Ch 5a Forces
Updated 1/7/04
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NEWTON'S
LAWS OF MOTION & FRICTION
1 a] Describe 5 types of forces:
Force | Range |
Description | Use
b] Draw the two and three body diagrams.
Force
| Range |
Description
| Use
---------------------------------------------------------------------------
GRAVITY | Long |Attraction between
any 2 objects| Holds planets in orbit
| |
|Causes nuclear reactions
---------------------------------------------------------------------------
ELECTRO-| Long | Opposite Charges
Attract |Holds
electrons in orbit
MAGNETIC| |
Like Charges Repel
|Holds matter together
---------------------------------------------------------------------------
| Short | Only in nucleus, holds | Creates Sunlight
NUCLEAR | |
repelling protons together
|
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FRICTION|Contact| Opposite object's motion | Allows you to walk
---------------------------------------------------------------------------
SPRING |Contact|Restoring, opposite
displacement| Support forces
---------------------------------------------------------------------------
BOTH
GRAVITATIONAL AND ELECTRICAL FORCES ARE INVERSE SQUARE LAWS
Fg = G M1 M2/d2
| Fe = k Q1 Q2/d2
Fg ATTRACTIVE ONLY
|Fe BOTH ATTRACTIVE & REPULSIVE.
G IS VERY SMALL 6.67 10-11
| k IS VERY LARGE 9 109
M1 M2 PRODUCT OF MASSES
|Q1 Q2 PRODUCT OF CHARGES
d = Distance
between the centers of gravity or the centers of
charge.
2
Concepts| DISTANCE | MASS | SPEED | ACCELERATION |
g | FORCE |
Metric | Meter |Kilogram|
m/s | m/s2 |9.8
m/s2|Newton
|
English |
Foot |
Slug | ft/s |
ft/s2 |32
ft/s2|Pound
|
3
Concept
| Measures
| Unit
| Instrument | Location
Mass |The amount of
matter |kg or Slug | Balance |
Independent
Weight |The pull of gravity
| N or lb |Spring
Scale| Dependent
Volume |The size of matter.
|Liter or ft3| Ruler
| Independent
4 a] INERTIA IS THE TENDENCY OF AN OBJECT IN MOTION (or at rest) TO STAY IN
MOTION. IT IS MEASURED BY ITS MASS [kg].
b] EXPLAIN WHAT HAPPENS WHEN A MOUSE COLLIDES WITH YOU OUT IN SPACE.
The mouse stops moving because its tendency to stay in motion (inertia) is
much less than your tendency to stay at rest (inertia).
c] EXPLAIN WHAT HAPPENS WHEN A COW COLLIDES WITH YOU OUT IN SPACE.
The cow moves you because its tendency to stay in motion is much more than
your tendency to stay at rest.
NEWTON'S
3 LAWS
1st LAW:
[LAW OF INERTIA] AN OBJECT IN MOTION (or at rest) WILL STAY IN MOTION (or
at rest); IN A STRAIGHT LINE, UNLESS AN UNBALANCED FORCE ACTS ON IT. INERTIA IS THE TENDENCY OF AN OBJECT IN MOTION (or at rest)
TO STAY IN MOTION. IT IS MEASURED
BY ITS MASS [kg].
EX: CAR TURNS A CORNER (changes
direction). THE PASSENGERS ARE
THROWN SIDEWAYS BECAUSE THEIR INERTIA IS KEEPING THEM MOVING IN A STRAIGHT LINE.
2nd
LAW: ACCELERATION IS DIRECTLY
PROPORTIONAL TO THE NET FORCE AND INVERSELY PROPORTIONAL TO THE OBJECT'S MASS.
a = F/m or F = ma
WHEN APPLIED TO GRAVITY: WEIGHT = MASSxACCELERATION OF GRAVITY, W= mg
EX: FOR THE SAME OBJECT IF YOU
DOUBLE THE FORCE YOU DOUBLE THE ACCEL.
