Download chapter 3 part 1

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1
3 .1
Force,weight,and gravitational mass
What is a Force?
Force is an action that can change motion.
A force is what we call a push or a pull, or any action that has the
ability to change an object’s motion.
•the force causes a change in velocity, or an acceleration
• Because velocity is a vector, having both magnitude and direction, a force
can alter an objects speed, its direction, or both.
• All forces have a strength (magnitude) and a direction. Therefore
• , forces are vector quantities.
•Forces are either balanced or unbalanced
•There are two types of forces that we can see
•Contact Forces
•At a distance forces
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Fkick
Fgrav
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Contact Force
Forces in which the two interacting objects are physically in contact with each
other.
EXAMPLES
•Hitting
•Pulling with a rope
•Lifting weights
•Pushing a couch
•Frictional Force
•Tensional Force
•Normal Force
•Air Resistance Force
•Applied Force
•Spring Force
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At A Distance Force (field force)
•Forces in which the two interacting objects are not in
physical contact with each other, but are able to exert a push
or pull despite the physical separation.
Example:
•Gravitational Force
•Electrical Force
•Magnetic Force
•Nuclear Forces
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Measuring Forces
• Forces are measured in
newtons
(kg . m/s2).
• Forces are measured
using a spring scale.
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Balanced Force
• A force that produces no
change in an object’s motion
because it is balanced by an
equal, opposite force.
• no change in velocity
The object shown in the diagram at rest since there is no
net force acting on it.
If you were to add these forces they would = 0
FALSE! A net force does not cause motion. A net
force causes a change in motion, or acceleration.
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Unbalanced Forces
• Are forces that results in an object’s
motion being changed.
+
• velocity changes (object accelerates)
Fnet
Fpull
Ffriction
N
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W
N
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example
On a horizontal, frictionless surface, the blocks above are being
acted upon by two opposing horizontal forces, as shown. What
is the magnitude of the net force acting on the 3kg block?
A. zero
B. 2N
C. 1.5 N
D. 1N
E. More information is
needed.
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example
• A woman is holding two dogs on a leash.
• If each dog pulls with a force of 80 N,
how much force does the woman have
to exert to keep the dogs from moving?
160 N to the left
example
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example
In a two-dimensional tug-of-war, Alex, Betty, and Charles pull
horizontally on an automobile tire at the angles shown in the
overhead view of the Fig.
• The tire remainsstationary in spite of the three pulls. Alex
pulls with force
 FA of magnitude 220 N, and Charles pulls
with force FC of magnitude 170 N. The direction of
 FC is
not given. What is the magnitude of Betty's force FB ?
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


