Download Newton`s Second Law

6/3/2013
The Second Law
Newton’s Second Law
Force = mass X acceleration
SF = ma
Sum of all the forces acting on a body
Vector quantity
Chapter 6
SF = 0
or
SF = ma
-Still object
-Accelerating object
-Obj. at constant velocity
The Second Law
The Second Law
Situation One:
Non-moving Object
• Still has forces
• Equilibrium
Situation Two: Moving Object: Constant Velocity
Force of
the
material
of the
rock
SF = 0
Fpedalling = Fair + Ffriction
Force of
gravity
Fpedalling
Fair
Ffriction
http://alfa.ist.utl.pt/~vguerra/Other/Rodin/thinker.jpg
The Second Law
The Second Law
Situation Two: Moving Object: Accelerating
Unit of Force = the Newton
SF=ma
SF = (kg)(m/s2)
1 N = 1 kg-m/s2  (MKS)
SF = ma
ma = Fpedalling – Fair - Ffriction
Fair
http://2003.tour-de-france.cz/images/foto/05-07-2003/armstrong4.jpg
Fpedalling
1 Newton can accelerate a 1 kg object from rest
to 1 m/s in 1 s.
Ffriction
http://2003.tour-de-france.cz/images/foto/05-07-2003/armstrong4.jpg
1
6/3/2013
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.
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?
?
100 N
200 N
Accelerating Systems
What Force is needed to accelerate a 5 kg
bowling ball from 0 to 20 m/s over a time
period of 2 seconds?
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
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 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.
2
6/3/2013
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)
– Metric: Newton
– English: 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
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
mg
a is upward
mg
a is zero
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
FN
FN
mg
a is down
3
6/3/2013
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
A 10.o kg present is sitting on a table. Calculate
the weight and the normal force.
FN
FN
Fg = W
mg
Now your friend lifts up with a
string (but does not lift the
box off the table). Calculate
the normal force.
Suppose someone leans on
the box, adding an
additional 40.0 N of force.
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
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)
ma = 2N
a = 2N/10.0 kg = 0.2 m/s2
Fp = 40.0 N
30o
FN
Fg = mg = 98.0 N
mg
4
6/3/2013
Friction
Friction
• Always opposes the direction of motion.
• Proportional to the Normal Force (more
massive objects have more friction)
FN
Ffr
FN
Fa
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
mg
mg
Rolling Friction
Much lower than fs and fk
Braking uses kinetic friction
– A new part of the wheel/tire is coming in contact
with the road every instant
Point A gets
drug across the
surface
B
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
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.
5
6/3/2013
Your little sister wants a ride on her sled. Should
you push or pull her?
Inclines
Inclines
FN
mg
q
What trigonometric function does this resemble?
Inclines
Inclines
FN
FN
Ffr
mgsinq
q mgcosq
q
mg
mgcosq
q
mgsinq
6
6/3/2013
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)
First we need to
resolve the force of
gravity into x and y
components:
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)
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
To find the speed after 4 seconds:
v = vo + at
v = 0 + (4.0 m/s2)(4.0 s) = 16 m/s
(note that this is independent of the skier’s
mass)
7
6/3/2013
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)
A dog sled and rider have a mass of 200.0 kg. It
takes the dogs 15.0 m to reach their cruising
speed of 5.00 m/s. The ropes are connected
upwards to the two dogs at 10.0o. Calculate
the tension in the ropes at the start of the
race. (mk = 0.06)
Since the speed is constant, acceleration =0
0 = gsin30o– (mk)(gcos30o)
(mk)(gcos30o) = gsin30o
mk= gsin30o
gcos30o
=
sin30o = tan30o =0.577
cos30o
Drag: Low Speeds
D ≈ -bv
D = drag force
b = constant
v = velocity
(works well for liquids)
Derive the formula for the terminal speed of a
body falling through a fluid. Remember that
the sum of all forces will be zero.
Setup, but do not solve, a differential equation
for the velocity of the particle before reaching
terminal velocity.
(140 N)
Drag: High Speeds
D ≈ ¼Av2 (works well for air)
D = drag force
A = Area
V = velocity
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)
c. Setup, but do not solve, a differential
equation for the velocity of the particle.
8
6/3/2013
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)
Formula Wrap-Up
SF=ma
Weight = mg (g = 9.8 m/s2)
Fmax = msFN
Ffr = mkFN
26. 0.56 N, 57 g
40. a) 15.7 N b) 2.9 m/s
42. 0.165
52. Stays at rest
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 from the ground,
one on each side connected to the dogs. (mk =
0.060)
a. Calculate the acceleration of the sled. (0.833
m/s2)
b. Calculate T1 and T2 during the acceleration
period. (140 N)
4. 508 N
8. 4 m/s2 (0-3s)
-6 m/s2 (3-6s)
8 N (at 1s)
-12 N (at 7s)
10. a) T = 0 N b) F = T = 0N c) 250 N
12. a) 3.96 N b) 2.32 N
14.a) 540 N
b) 89 N
16. 1035 N, 740 N, 590 N
20. 10, 154 N
26. 0.56 N, 57 g
30. a) 0.0036 N
b) 0.0104 N
c) 4.4 m/s
9