Transcript Newton`s Second Law

Newton’s Second Law
Chapter 6
The Second Law
Force = mass X acceleration
SF = ma
Sum of all the forces acting on a body
Vector quantity
SF = 0
or
SF = ma
-Still object
-Accelerating object
-Obj. at constant velocity
The Second Law
Situation One:
Non-moving Object
• Still has forces
Force of
the
material
of the
rock
Force of
gravity
http://alfa.ist.utl.pt/~vguerra/Other/Rodin/thinker.jpg
The Second Law
Situation Two: Moving Object: Constant Velocity
SF = 0
Fpedalling = Fair + Ffriction
Fair
Fpedalling
Ffriction
http://2003.tour-de-france.cz/images/foto/05-07-2003/armstrong4.jpg
The Second Law
Situation Two: Moving Object: Accelerating
SF = ma
ma = Fpedalling – Fair - Ffriction
Fair
Fpedalling
Ffriction
http://2003.tour-de-france.cz/images/foto/05-07-2003/armstrong4.jpg
The Second Law
Unit of Force = the Newton
SF=ma
SF = (kg)(m/s2)
1 N = 1 kg-m/s2  (MKS)
1 Newton can accelerate a 1 kg object from rest
to 1 m/s in 1 s.
Equilibrium
1. No motion
2. Constant velocity
BOTH INDICATE NO ACCELERATION
SF=0
Three ropes are tied together for a wacky tug-ofwar. One person pulls west with 100 N of
force, another south with 200 N of force.
Calculate the magnitude and direction of the
third force.
?
100 N
200 N
A car with weight 15,000 N is being towed up a
20o slope (smooth) at a constant velocity. The
tow rope is rated at 6000 N. Will it break?
Accelerating Systems
• SF=ma
• Must add up all forces on the object
What Force is needed to accelerate a 5 kg
bowling ball from 0 to 20 m/s over a time
period of 2 seconds?
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
v2 = vo2 + 2a(x-xo)
a = (v2 - vo2)/2(x-xo)
a = 02 – (28 m/s)2/2(55m) = -7.1 m/s2
A 1500 kg car is pulled by a tow truck. The
tension in the rope is 2500 N and the 200 N
frictional force opposes the motion. The car
starts from rest.
a. Calculate the net force on the car
b. Calculate the car’s speed after 5.0 s
A 500.0 gram model rocket (weight = 4.90 N) is
launched straight up from rest by an engine
that burns for 5 seconds at 20.0 N.
a. Calculate the net force on the rocket
b. Calculate the acceleration of the rocket
c. Calculate the height and velocity of the
rocket after 5 s
d. Calculate the maximum height of the rocket
even after the engine has burned out.
Calculate the sum of the two forces acting on
the boat shown below. (53.3 N, +11.0o)
Mass vs. Weight
Mass
– The amount of matter in an object/INTRINSIC
PROPERTY
– Independent of gravity
– Measured in kilograms
Weight
– Force that results from gravity pulling on an object
– Weight = mg (g = 9.8 m/s2)
Mass vs. Weight
• Weight = mg is really a re-write of F=ma.
– Weight is a force
– g is the acceleration (a) of gravity
• Metric unit of weight is a Newton
• English unit is a pound
A 60.0 kg person weighs 1554 N on Jupiter. What is
the acceleration of gravity on Jupiter?
Elevator at Constant Velocity
a= 0
SF = FN – mg
ma = FN – mg
0 = FN – mg
FN = mg
Suppose Chewbacca has a mass
of 102 kg:
FN = mg = (102kg)(9.8m/s2)
FN = 1000 N
FN
mg
a is zero
Elevator Accelerating Upward
a = 4.9 m/s2
SF = FN – mg
ma = FN – mg
FN = ma + mg
FN = m(a + g)
FN=(102kg)(4.9m/s2+9.8 m/s2)
FN = 1500 N
FN
mg
a is upward
Elevator Accelerating Downward
a = -4.9 m/s2
SF = FN – mg
ma = FN – mg
FN = ma + mg
FN = m(a + g)
FN=(102kg)(-4.9m/s2+9.8 m/s2)
FN = 500 N
FN
mg
a is down
At what acceleration will he feel weightless?
FN = 0
SF = FN – mg
ma = FN – mg
ma = 0 – mg
ma = -mg
a = -9.8 m/s2
Apparent weightlessness
occurs if a > g
FN
mg
A 10.o kg present is sitting on a table. Calculate
the weight and the normal force.
FN
Fg = W
Suppose someone leans on
the box, adding an
additional 40.0 N of force.
Calculate the normal force.
Now your friend lifts up with a
string (but does not lift the
box off the table). Calculate
the normal force.
What happen when the person pulls upward
with a force of 100 N?