FA  FB  FC  m(0)  0



FB   FA  FC
FBy  FAy  FCy
FB sin (  90 )   FA sin 133  FC sin  ,
 FB   ( 220 N ) (sin 133 )  (170 N ) sin 
Force components in the x-axis :
FBX  FAX  FCX
FB cos (  90 )   FA cos 133  FC cos 
0   ( 220 N) (cos 133 )  (170 N) cos 
  cos
1
(220 N) (cos 133 )
 28.04
170 N
Therefore,
FB  241 N
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Weight
Weight = Mass x Gravity
W  m g
The weight of an object is the gravitational force that the
planet exerts on the object. The weight always acts
downward, toward the center of the planet.
SI Unit of Weight: : newton (N)
– m: mass of the body (units: kg)
– g: gravitational acceleration (9.8m/s2,
• As the mass of a body increases, its’
weight increases proportionally
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• Weight
– the force of gravity on an object
W = mg
MASS
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always the same
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(kg)
W: weight (N)
m: mass (kg)
g: acceleration due
to gravity (m/s2)
WEIGHT
depends on gravity
(N)
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Mass …
• is a measure of how much matter there is in something - usually measured in kilograms in science
• causes an object to have weight in a gravitational field
•A measure of the resistance of an object to changes in its
motion due to a force (describes how difficult it is to get an
object moving)
•Scalar
Note that mass is involved in the force of gravity! This is a
separate property from that of inertia, so we give this
property the name gravitational mass.
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Example :
• what is the weight of a 2 kg mass?
• W = Fg = mxg = 2 kg x 9.8 m/s2
= 19.6 N
Example :
• What is the mass of a 1000 N person?
• W = Fg = mxg
m = Fg/g = 1000 N / 9.8 m/s2
=102 kg
Example :
• A girl weighs 745 N. What is his mass
m=F÷g
m = (745 N) ÷ (9.8 m/s2)
m = 76.0 kg
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3.2 Density
• A measure of how much mass is in a certain volume of space
• The density of a substance is equal to its mass divided by its
volume
Relative Density
•The density of a material or substance, relative to another substance
•Water is the substance to which we generally compare other
substances
•ALSO known as SPECIFIC GRAVITY
•For water: Absolute Density = 1000kg/m3
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Example :
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Example :
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Ex -3-17
Find the density of gasoline if 5 Kg has a volume of 7.35 x10 -3 m 3
And its relative density (specific gravity)
Relative density =( 0.68 x 10 3 Kg / m 3) / 10 3 Kg / m 3
= 0.68
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3.3
Newton's first law
Every object continues in a state of rest , or of uniform motion in
a straight line , unless it is compelled to change that state by
forces acting upon it.
An equivalent statement of the first law is that :
An object at rest will stay at rest, and an object in
motion will stay in motion at constant velocity, unless
acted upon by an unbalanced force.
This, at first, does not seem obvious. Most things on earth tend
to slow down and stop. However, when we consider the
situation, we see that there are lots of forces tending to slow
the objects down such as friction and air resistance.
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Newton’s First Law
• Newton’s First Law of Motion
– “Law of Inertia”
• Inertia
– tendency of an object to resist any change in its motion
– increases as mass increases
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Mass = Inertia – an object’s
resistance to motion
Would it be more difficult to pull an elephant or a
mouse?
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The net force acting on an object is the
vector sum of all the forces acting on it.
Examples:
9N
8N
8N
6N
4N
7N
12 N
8N
?
5N
4N
3N
7N
4N
If an object is remaining at rest, it is incorrect
to assume that there are no forces acting on
the object.
We can only conclude that the net force on
the object is zero.
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• The condition of zero acceleration is called
equilibrium.
• In equilibrium, all forces cancel out leaving
zero net force.
• Objects that are standing still are in
equilibrium because their acceleration is
zero.
• Objects that are moving at constant speed
and direction are also in equilibrium.
• A static problem usually means there is no
motion.
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3.5Newton’s Third Law of Motion
There is one further important aspect of motion that Newton
identified: the distinction between forces that act on an object
and forces that act by the object. This leads to his Third Law of
Motion: For every force by a first object on a second object,
there is a force by the second object on the first object with the
same magnitude but in the opposite direction.
This is sometimes called the law of action and reaction.
I like to call it: you can’t push yourself! You can only push
on an object and hope that it pushes back.
Example: when you walk up a stairs, you use your muscles to
push down on the stairs and you trust that the stairs will
push back up on you lifting you up the stairs.
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Newton’s Third Law
• Newton’s Third Law of Motion
– When one object exerts a force on a second object,
the second object exerts an equal but opposite force
on the first.


F AB   F BA
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• The action and reaction forces are EXACTLY 180o apart in
direction.
how come they don’t always cancel since they push on
different objects., making net force and acceleration
impossible?
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Newton’s Third Law
r
F12   F21
Force on “1” due to “2”
• Single isolated force cannot exist
• For every action there is an equal and opposite reaction