(A
teenager can throw a ball faster than a child)
FOR THE SAME FORCE IF YOU DOUBLE THE MASS YOU HALF THE ACCEL.
(You can throw a ball faster than you can throw a watermelon.)
3rd
LAW: FORCES ALWAYS ACT IN PAIRS:
THE FORCE ON OBJECT A IS EQUAL AND OPPOSITE TO THE FORCE ON OBJECT
B: Fa = -Fb
EX: A PERSON JUMPS FROM THE BOAT TO
THE DOCK & THE BOAT MOVES AWAY.
FIRST
LAW DESCRIBES MOTION WHEN A BALANCED FORCE ACTS IT.
SECOND LAW DESCRIBES MOTION WHEN AN UNBALANCED FORCE ACTS ON IT.
THIRD LAW EXPLAINS WHY FORCES ACT
IN PAIRS.
5 a] WHAT IS A NEWTON (IN WORDS & IN AN EQUAITON)?
A Newton is the force needed to accelerate one kg at 1 m/s/s
1 Newton = 1 kg x 1 m/s2
b] What does 4 ft/s2 mean?
An object increases its speed by 4 ft/s every second.
Ex: t | 0 | 1 | 2 | 3 | 4 |
Speed| 0 | 4 | 8 | 12| 16|
c] When does a = 9.8 m/s2
a = 9.8 m/s2 when an object moves through air or space under
the influence of gravity only (No air friction, lift, or thrust).
d] Find the weight of a 50 kg object in Newtons and pounds
m = 50 kg W = mg
W = ? = 50(9.8 m/s2) = 490 N
490 N x 1lb/4.445 N = 110 lb
e] Find the weight of a 130 lb girl in Newtons.
130 lb x 4.445 N/lb = 578 N
f] WEIGHT = mg (W = mg); WHY DO WE USE g WHEN AN OBJECT IS NOT MOVING?
This equation is based on the 2nd Law: Fnet = ma or a = Fnet/m or g = W/m
Both forms of this equation are true:
g = W/m is true when gravity is the only force acting.
That is, there is no air resistance or support force.
a = Fnet/m is the more general equation and applies when more than one force
is acting on an object.
Ex#1: a book on a desk has gravity and a support force acting, Fnet = Zero
Ex#2: an object falling through the air has gravity and air friction actin
and Fnet = Weight - Air Resistance
6 a] NAME & DEFINE 2 TYPES OF FRICTION
b] WHAT IS THE FRICTION EQUATION? NAME & DEFINE ALL TERMS.
c] WHAT IS THE CAUSE OF FRICTION.
d] WHAT RELATIONSHIP EXISTS BETWEEN STATIC AND SLIDING FRICTION?
e] WHAT ARE THE IMPLICATIONS FOR STOPPING A CAR?
f] IS THE FORCE OF FRICTION GREATER FOR A WIDE TIRE OR A NARROW TIRE?
g] A BOOK SITS ON A TABLE. THE FORCE OF GRAVITY AND THE SUPPORT FORCE ACT ON IT.
DOESN'T THE FORCE OF FRICTION ACT AS WELL?
THE
FRICTION EQUATION: Ff = u F|
Ff = FRICTION FORCE--OPPOSES MOTION, PARALLEL TO THE SURFACES IN CONTACT, NEEDED
TO MAINTAIN A CONSTANT SPEED OF A SLIDING OBJECT.
F| = NORMAL FORCE--PRESSES 2 SURFACES TOGETHER, PERPENDICULAR TO THEM.
u = COEFFICIENT OF
FRICTION--MEASURES ROUGHNESS OF BOTH SURFACES.
STATIC FRICTION: FORCE NEEDED TO
START MOVING
SLIDING FRICTION: FORCE NEEDED TO
KEEP MOVING
7
A 90 kg BOY AND A 60 kg GIRL HAVE A TUG-OF-WAR ON AN ICY
FRICTIONLESS
SURFACE.
IF THE ACCELERATION OF THE GIRL TOWARD THE BOY IS 3 m/s2,
FIND THE BOY'S ACCELERATION, MAGNITUDE & DIRECTION.
