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FORCES – SUMARY – REVIEW for TEST
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Inertia is resistance an object has to a change of velocity
• tendency of a body to maintain its state of rest or uniform motion in a straight line
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Mass is numerical measure of the inertia of a body / is a measure of the amount of matter in the object unit: kg
• doesn’t depend on the location of the object . Object of mass of 1 kg here on earth would have the mass
of 1 kg on the moon, even though it would weigh only one-sixth as much.
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Weight is the gravitational force acting on an object .
• W = mg
weight = (mass)(gravitational acceleration)
direction: downward
Force is an influence on an object that causes the object to accelerate.
• Forces are vector quantities, having both direction and magnitude.
• 1 N is the force that forces a 1-kg object to accelerate 1 m/s2.
unit: Newton (N)
unit: Newton (N)
Net force, F
, is the vector sum of all forces acting on an object
net
Free Body Diagram/ Force diagram is a sketch of a body and all forces acting on it.
Newton’s first law: An object continues in motion with constant speed in a straight line (constant velocity)
or at rest unless acted upon by a net external force.
• Newton's 1. Law says that if the net force on an object is zero it won't accelerate.
If
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Fnet = 0, a = 0
 no change in velocity
Translational equilibrium: If an object is at rest, or if it is moving with constant velocity,
then the object is in equilibrium. Net force acting on the object in equilibrium is zero.
Translational equilibrium math:
 if net F = 0 then a = 0, and velocity is constant or zero
 if velocity is constant or zero, then a = 0, and Fnet = 0
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Newton’s second law:
If a net force is acting on an object of mass m, object will acquire acceleration proportional to the net force
and inversely proportional to the mass of the object. Direction of acceleration is direction of the net force.
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If net force is zero, acceleration is zero, velocity is constant (or zero). The object is in equilibrium.
Newton’s third law:
Whenever object A exerts a force on object B, object B exerts an equal in magnitude but
opposite in direction force on object A
FA - force object A exerts on object B
FB - force object B exerts on object A
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Tension
T
We are talking about forces
acting on two different bodies.
is a force that the end of the rope exerts on whatever is attached to it.
Direction of the force is along the rope.
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Normal force
Fn
is the force which is preventing an object from falling through the surface of another body .
That’s why normal force is always perpendicular (normal) to the surfaces in contact.
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Friction force
Ffr
is the force that opposes slipping (relative motion ) between two surfaces in contact;
it acts parallel to surface in direction opposed to slipping.
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Friction depends on type and roughness of surfaces and normal force.
Ffr = μ Fn
μ is called coefficient of friction
• μ has no units
• it is a measure of surface-to-surface roughness
• depends on characteristics of both surfaces
• different values for static and kinetic coefficient of friction (tables)
• kinetic μ is smaller than static μ. You probably noticed that once you moved
something from rest it becomes easier to push around.
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Air Drag and Terminal Velocity
 When an object moves through air or any other fluid, the fluid exerts a frictionlike force on the moving object.
The force is called drag or air resistance or … It depends on the speed of the object, becoming larger as the speed
increases. It also depends upon the size and the shape of the object and the density and kind of fluid.
 Object dropped from the certain height has free fall acceleration initially. As the object accelerates, however,
its speed increases and the drag on it becomes greater and greater until it is equal to the weight of the object.
At that point, the net force on the falling object is zero, so it no longer accelerates. Its speed now remains
constant and is called terminal speed.
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Newton’s Law of Universal Gravitation
The force between two objects is proportional to the product of their masses
and inversely proportional to the square of the distance between their centers.
The force acts along the line joining the two objects.
F=G
m m
r
G = 6.67x10-11 Nm2/ kg2 – “Universal gravitational constant”
the same value anywhere in the universe - very small value – no significant forces of attraction
between ordinary objects.
F=G
m
= mg
g=G
= 9.80 m/s2
R – radius of the Earth is distance between Earth and m
M – mass of the Earth
 Forces are usually divided into two types or classes.
1. Contact forces – arising because of physical contact between objects. For example when you push
on a door to open it or throw or kick a ball, you exert a contact force on the door or ball.
2. Field forces (Fundamental forces) – they act (push or pull) “on distance through space” - the presence of
an object effects the space around it so, and that region is called a field (for example gravitational field of the earth).
Contact Forces
Frictional Force
Tension Force
Normal Force
Air Resistance Force – Drag Force
Applied Force
Spring Force
Field Forces
Gravitational force
Electromagnetic force
Strong nuclear force
Weak nuclear force
attraction between objects due to their masses
attraction or repulsion between charges
keeps nucleus together
arise in certain radioactive processes
At the atomic level – all contact forces are result of repulsive electromagnetic forces (at very small distances)
That means that objects have no actual contact, but their electric fields (outer electrons repel each other)
STEPS:
1. Draw Free Body Diagram/Force diagram (sketch of an object and all forces acting on that object)
2. Label the forces
3. Resolve forces into components
4. Apply Newton’s second law, Fnet = ma, in the x and y direction (in the case of incline,
apply Newton’s second law in the direction along the incline and perpendicular to incline)
5. Solve equations for unknowns.
EXAMPLES:
1. Howard, the soda jerk at Bea’s diner, slides a 0.60-kg root beer from the end of the counter to a
thirsty customer. A force of friction of 1.2 N brings the drink to a stop right in front of the customer.
a) What is the acceleration of root beer?
b) What is the coefficient of kinetic friction between the glass and the counter?
c) If the glass encounters a sticky patch on the counter, will this spot have a higher or lower coefficient of friction?