• “For every action force, there is … a reaction force” means:
– Forces ALWAYS occur in pairs.
– Single forces NEVER happen.
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Newton’s Third Law
• example:
 How can a horse pull a cart if the cart is pulling back
on the horse with an equal but opposite force?
 Aren’t these “balanced forces” resulting in no
acceleration?
 Explanation:
NO!!!
– forces are equal and opposite but act on different objects
– they are not “balanced forces”
– the movement of the horse depends on the forces acting on
the horse
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Newton’s Third Law
• Action-Reaction Pairs
 The hammer exerts a force
on the nail to the right.
 The nail exerts an equal
but opposite force on the
hammer to the left.
Flying gracefully through the air, birds
depend on Newton’s third law of motion.
As the birds push down on the air with
their wings, the air pushes their wings up
and gives them lift.
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Newton’s Third Law
• Action-Reaction Pairs
 The rocket exerts a downward
force on the exhaust gases.
 The gases exert an equal but
opposite upward force on the
FG
rocket.
FR
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According to Newton, whenever objects
A and B interact with each other, they
exert forces upon each other. When you
sit in your chair, your body exerts a
downward force on the chair and the
chair exerts an upward force on your
body.
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Other examples of Newton’s Third Law
• The baseball forces the bat to
the left (an action); the bat
forces the ball to the right (the
reaction).
• Locomotion is the act of moving or the ability to move from one
place to another.
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What happens if you are standing on a
skateboard or a slippery floor and push against
a wall? You slide in the opposite direction
(away from the wall), because you pushed on
the wall but the wall pushed back on you with
equal and opposite force.
Why does it hurt so much when you stub
your toe? When your toe exerts a force on a
rock, the rock exerts an equal force back on
your toe. The harder you hit your toe against
it, the more force the rock exerts back on your
toe (and the more your toe hurts).
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Calculate force
• Three people are each applying 250 N of force to try to move a heavy
cart.
• The people are standing on a rug.
• Someone nearby notices that the rug is slipping.
• How much force must be applied to the rug to keep it from slipping?
• Sketch the action and reaction forces acting between the people and the
cart and between the people and the rug.
The third law says that each of the forces applied creates a reaction force.
So Each person applies a force to the cart and the cart applies an equal and opposite
force to the person.
The force on the rug is the sum of the reaction forces acting on each person.
The total force that must be applied to the rug is 750 N in order to equal the reaction
forces from all three people.
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• Locomotion is the act of moving or the ability to move from one
place to another.
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Action-eaction Pairs
r
n   n
Fg  Fg'
Define the OBJECT (free body)
• Newton’s Law uses the forces
acting ON object
• N and Fg act on object
• n’ and Fg’ act on other objects
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Normal Force Is Not Always Equal to the Weight
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Normal Force
The normal force is the
force exerted by a surface on
an object.
and always perpendicular to
the surface that produces
it.
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Normal Force
The normal force may be equal to, greater
than, or less than the weight.
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The Other Logical Problem
•
If the Newton’s Third Law action and reaction forces are
always equal and opposite, how do two objects of different
sizes get different accelerations in the same interaction?
(When a bug hits a windshield, different things happen to the
bug and windshield.)
• If the action and reaction forces are the same size, how can
two objects push on each other and get different
accelerations?
• Newton’s Second Law says that the acceleration of an object
depends not only on the force on it, but on the object’s mass.
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3.6 Newton’s Second Law
• Newton’s Second Law of Motion
• The force F needed to produce an acceleration a is
F = ma
– The acceleration of an object is directly proportional to the
net force acting on it and inversely proportional to its mass.
• The same force acting on objects of different mass
will produce different accelerations!
Same Force
Fnet
m
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a
Fnet
=
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m
=a
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Newton’s Second Law
• The acceleration of an object is equal to the force you apply
divided by the mass of the object.
•Force and acceleration are vectors . They both have:
Magnitude and direction if you exert a force east . The object will
accelerate east
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• The acceleration of an object is equal to the force you apply
divided by the mass of the object.
Newton’s Third Law
• Action-Reaction Pairs
 Both objects accelerate.
 The amount of acceleration depends on the mass of the object.
a 
F
m
 Small mass  more acceleration
 Large mass  less acceleration
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• If you apply more
force to an object, it
accelerates at a
higher rate.
• If an object has more
mass it accelerates at a
lower rate because
mass has inertia.
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Newton’s second Law
F =ma
a
F