Mboy =
90 kg
3rd Law:
Fboy
= -Fgirl
Mgirl = 60 kg
2nd Law, F = ma:
MboyAboy = -MgirlAgirl
Agirl = 3 m/s2
90 kg Aboy = -60 kg 3 m/s2
Aboy =
?
Aboy
= -60 kg 3 m/s2
90 kg
Aboy = -2 m/s2
Minus sign indicates the direction is opposite
the girl's acceleration (toward the girl)
8
WHICH OF THE FOLLOWING OBJECTS IS MORE MASSIVE?
|----| a = 50 m/s2
|----| a = 30 m/s2
|----| a = 20 m/s2
| A |---------->
|
B |----------> |
C |---------->
|----| F = 200 N
|----|
F = 150 N
|----|
F = 120 N
Use the 2nd Law: F = ma ---> m =
F/a
A: m = F/a = 200 N/50 m/s2
= 4 kg
B: m = F/a = 150 N/30 m/s2
= 5 kg
C: m = F/a = 120 N/20 m/s2
= 6 kg--->Most massive.
9
a] DEFINE TERMINAL VELOCITY, AND DRAW & LABEL THE DIAGRAM.
Terminal
Velocity: The final velocity of an
object in free fall.
It occurs when the Fnet =0,
i.e., Air Resistance = Weight (R = W)
/|\ Air Resistance (R)
|
+------+
|Mass m|
Terminal Velocity occurs when
+------+
Air Resistance = Weight
|
R = W
\|/ Weight (W)
b] DERIVE THE RELATIONSHIP BETWEEN TERMINAL VELOCITY AND MASS.
Fnet = Weight - Air
Resistance or Fnet = W - R
2nd Law: Fnet = ma & W =
mg---> ma = mg - R Dividing
by m
a = g - R/m
This says that the acceleration is reduced from the g
(the maximum accel) by the ratio of Air Resistance/Mass
That is accel increases as mass increases.
Thus, the Vf increases as mass increases.
Note: Air resistance is directly
proportional to the area of the falling object and the square of its speed.
Ex: Parachutes increase air
resistance by presenting a large area.
c] COMPARE THE TERMINAL VELOCITIES OF LIGHT AND HEAVY OBJECTS.
In short, heavy objects will
have a higher terminal velocity an light ones.
d] GIVE AN EXAMPLE OF AN IMPLICATION.
Ex #1: If a squirrel and a cat fall
out of a tree the cat will hit the ground at a higher speed.
Ex #2: If a spider falls off a tall
building it will hit the ground at such a slow speed it will not get hurt.
10
a] WHAT FORCE IS NEEDED TO ACCELERATE A 2000 kg CAR FROM 10 m/s TO 40
m/s
IN 20 s?
F = ?
F = ma
m = 2000 kg
F = m(Vf - Vi) Definition
of a
Vi = 10 m/s
t
Vf = 40 m/s
F = 2,000 kg (40 - 10)m/s
t =
20 s
20
s
3,000
kg m/s/s = 3,000 N
b] A 2000 N FORCE IS APPLIED TO A 1000 kg CAR MOVING AT 10 m/s.
FIND THE FINAL VELOCITY AFTER 20 s.
F = 2,000 N
F = ma
m = 1,000 kg
F = m(Vf - Vi) Definition
of a
Vi = 10 m/s
t
t = 20 s
2,000 N = 1,000 kg (Vf - 10)m/s Divide by 1000
Vf = ?
20 s
2 =
(Vf - 10)
Multiply by 20
20
40 = Vf - 10 --->Vf = 50 m/s
11
A 500 kg SPEED BOAT STARTS
FROM REST AND TRAVELS 200 m IN 3.0 s.
FIND THE NET FORCE ON THE BOAT. [D = Vit + at2/2]
m = 500 kg
F = ma but we need a
Vi = 0 m/s
D = Vit + at2/2
d = 200 m 200
m = 0(3 s) + a (3 s)2/2
t = 3.0 s
200 m = 0 + a 9 s2/2
F = ?