SF = FN+ Fp – mg
SF = 0 +100.0N – 98N = 2.0N
Fp = 100.0 N
ma = 2N
a = 2N/10.0 kg = 0.2 m/s2
Fg = mg = 98.0 N
Free Body Diagrams: Ex. 3
A person pulls on the box (10.0 kg) at an angle
as shown below. Calculate the acceleration of
the box and the normal force. (78.0 N)
Fp = 40.0 N
30o
FN
mg
Friction
• Always opposes the direction of motion.
• Proportional to the Normal Force (more
massive objects have more friction)
FN
Ffr
FN
mg
Fa
mg
Friction
Static -opposes motion before it moves (ms)
• Generally greater than kinetic friction
• Fmax = Force needed to get an objct moving
Fmax = msFN
Kinetic - opposes motion while it moves (mk)
• Generally less than static friction
Ffr = mkFN
Friction and Rolling Wheels
Rolling uses static friction
– A new part of the wheel/tire is coming in contact
with the road every instant
B
A
Braking uses kinetic friction
Point A gets
drug across the
surface
A
A 50.0 kg wooden box is pushed across a
wooden floor (mk=0.20) at a steady speed of
2.0 m/s.
a. How much force does she exert? (98 N)
b. If she stops pushing, calculate the
acceleration. (-1.96 m/s2)
c. Calculate how far the box slides until it stops.
(1.00 m)
A 100 kg box is on the back of a truck (ms = 0.40).
The box is 50 cm X 50 cm X 50 cm.
a. Calculate the maximum acceleration of the truck
before the box starts to slip.
Your little sister wants a ride on her sled. Should
you push or pull her?
Inclines
What trigonometric function does this resemble?
Inclines
FN
mg
q
Inclines
FN
q mgcosq
q
mg
mgsinq
Inclines
FN
Ffr
mgsinq
mgcosq
q
A 50.0 kg file cabinet is in the back of a dump
truck (ms = 0.800).
a. Calculate the magnitude of the static friction on
the cabinet when the bed of the truck is tilted
at 20.0o (170 N)
b. Calculate the angle at which the cabinet will
start to slide. (39o)
Given the following drawing:
a. Calculate the acceleration of the skier. (snow
has a mk of 0.10) (4.0 m/s2)
b. Calculate her speed after 4.0 s? (16 m/s)
First we need to
resolve the force of
gravity into x and y
components:
FGy = mgcos30o
FGx = mgsin30o
The pull down the hill is:
FGx = mgsin30o
The pull up the hill is:
Ffr= mkFN
Ffr= (0.10)(mgcos30o)
SF = pull down – pull up
SF = mgsin30o– (0.10)(mgcos30o)
ma = mgsin30o– (0.10)(mgcos30o)
ma = mgsin30o– (0.10)(mgcos30o)
a = gsin30o– (0.10)(gcos30o)
a = 4.0 m/s2
(note that this is independent of the skier’s
mass)
To find the speed after 4 seconds:
v = vo + at
v = 0 + (4.0 m/s2)(4.0 s) = 16 m/s
Suppose the snow is slushy and the skier moves
at a constant speed. Calculate mk
SF = pull down – pull up
ma = mgsin30o– (mk)(mgcos30o)
ma = mgsin30o– (mk)(mgcos30o)
a = gsin30o– (mk)(gcos30o)
Since the speed is constant, acceleration =0
0 = gsin30o– (mk)(gcos30o)
(mk)(gcos30o) = gsin30o
mk= gsin30o
gcos30o
=
sin30o = tan30o =0.577
cos30o
Drag
D ≈ ¼Av2
D = drag force
A = Area
V = velocity
Fails for
• Very small particles (dust)
• Very fast (airplanes)
• Water and dense fluids
Finding Acceleration with Drag
Derive the formula for the acceleration of a
freefalling object including the drag force.
Terminal Speed
a. Find the formula for terminal speed (a=0) for
a freefalling body
b. Calculate the terminal velocity of a person
who is 1.8 m tall, 0.40 m wide, and 75 kg.
(64 m/s)
A 1500 kg car is travelling at 30 m/s when the
driver slams on the brakes (mk = 0.800).
Calculate the stopping distance:
a. On a level road. (57.0 m)
b. Up a 10.0o incline (48.0 m)
c. Down a 10.0o incline (75.0 m)
A dogsled has a mass of 200 kg. The sled reaches
cruising speed, 5.0 m/s in 15 m. Two ropes are
attached to the sled at 10.0o, one on each side
connected to the dogs. (mk = 0.060)
a. Calculate the acceleration of the sled. (0.833
m/s)
b. Calculate T1 and T2 during the acceleration
period. (140 N)
Formula Wrap-Up
SF=ma
Weight = mg (g = 9.8 m/s2)
Fmax = msFN
Ffr = mkFN