m
F
m a
m: mass (kg)
a:acceleration (m/s2)
1 N = 1 kg ·m/s2
Three forms of the second law:
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More about F = ma
If you double the mass, you double the force. If you
double the acceleration, you double the force.
What if you double the mass and the acceleration?
(2m)(2a) = 4F
Doubling the mass and the acceleration quadruples the
force.
So . . . what if you decrease the mass by half? How
much force would the object have now?
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Example :
A railway engine pulls a wagon of mass 10 000 kg along a straight track at a steady
speed. The pull force in the couplings between the engine and wagon is 1000 N.
a.What is the force opposing the motion of the wagon?
b.If the pull force is increased to 1200 N and the resistance to movement
of the wagon remains constant, what would be the acceleration of the wagon?
Solution
a)
When the speed is steady, by Newton’s first law, the resultant force must be zero.
The pull on the wagon must equal the resistance to motion. So the force resisting
motion is 1000 N.
b)
The resultant force on the wagon is 1200 – 1000 = 200 N
By Equation
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Example :
a) Find the acceleration of a 20 kg crate along a horizontal
floor when it is pushed with a resultant force of 10 N parallel
to the floor.
b) How far will the crate move in 5s (starting from rest)?
Solution
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Example :
A 1kg stone fall freely from rest from a bridge.
a -What is the force causing it to accelerate?
b -What is its speed 4s later?
c -How far has it fallen in this time?
Solution
The force causing it to fall is its weight. As it is falling with acceleration due to gravity
b)
c)
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Equilibrium
• When the net force acting on an object is zero, the
forces on the object are balanced.
• We call this condition equilibrium.
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Equilibrium
• A moving object continues to move with the same speed and
direction.
• Newton’s second law states that for an object to be in
equilibrium, the net force, or the sum of the forces, has to be
zero.
• Acceleration results from a net force that is not equal
to zero.
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Example :
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Balancing Forces (Statics)
There are often situations where a number of forces are acting on something,
and the object has no motion – it is STATIC or in EQUILIBRIUM or at rest.
• This means the NET FORCE on the object is zero, or in other words the forces
balance each other out.
•
•ΣF = m a
a =0
ΣF = 0
Objects in equilibrium (no net external force)
also move at constant velocity
In a two-dimensional problem, we can separate this equation into its two
component, in the x and y directions:
Σ Fx = 0
Σ Fy = 0
unbalanced Forces
(Dynamics)
•There are other situations where all the forces acting on something do not cancel each
other out completely.
•This means the NET FORCE on the object is not zero, the object will change its motion and
accelerate proportional to the object’s mass.
•ΣF = m a
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Example :
Calculating the net force from four forces
Four people are pulling on the same 200 kg box with the forces
shown. Calculate the acceleration of the box.
1. Use a = F ÷ m.
2. First add the forces to find the net force.
F = - 75N - 25N + 45N + 55N = 0 N,
a=0
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Example :
Using equilibrium to find an unknown force
Two chains are used to lift a small boat. One of the
chains has a force of 600 N. Find the force on the
other chain if the mass of the boat is 150
kilograms.
1.
2.
3.
Use: net force = zero,
Fw = mg
and g = 9.8 N/kg.
Fw = mg = (150 kg)(9.8 N/kg) = 1,470 N.
Let F be the force in the other chain, equilibrium
requires:
– F + (600 N) = 1,470 N F = 1,470 N – 600 N
F = 870 N.
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Applications of equilibrium
• Real objects can move in three
directions: up-down, right-left, and
front-back.
• The three directions are called
three dimensions and usually given
the names x, y, and z.
• When an object is in equilibrium,
forces must balance separately in
each of the x, y, and z dimensions.
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example
• Three people are pulling on a wagon applying forces of
100 N,150 N, and 200 N.
• The wagon has a mass of 25 kilograms.
• Determine the acceleration and the direction the wagon
moves.
a =200 –(100+150) / 25
= - 50 / 25 = - 2 m / s2
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Example :
 What force would be required to accelerate
a 40 kg mass by 4 m/s2?
F = ma
F = (40 kg)(4 m/s2)
F = 160 N
Example :
 A 4.0 kg shotput is thrown with 30 N of force. What is
its acceleration?
a=F÷m
a = (30 N) ÷ (4.0 kg)
a = 7.5 m/s2
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Example :
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Example
A dragster accelerates from 0 to 60 m/s in 6 seconds.
Calculate the force on the car if the mass is 103 kg.
F = ma
and
a = ∆v / ∆t
Answer: 104 N
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Solving Force Problems
A 5 Kg rifle fires a 9 gram bullet with an acceleration of 30,000 m/s2.
(a) What force acts on the bullet? (b) What force acts on the rifle? (
c) If the bullet is in the rifle for 0.007 seconds, what is its muzzle
velocity?
a = 30,000 m/s2
(a) F = ma, F = 0.009 kg x 30,000 m/s2 = 270 N(grams must be
converted to kilograms in order to use the MKS system and get
newtons as force unit answer)
(b) Newton’s 3rd Law (Action / Reaction) – the rifle is “shot” in the
opposite direction with the same force as the bullet or – 270 N
(c) (muzzle velocity means the velocity at which the bullet leaves
the rifle barrel) Using the equation V = Vo + at, V = o + 30,000 x
0.007 = 210 m/s (Vo = 0 since the bullet is not moving before it
is fired)
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Solving Force Problems
(a) How much force is needed to reduce the velocity of a 6400 Kg
truck from 20 m/s to 10 m/s in 5 seconds? (b) What is its stopping
distance?
time = 5 sec
10 mph
20 mph
• (a) F = ma, we must first find mass. The weight is 6400 Kg. Wt =
mass x gravity therefore, m = w/g,
mass =
6400 Kg /9.8 m/s2 = 653.06 Kg.
• Now, to find acceleration, V = Vo + at or a = (V – Vo) / t
• a = (20 – 10)/ 5 = - 2 m/s2 and F = 200 x (-2) = - 400 N (negative
means the force is opposing the motion)
• (b) Δx = Vot + ½ at2 we get Δx = (20 x 5) + ½ (-2) 52 = 75 m is the
distance traveled during stopping.
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Example :
A force which gives a 2.0 kg object an acceleration of 1.6 m/s2
would give an 8.0 kg object what acceleration ?
(A) 0.2 m/s2 (B) 0.4 m/s2 (C) 1.6 m/s2 (D) 6.4 m/s2
Example :
A car slows form 50 m/s to 15 m/s in 10 seconds when the brakes
exert a force of 200 N. What is the weight of the car ?
(A) 57 N (B) 560 N
(C) 1830 lbs (D) 22,400N
Example :
A cable supports an elevator which is 2000 Kg. The tension in the cable
lifting the elevator is 25 KN. What is the acceleration ?
(A) 2.7 m/s2 (B) 0.4 m/s2 (C) 1.6 m/s2 (D) 8.4 m/s2
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Example :
m = 1850 kg
With two guys pushing…….what are The net forces???
 F  + 275 N + 395 N  560 N = +110 N
a 
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F
m