200 m x2/9 s2=
a
a = 44.4 m/s2
F = ma
F = 500 kg 44.4 m/s2 = 22,200 N
12 A
CERTAIN FORCE F GIVES A MASS M AN ACCELERATION OF A. DERIVE THE EQUATION TO
FIND THE ACCELERATION [A'] FOR A FORCE 2F & A MASS
3M.
Old: A =
F/M
A' = F'/M'
New: A' =
?
A' = 2F/3M
F' = 2F A'
= 2/3A
M' = 3M The
new acceleration = 2/3 the old
1
a] FIND THE MAGNITUDE & DIRECTION
15 N----->|------|-----> 30 N
OF THE MISSING FORCE.
| 8 kg |
b] SHOW 3 STEP CHECK.
3 N<------|------|<---?--->
a = 5 m/s2 ---->
Positive Direction is in the direction of acceleration
F = ma is for 1 force,
Fnet = Algebraic Sum of all Forces = ma is for more
than one force.
Sum F = ma
+15 +30 -3 + X = 8 kg(5 m/s2)
Note: always make X positive
42 + X
=
40
If it turns out negative the direction
-42
-42 &nbbsp;
is opposite the accleertion.
X = -2 N, or 2 N Left
Check:
Step 1 Draw the diagram using the answer for X
15 N----->|------|-----> 30 N
| 8 kg |
3 N<------|------|<---2 N
a = 5 m/s2 ---->
Step 2 Add the forces on each side (note the
direction).
|------|
12
N----->|
8 kg |-----> 28 N
|------|
a = 5 m/s2 ---->
Step 3 Add the forces (note the direction) on
opposite sides.
|------|
| 8 kg |-----> 40 N ?=? ma
|------|
?=? 8 kg (5 m/s2)---->
a = 5 m/s2
----> &nnbsp; 40 N
---> It Checks.
Practice:
15 N----->|------|-----> 30 N
| 8 kg |
Note: the only change is in
3 N<------|------|<---?--->
the direction of the
<---a = 5 m/s2
acceleration
Positive Direction is in the direction of acceleration
Sum F = ma
-15 -30 +3 + X = 8 kg(5 m/s2)
Note: always make X positive
-42 + X
=
40
If it turns out negative the direction
+42
+42
is opposite the accleertion.
X = +82 N, or 82 N Left
Check:
Step 1 15
N----->|------|-----> 30 N
| 8 kg |
3 N<------|------|<---82 N
<----a = 5 m/s2
Step 2
|------|
12 N----->| 8 kg |<----- 52 N
|------|
<----a = 5 m/s2
Step 3
|------|
| 8 kg |<----- 40 N ?=? ma
|------|
?=? 8 kg (5 m/s2)<----
<----a = 5 m/s2
40 N <--- It Checks.
2
FIND THE FOLLOWING:
SHOW WORK
/|\
5 N
a]
NORMAL FORCE
------>
|
b]
NET PARALLEL FORCE
a = 2 m/s2
|------|
c]
FRICTION FORCE
| 9 kg |----> 30 N
d]
PARALLEL FORCE
Ff <-----|------|
.
e]
m
/////////////////////(immovable surface)
a] Fn = F| = Force pushing the 2 surfaces together.
Note the 5 N force lifting up. This occurs when you lift heavy
furniture. You pick up so even if you don't lift it all the way
you prevent the legs from scraping the floor too hard.
Fn = W - 5 N
mg - 5 N
9 kg(9.8 m/s2)
- 5 N
88.2
- 5 N
83.2 N
b] Fnet = ma
= 9 kg(2 m/s2)
= 18 N
c] Sum F = ma
30 N - Ff = 9 kg(2 m/s2)
30 N - Ff = 18 N
-30 N -30 N
-Ff =
-12 N
Ff
= 12 N
Note: The forces perpendicular to each other are
independent.
That is they
don't influence each other.
