= 120N
= 123N (to left along x-axis)
(to left along x-axis)
Fg,y = Fg · cos = 245N · cos30
= 212N (down along y-axis) Label Forces on your sketch
Solve the Problem
Solve the Problem
FNet,x = Fa – Ff – Fg,x
= 245N – 120.N – 123N
= 2N
FNet,x = max
2N = 25.0kg · ax
ax = .08 m/s2
NOTE: The box may be moving up the incline at any
velocity. However, at the specified conditions it will
be accelerating.
Example Problem (Connected Objects – Flat Surface)
FT
Fa
Two similar objects are pulled across a horizontal surface at
constant velocity. The required Fa is 350.N. The mass of the
leading object is 125kg while the mass of the trailing object
is 55kg. The values for k are the same for each object.
Calculate k and calculate the Force of “Tension” in the
connecting rope.
NOTE: FT = Force of Tension is not a new type of force.
It is just a specific type of applied force.
• Label the forces.
• Calculate the magnitude of the forces.
• Solve the problem(s).
FN,2
m2 = 55kg
Ff,2
FN,1
FT
m1 = 125kg
Ff,1
Fg,1
Fg,2
Fg,1 = m1g = 125kg · 9.80m/s2
= 1230N (down)
Fg,2 = m2g = 55kg · 9.80m/s2
= 540N (down)
FN,1 = 1230N (up)
FN,2 = 540N (up)
Ff,1 = k · 1230N (left)
Ff,2 = k · 540N (left)
FNet,x = 0N Constant Velocity a = 0m/s2
= ma
Fa = Ff,1 + Ff,2 = k · 1230N + k · 540N
Fa = 350.N = k (1230N + 540N)
FT = k · 540N
FT = 110N
k = .20
Fa=350.N
Example Problem (Elevators)
m1
m2
Two weights are connected
across a frictionless pulley by
weightless string. Mass of object
1 is 25.0kg. The mass of object 2
is 18.0kg. Determine the
acceleration of the two objects.
Fg,1 = m1g = 25.0kg · 9.80m/s2
= 245N (down on right)
Fg,2 = m2g = 18.0kg · 9.80m/s2
= 176N (down on left)
Fg,net = 245N – 176N
2
a
=
1.60
m/s
= 69N (down on right)
m1 accelerates down
69N = ma = (m1 + m2) · a
m2 accelerates up
69N = (25.0kg + 18.0kg) · a