 110 N
 0.059 m/s 2
1850 kg
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If the airplane’s mass is 13 300 kg, what is the magnitude of
the net force that the catapult and jet engine exert on the
plane?
2
F

ma

(13
300
kg)(31
m/s
)

= 4.1105 N
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Where does the calculus fit in?


dv
d 2x
F  ma  m  m
dt
dt
There could be situations where you are
given a displacement function or velocity
function. The derivative will need to be
taken once or twice in order to get the
acceleration. Here is an example.
You are standing on a bathroom scale in an elevator in a tall
building. Your mass is 72-kg. The elevator starts from rest
and travels upward with a speed that varies with time
according to:
2
v(t )  3t  0.20t
When t = 4.0s , what is the reading on the bathroom scale
(a.k.a. Force Normal)?
dv d (3t  0.20t )
a

 3  0.40t
dt
dt
a (4)  3  0.40(4)  4.6 m / s2
2
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Fnet  ma
N  mg  ma 
N  ma  mg
N  (72)(9.8)  (72)( 4.6)  1036.8 N
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Example on Newton’s 2. and 3. Law
Suppose you are an astronaut in outer space giving a brief push to a
spacecraft whose mass is bigger than your own
1) Compare the magnitude of the force you exert on the spacecraft,
FS, to the magnitude of the force exerted by the spacecraft on you,
FA, while you are pushing:
1. FA = FS
2. FA > FS
3. FA < FS
correct
Third Law!
2) Compare the magnitudes of the acceleration you experience, aA,
to the magnitude of the acceleration of the spacecraft, aS, while
you are pushing:
1. aA = aS
a=F/m
2. aA > aS
correct
F same  lower mass gives larger a
3. aA < aS
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Example:
You are stranded in space, away from your
spaceship. Fortunately, you have a propulsion
unit that provides a constant net force F for
3.0 s. You turn it on, and after 3.0 s you have
moved 2.25 m.
If your mass is 68 kg, find F.
x  v0t  12 at 2  12 at 2
2x 2(2.25 m)
2
a 2 

0.50
m/s
t
(3.0 s)2
a  axiˆ  0.50 m/s2iˆ
F  ma  (68 kg)(0.50 m/s 2 )iˆ  34 Niˆ
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Example: Three Forces
Moe, Larry, and Curley push on a 752 kg boat, each exerting a 80.5 N force
parallel to the dock.
(a) What is the acceleration of the boat if they all push in the same direction?
(b) What is the acceleration if Moe pushes in the opposite direction from Larry
and Curley as shown?
F1  FM  FL  FC  3(80.5 N)  241.5 N
a1  F1 / m  (241.5 N) / (752 kg)  0.321 N/kg  0.321 m/s 2
F2  FM  FL  FC  80.5 N
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a2  F2 / m  (80.5 N) / (752 kg)  0.107 m/s 2
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Newton’s Second Law of Motion
Note that this is a vector equation, and
should really be worked in component
form:
S Fx = max
S Fy = may .
We can now see that Newton’s First Law of
Motion is really just a special case of his
Second Law of Motion.
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example
In the Figs. one or two forces act on a puck that moves over
frictionless ice and along an x axis, in one-dimensional

 motion.
• The puck's mass is m = 0.20 kg. Forces F 1 and F 2 are
directed along the axis
 and have magnitudes F1 = 4.0 N and
F2 = 2.0 N. Force F 3 is directed at angle  = 30° and has
magnitude F3 = 1.0 N. In each situation, what is the

acceleration of the puck?
F
 ma .
net , x
x
F1
4.0 N

 20 m / s 2 .
m 0.20 kg
F F
4.0 N  2.0 N
ax  1 2 
 10 m / s 2 .
m
0.20 kg
ax 
F3, x  F2 F3 cos   F2