So the 5 N
vertical force is not used in the sum of the forces
parallel to
the surface.
d] The parallel force (F||) is the force parallel to the
surface.
F|| = 30 N by inspection (Simply look
at the diagram, no calc needed.)
e] Ff = mF|
12 N = m83.2
N
m
= 12 N
83.2 N Note: The Newtons cancel.
m
= 0.144 m
is always unitless, a pure number.
3
THE FRICTION ON A 9 kg CRATE SLIDING AT CONSTANT VELOCITY IS 100 N.
a] DIAGRAM
b] FIND THE ACCELERATION.
c] FIND THE NET FORCE ACTING ON THE CRATE?
d] FIND THE FORCE APPLIED
|------|
| 9 kg |----> Fa = ?
100 N = Ff <-----|------|
.
/////////////////////(immovable surface)
b] Constant V==> a = 0 m/s2
c] Fnet = ma
Fnet = 9kg(0 m/s2)
Fnet
= 0 N
d] Sum F = ma
Fa - 100 N = 0
Fa = 100 N
Note: any time a = 0 the opposite forces will be equal.
4A
IF A FORCE OF 30.0 N IS NEEDED TO SLIDE A 10 kg CRATE AT CONSTANT
VELOCITY.
DIAGRAM
& FIND:
a]
Ff; b] F|; & c] m.
|-------|
| 10 kg |----> Fa = 30.0 N
? = Ff <-----|-------|
.
/////////////////////(immovable surface)
a] Constant velocity ===> a =
0 |
Sum F =
ma
\|/ F| = mg
30 N - Ff =
0
Ff = 30 N
b] F| = mg
= 10 kg (9.8 m/s2)
=
98 N
c] Ff = mF|
===> m
= Ff/F|
= 30 N
.
98 N
m
= 0.306
B
A FORCE OF 100 N IS NEEDED TO ACCELERATE A 30 kg CRATE HORIZONTALLY AT 2
m/s2.
DIAGRAM & FIND:
a]
Ff; b] F|; & c] m.
a = 2
m/s2
---> |-------|
| 30 kg |----> Fa = 100.0 N
? = Ff <-----|-------|
.
/////////////////////(immovable surface)
a] Sum F =
ma
|
100 N - Ff = 30 kg(2
m/s2)
\|/ F| = mg
100 N - Ff = 60 N
-100 N
-100 N ;
- Ff =
-40 N
Ff = 40 N
b] F| = mg
= 30 kg (9.8 m/s2)
=
294 N
c] Ff = mF|
===> m
= Ff/F|
= 40 N
.
294 N
m
= 0.136
5
IT TAKES 745 N TO LIFT A BEAM WITH AN ACCELERATION OF 20 m/s2.
a] DIAGRAM
/|\
b] FIND ITS MASS.
| 745 N
+-------+
/|\
Sum F =
ma
| Beam | a = 20
m/s2
|
745 N - W =
ma
+-------+
745 N - m 9.8 m/s2
= m 20
m/s2
|
+ m 9.8 m/s2
+ m 9.8 m/s2
\|/ W = mg
745
N
= m 29.8
m/s2
745 N
= m
29.8
m/s2
25 kg = m
6
A 5 kg MASS AND A 2 kg MASS HANG OVER A FRICTIONLESS PULLEY.
a] DIAGRAM THE WHOLE SYSTEM.
/////////
|
+---------+
| Pulley |
+---------+
| |
| |
| +------+ |
| | 5 kg | a | Note: Both accelerations are of
equal
| +------+ \|/
size.
|
|
/|\ +-----+
a | |2 kg |
| +-----+
b] UNDER WHAT CONDITIONS WILL a = 9.8 m/s2 & 0.0 m/s2
a
= 9.8
m/s2
if The Light Mass = 0
a = 0.0 m/s2
if the Light Mass = Heavy Mass
c] DIAGRAM & FIND THE ACCELERATION OF THE HEAVY
MASS?