Vertical direction:
Horizontal direction:
Vertical acceleration = 0
Net force = friction force: Fnet = Ffr =1.2 N
Vertical net force = 0
Fnet = ma
Fn = mg = 6.0 N
1.2 = 0.60 a
Ffr =
a = 2.0 m/s2
Fn
Ffr /
c. higher
= 1.2/6.0 = 0.20 (no units)
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2. A boy exerts a 36-N horizontal force as he pulls a 52-N sled across a cement sidewalk at constant speed. What is the
coefficient of kinetic friction between the sidewalk and the metal sled runners? Ignore air resistance.
W = mg = 52 N
m = 5.2 kg
F = 36 N
Vertical direction:
Horizontal direction:
Vertical acceleration = 0
v is constant,
Vertical F
a = 0 and F
net
=0
net
F = mg = 52 N
=0
F = F = 36 N
n
fr
F =μF
fr
n
F /
fr
= 36/52 = 0.69
3. A force of 40.0 N accelerates a 5.0-kg block at 6.0 m/s2 along a horizontal surface.
a. How large is the frictional force?
b. What is the coefficient of friction?
m = 5.0 kg
F = 40.0 N
a = 6.0 m/s2
Vertical direction:
a = 0, so Fnet = 0
Fn = mg = 50 N
→
horizontal direction:
F
net
Ffr = μ Fn = 50 μ
a = 6.0m/s2
= ma
F – F = ma
fr
40.0 – F = 30
fr
F = 10 N
fr
Ffr = μ Fn
Ffr /Fn = 10/50 = 0.2
4. Luke Skywalker starts to pull a sled with Princess Leia across a large ice pond with the force of 100 N at an angle of
30.0° with the horizontal (with nails on his shoes). Find normal force and initial acceleration if the weight of sled and
Princess Leia is 800 N and the friction force is 40 N.
free body diagram
components
 = 300
mg = 800 N
m = 80 kg
vertical direction :
F sin θ + Fn = mg
50 + Fn = 800
Fn = 750 N
Horizontal direction:
F cos θ – F = ma
fr
86.6 – 40 = 80 a
a = 0.58 m/s2
F = 100 N
Ffr = 40 N
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5. A cute bear, m = 60 kg, is sliding down an iced incline 30 0. The ice can support up to 550 N.
Will bear fall through the ice? If the coefficient of the friction is 0.115, what is the acceleration of
the bear?
m = 60 kg
Ffr = 300 N
θ = 300
g = 10 m/s2
vertical direction :
F sin θ + F
n
Horizontal direction:
= mg
F cos θ – F = ma
fr
50 + F = 800
86.6 – 40 = 80 a
n
F = 750 N
n
6. A 1000-kg elevator is rising and its speed is increasing at 3 m/s2. The tension in the elevator cable is:
a = 3 m/s2 ↑
Fnet = ma
T – 10000 = (1000)(3)
T = 13000 N
7. A monkey of 40 kg is standing on a scale in an elevator moving upward and slowing down at 3 m/s2.
What would scale show (normal force)? How does monkey feel, lighter or heavier?
a = 3 m/s2 ↓
Fnet = ma
Fn – 400 = (40)(-3)
Fn = 320 N
the scale would show less, and monkey would feel lighter
apparent weight < weight
8. A system of two cables supports a 150-N ball as shown.
a) What is the tension in the right-hand cable?
b) What is the tension in the horizontal cable?
y:
T1 sin 300 = 150
T1 = 300 N
x:
T2 = T1 cos 300
T2 = 260 N
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WHEN THERE ARE TWO BODIES YOU HAVE TO DRAW TWO BODY DIAGRAMS !!!!!!
10. Two blocks are connected by a string and pulley as shown. Assuming that the string and pulley are massless, find
a) the magnitude of the acceleration of each block
b) Tension force on the blocks
the same string – the same tension
the same acceleration, except that 110 g accelerate
down, and 90 g accelerate up.
Fnet = ma
a is up
T – mg = ma
T – 0.9 = 0.09a
first equation
Fnet = ma
a is up
mg – T= ma
1.1 – T = 0.11a
second equation
two equations with two unknowns
T – 0.9 = 0.09a
T = 0.09a + 0.9
1.1 – T = 0.11a
T = 1.1 – 0.11a
0.09a + 0.9 = 1.1 – 0.11a ⟹
0.2a = 0.2
2
a = 1 m/s
T = 1.1 – 0.11a = 1.1 – 0.11(1) = 0.99 N
11. A 10-kg block is connected to a 40-kg block as shown in the figure. The surface on which the blocks slide is
frictionless. A force of 50 N pulls the blocks to the right.
a) What is the magnitude of the acceleration of the 40-kg block?
b) What is the magnitude of the tension T in the rope that connects the two blocks?
As these two objects are connected with the same rope, tension is
the same and acceleration is the same for two objects.
Fnet = ma
T = 10a
50 – T = 40a
50 – 10a = 40a
a = 1 m/s2
T = 10a = 10 N