m
m
(1.0 N) ( cos 30 )  2.0 N

 5.7 m / s 2
0.20 kg
ax 
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example
Solving Problems with Newton’s Laws: Free-Body Problems
• Newton’s second law tells us that the acceleration of an
object is proportional to the net force acting on an object.
• The net force is the vector sum of all the forces acting on the
object.
FR = 141 N
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example
Adding force vectors.
Calculate the sum of the two forces shown acting on the boat.
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Example :
a=3.02 m / s
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Example :
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Example :
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Free-body diagrams
• To keep track of the number and
direction of all the forces in a
system, it is useful to draw a
free-body diagram.
• A free-body diagram makes it
possible to focus on all forces
and where they act
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Free-body diagrams
• Forces due to weight or acceleration
may be assumed to act directly on an
object, often at its center.
• A reaction force is usually present at
any point an object is in contact with
another object or the floor.
• If a force comes out negative, it means
the opposes another force.
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Applications of equilibrium
What is the upward force in
each cable?
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• If an object is not moving, then
you know it is in equilibrium
and the net force must be zero.
• You know the total upward
force from the cables must
equal the downward force of
the sign’s weight because the
sign is in equilibrium.
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Tension force
The force exerted by the string, at either end:
• direction is parallel to the string
• magnitude (same at both ends) is called the tension
Example. Suppose a string can withstand string
tension 500 N without breaking. What is the
maximum mass M that it can hold suspended in
Earth’s gravity?
Weight W = Mg. Static equilibrium implies T = W. So
m = W/g = T/g
(500 N) / (9.8 m/s^2) = 51.0 kg.
The forces on either mass are weight and string tension. But the net
force is 0; the two forces cancel
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Rules for Ropes and Pulleys
• Force from rope points AWAY from
object
– (Rope can only pull)
• Magnitude of the force is Tension
– Tension is same everywhere in the
rope
• Tension does not change when
going over pulley
Approximations: Neglect mass of rope and pulley,
neglect friction in pulley
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example:
to ceiling
box hangs from a rope attached
SFy = may
y
T
T - W = may
T = W + may
W
In this case ay = 0
So T = W
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The tension force
•
•
•
•
The tension force pulls on a body.
The cord is often assumed to be massless.
We usually assume pulleys to be frictionless
A string has a single tension force (magnitude). The
direction depends on the body on which this force acts
upon. The tension forces on two sides of a frictionless
pulley are the same in magnitude.
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3 . 7 the significance of Newton's laws of motion
3 .7 some examples of Newton's laws
Make a diagram (conceptualize)
Categorize: no acceleration (at rest )
F  0
 F  ma
Isolate each object and draw a free body diagram for each
object. Draw in all forces that act on the object.
Establish a convenient coordinate system.
Write Newton’s law for each body and each coordinate
component.  set of equations.
Finalize by checking answers.
F  0
accelerating object:
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example
the Figure shows a block S (the sliding block) with mass M =
3.3 kg. The block is free to move along a horizontal frictionless
surface such as an air table.
Find (a) the acceleration of the
sliding block, (b) the acceleration
of the hanging block, and (c) the
tension in the cord.
Consider mass M :
The positive direction of a is to the right.
Consider mass m :
S F  Ma
T  Ma
T  mg  - ma
m
g  3.8 m / s 2
Mm
Same acceleration for
both (they move
Mm
T
g  13 N
together)
Mm
a
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example
• In Fig. a, a block B of mass M =
15.0 kg hangs by a cord from a
knot K of mass mK, which hangs
from a ceiling by means of two
other cords. The cords have
negligible mass, and the
magnitude of the gravitational
force on the knot is negligible
compared to the gravitational
force on the block. What are the
tensions in the three cords?
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At a knot, the tensions are different. The tensions on both
sides of a frictionless pulley are the same.
T3  Mg  M (0)  0
T3  147 N
At the knot,



T1  T2  T3  0
 T1 cos 28  T2 cos 47   0  0
T1 sin 28  T2 sin 47   147 N  0
T1  104 N
and
T2  134 N
The tensions are 104 N in cord 1, 134 N in cord 2, and 147 N in cord 3
(b) We now cut the cord. As the block then slides down the inclined
plane, does it accelerate? If so, what is its acceleration?
mg sin   ma
Positive direction is down the inclined plane
a  g sin 
a  (9.8 m / s 2 ) (sin 27 )  4.4 m / s 2
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example

• In Fig. a, a constant horizontal force Fap of magnitude 20 N is
applied to block A of mass mA = 4.0 kg, which pushes
against block B of mass mB = 6.0 kg. The blocks slide over a
frictionless surface, along an x axis. (a) What is the
acceleration of the blocks?
Fap  FAB  m A a
Fap  (m A  m B )a
Fap
20 N
a

 2.0 m / s 2
m A  m B 4.0 kg  6.0 kg
The acceleration of block A
and block B is the same

(b) What is the force FBA on block B from block A (Fig. c)?
FBA  mB a
FBA  (6.0 kg) (2.0 m / s 2 )  12 N
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Free body diagram
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Example :
A lift with its load has a mass of 2000 kg. It is supported by a
steel cable. Find the tension in he cable when it:
a -is at rest
b - accelerates upwards uniformly at 1m/s2
C - move upwards at a steady speed of 1 m/s
d - moves downwards at a steady speed of 1 m/s
e - accelerates downwards with uniform acceleration of 1 m/s2
a) When at rest we can use Newton’s first law which says that the
resultant force on the lift is zero.
Force acting down is the lifts weight, the force acting up is the tension in
the cable. These two must be equal and opposite to give a resultant
force of zero.
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b) As the lift is accelerating upwards so T must exceed the
weight mg. So the resultant acceleration force
by Newton second law, F = ma, so
c) As in (a), by Newton’s first law, the resultant force on the
lift must be zero, so
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d) As in (c) the tension in the cable will still equal mg since the
change in direction of motion does not alter the fact that there
is no acceleration.
e) If the lift accelerates downwards, then mg must exceed
the tension T. So the resultant accelerating force is
By Newton’s second law F = ma, so
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example
• In Fig. a, a cord holds stationary a block of mass m = 15 kg,
on a frictionless plane that is inclined at angle
  = 27°.
(a) What are the magnitudes of
 the force T on the block from
the cord
 and the normal force N on the block from the plane?
T  N  Fg  0
T  0  mg sin   0
T  mg sin   (15 kg) (9.8 m / s 2 )(sin 27 )  67 N
N  mg cos   0
N  mg cos 
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 (15 kg) (9.8 m / s 2 )(cos 27 )
 131 N
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Example
Find the acceleration of the boxes when the system
below is released. Friction is negligible.
Same acceleration for both (they
move together)
2m
m
1. Draw free body diagram for both boxes.
35o
2. Select axes
3. Write Newton’s 2nd law
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N
T
y
2m
y
T
x
m
2mg
mg
T  mg  ma
2mg sin  T  2ma