/|\ Wlight(WL) = mg
| WL = 2 kg( 9.8
m/s2)
| = 19.6
N
+-------+ |
| 5 kg | a |
+-------+ \|/
|
|
\|/ Wheavy (Wh) = mg
=
5 kg( 9.8 m/s2)
= 49 N
Sum F = ma
49 N - 19.6 N = (2+5 kg)a
Note: The net force on the Wh accelerates both
29.4 N
=
7a
masses. So both masses (2+5) are used.
29.4 N/7 kg = a
4.20 m/s2
= a
d] DIAGRAM & FIND THE TENSION FORCE ON THE ROPE?
/|\ T = ?
|
| /|\
+-------+ |
| 2 kg | a | Note: The Tension has to be greater
than
+-------+
| 19.6 N for it to
move up.
|
|
\|/ WL = 19.6 N
Sum F = ma
T - 19.6 N = (2 kg) 4.2 m/s2
T - 19.6 N = 8.4
N
+ 19.6 N = +19.6 N
T
= 28 N
7
A 20 kg MASS ON A FRICTIONLESS TABLE IS PULLED BY A 5 kg MASS HANGING
FROM THE
TABLE.
[3 DIAGRAMS NEEDED]
(Table Mass) Wt |-------|
|
20 kg |--------0
|-------|
/|
.
//////////////////////
|
Table
|
+-------+
| 5 kg | Wh
(Hanging Mass)
+-------+
a] UNDER WHAT CONDITIONS WILL THE ACCELERATION BE 9.8
m/s2 & 0.0 m/s2
a = 9.8
m/s2 if Wt = 0.0 kg
a = 0.0 m/s2 if Wh = 0.0 kg
OR if Ff = Wh
b] FIND THE ACCELERATION OF THE 20 kg.
|-------|
|
20 kg |----> Wh = 5 kg(9.8
m/s2) = 49 N
Ff = 0 <-----|-------|
.
//////////////////////
Sum F = ma Note:
The net force on the Wh accelerates both masses.
Wh - Ff = ma
So both masses (20+5) are used.
49 N - 0 = (20
kg + 5 kg)a
49 N/25 kg = a
1.96 m/s2 = a
c] FIND THE TENSION ON THE CORD
/|\ T = ?
|
|
+-------+ |
| 5 kg | a | Note: The Tension has to be less
than
+-------+ \|/
49 N for it to move down.
|
|
\|/ WL = 49.0 N
Sum F = ma
49 N - T = (5 kg) 1.96 m/s2
49 N - T = 9.8
N
-49 N -49.0 N
.
- T = -39.2 N
T = 39.2 N
Pract #1:
25 kg on table, m
= 0.25, 10 kg hanging.
Pract #2:
15 kg on table, m
= 0.20,
3 kg hanging.
Pract #3:
30 kg on table, m
= 0.45, 10 kg hanging.
8a
A 80 kg ASTRONAUT IS STANDING ON A SCALE IN A SPACESHIP.
WHILE THE SHIP IS
MOVING IN A STRAIGHT LINE WITH CONSTANT VELOCITY OF
100.0 m/s NEAR A LARGE
PLANET, THE SCALE READS 300.0 N.
THE SHIP ACCELERATES UPWARD AT 7.0 m/s2.
WHAT DOES THE SCALE READ? [Fnet = MAnet]
b
A 60 kg PERSON IS STANDING ON A SCALE IN AN ELEVATOR.
THE ELEVATOR
ACCELERATES DOWNWARD AT 7.0 m/s2.
DIAGRAM & FIND THE SCALE READING.
c
A 60 kg PERSON IS STANDING ON A SCALE IN AN ELEVATOR.
THE ELEVATOR
ACCELERATES UPWARD AT 7.0 m/s2.
DIAGRAM & FIND THE SCALE READING.
ANS:
[7] 1.96 m/s2, T = 39.2 N; [P1] Ff = 61.25 N; 1.05 m/s2,
T = 85.7 N
[8a] 860 N;
[b] 168 N;
[c] 1008 N;
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