N  2mg cos  0
35o
T  mg  a 
2mg sin  m  g  a   2ma
2g sin  g  a  2a
g
g

a  2sin   1  3 2sin35  1  0.049 g
3
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 0.48 m/s2
101
a
y
2m
If θ <30o, a < 0
y
x
m
θ
Write Newton’s 2nd law for the whole system
2mg sin  mg  3ma
2sin   1
a g
3
y
N
T
y
2m
T
x
m
2mg
35o
mg
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Example :
a 2.0 kg cookie tin is accelerated at 3.0 m/s2 in the direction
shown by a , over a frictionless horizontal surface.
• The acceleration is caused by three
horizontal

 forces, only two of which are
shown: F 1 of magnitude 10 N and F 2
of magnitude 20 N. What is the third
force F 3 in unit-vector notation and as a
 magnitude and an angle?



F1  F2  F3  ma ,


 
F3  ma  F1  F2 .
• Along the x axis we have
F3, x  ma x  F1, x  F2, x
 m (a cos 50 )  F1 cos (150 )  F2 cos 90
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F3, x  (2.0 kg) (3.0 m / s 2 ) cos 50   (10 N) cos ( 150  )
 (20 N) cos 90 
 12.5 N
F3, y  ma y  F1, y  F2, y
 m ( a sin 50 )  F1 sin(  150 )  F2 sin 90
 (2.0 kg) (3.0 m / s 2 ) sin 50  ( 10 N) sin ( 150 )
 (20 N) sin 90
 10.4 N

F3  F3, x î  F3, y ĵ  (12.5 N) î  (10.4 N) ĵ
F3 
F3, x  F3, y  16 N
  tan
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2
1
F3, y
F3, x
2
  40
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Example :
• Ali pulled a railroad cars (with his teeth) on his end of the
rope with a constant force that was 2.5 times his body
weight, at an angle  of 30° from the horizontal. His mass
m was 80 kg. The weight W of the cars was 700 kN, and he
moved them 1.0 m along the rails. Assume that the rolling
wheels encountered no retarding force from the rails. What
was the speed of the cars at the end of the pull?
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SOLUTION:
We first find the acceleration of the cars in the x direction :
Fnet ,x  Ma
T cos   Ma
T  2.5 mg  (2.5) (80 kg) (9.8 m / s 2 )  1960 N
The mass M of the cars is :•
W 7.0 x 10 5 N
4
M


7
.
143
x
10
kg
2
g
9.8 m / s
T cos  (1960 N) (cos 30  )
2
a


0
.
02376
m
/
s
M
7.143 x 10 4 kg
v  v0  2 a ( x  x 0 )
2
2
v 2  0  2 ( 0.02376 m / s 2 ) (1.0 m)
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v  0.22 m / s
106
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Example:
A body of mass 5kg lies on a smooth horizontal table. It is connected by a light
inextensible string, which passes over a smooth pulley at the edge of the table, to
another body of mass 3kg which is hanging freely. The system is released from
rest. Find the tension in the string and the acceleration.
Since the string is inextensible and the two bodies are connected, they will both accelerate
at the same rate. If the body hanging freely accelerates at a certain rate, the string will pull
the other body so that it accelerates at the same rate.
The tension in the string on both sides of the pulley will also be the same. This is because
the pulley is smooth.
Remember, W = mg. The mass of the first body is 5 kg and so its weight is 5g N. The mass
of the second body is 3 kg and so its weight is 3g N.
Remember to mark in the normal reaction force, a consequence of Newton's Third Law.
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Example:
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Example:
When two objects with unequal masses are hung vertically over a
light, frictionless pulley as in the figure,. Calculate the magnitude
of the acceleration of the two objects and the tension in the string.
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Example:
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Solution (a) Both blocks must experience the same acceleration
because they are in contact with each other and keep so. Thus we
model the system of both blocks as a particle under a net force.
For the force is horizontal, we have
Fx  F  ( m1  m2 ) a
(1)
F
So the magnitude of the acceleration reads a 
.
(m1  m2 )
(b) It is obviously the contact force exerted on m2 is in the positive
x direction, and the net force acting on the m2 is
F2x  F2c
(2)
From Newton's second law, we get
m2 F
F2x  F2c  m2 a 
.
(m1  m2 )
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Example:
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Example:
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Example:
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Example:
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Example:
Equilibrium and Forces
• It is much more difficult for a gymnast
to hold his arms out at a 450 angle.
• To see why, consider that each arm
must still support 350 N vertically to
balance the force of gravity.
• Use the y-component to find the total force in
the gymnast’s left arm.
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Forces in Two Dimensions
• The force in the right arm must also be 495 N because it also
has a vertical component of 350 N.
• When the gymnast’s arms are at an angle, only
part of the force from each arm is vertical.
• The total force must be larger because the
vertical component of force in each arm must
still equal half his weight.
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Example:
θ = 23.5o
Find the component of gravity acting into the plane, and the
component acting down along the plane:
N = mg cos(θ) = (4.5 kg)(9.81 N/kg) cos(23.5o) = 40.48 N
Fx = mg sin(θ) = (4.5 kg)(9.81 N/kg)sin(23.5o) = 17.60 N
The plane is frictionless, so the component of gravity parallel to the
plane is unopposed. What is the acceleration of the block down the
plane?
F = ma,
-17.60 N = (4.50 kg)a
a = F/m = Fx/m = (-17.58 N)/(4.50 kg) = -3.91 m/s2
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Example:
+
θ = 23.5o
What force would make the block accelerate up the plane at
2.10 m/s2?
N = 40.48 N, Fx= 17.60 N, F = ma,
-17.60 N + F = (4.50 kg)(+2.10 m/s2)
F = 27.053 = 27.1 N
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Example:
+
θ = 23.5o
Suppose it accelerates down the plane at 2.71 m/s2. What
other force is acting on the block? What is the direction?
N = 40.48 N,
Fx = 17.60 N,
-17.60 N + F = (4.50 kg)(-2.71 m/s2)
F = 5.41 N (up the plane)
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F = ma,
128
+
θ = 23.5o
The block starts from rest and accelerates through a
distance of 1.24 m down the plane in .697 s. What other
force must be acting along the plane besides gravity?
N = 40.48 N,
Fx = 17.60 N,
F = ma,
Δx = v t + 1/2at2 , a = 2 Δx /t2 = -5.1049 m/s2
-17.60 N + F = (4.50 kg)(-5.1049 m/s2)
F = -5.3692 N = -5.37 N (down the plane)
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example
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example
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Example
The maximum load that can safely be supported by a rope in an
overhead hoist is 400 N. What is the maximum acceleration that
can safely be given to a 25-kilogram object being hoisted
vertically upward?
Free Body Diagram
Frope=400 N
W = mg = 25 kg * 9.8 m/s2
Frope
a = Fnet/m = (400 N – 245 N)/25 kg
a
= 6.2 m/s2
W
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Example
An object with a mass of 4.0 kg travels with a constant velocity of 4.8
m/s northward. It is then acted on by a force of 6.5 N in the
direction of motion and a force of 9.5 N to the south, both of which
continue even after the mass comes momentarily to rest. (a) How
far will the object travel before coming to rest? (b) What will be its
position 2.5 s after the object comes momentarily to rest?
M = 4 kg, vo = 4.8 m/s,
F2 = 9.5 N south.
F1 = 6.5 N north,
Solve for x when v = 0:
v
2
2
2
o
x = vot + ½ at
or
v = vo + 2ax.
Fnet = ma
6.5 N – 9.5 N = 4 kg x a,
a = -0.75 m/s2
This means that the acceleration is acting against the
initial velocity!
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F1
M
F2
135
’ when v = 0.
0 = (4.8 m/s)2 + 2(-0.75 m/s2)x
X = 15 m
(b) Where will the object be 2.5 s later. We
know its at x =15 with no velocity and -0.75
m/s2 acceleration.
x = vot + ½ at2
= 0 m/s x 2.5 s + ½ (-0.75 m/s2)(2.5 s)2
= -2.3 m,
So relative to its starting point its at x = 13 m
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Example
A 3.50-kg block on a smooth (frictionless) tabletop is attached by a
string to a hanging block of mass 2.80 kg, as shown in the Figure.
The blocks are released from rest and allowed to move freely.
• Find the acceleration of the blocks and the tension in the
string.
+
T
T
Mg
N
T=Ma
NMg=0
mg
Tmg=- ma
mg=ma+T=(m+M)a
a = mg/(m+M)
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Example
I pull a 5 kg mass up with a rope, so that it
accelerates 2 m/s2. What is the tension in the rope?
T = 59 N
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Example
Two blocks sit on a frictionless table. The masses are
M1=2 kg and M2=3 Kg. A horizontal force F=5 N
is applied to Block 1.
1. What is the acceleration of the blocks?
2. What is the force of block 1 on block 2?
F
M1
M2
1. a = 1 m/s2
2. F21= 3 N
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Example
a) Find acceleration
b) Find T, the tension in the string
7 kg
5 kg
a) a = g/6 = 1.635 m/s2
b) T = 57.2 N
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