Newton’s Laws of Motion. HFinks '072 6/2/2015 Basic Concepts Force – push or pull on an object - Vector quantity Mass – amount of matter in a body.
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HFinks '07 204/18/23
Basic Concepts
Force – push or pull on an object
- Vector quantity
Mass – amount of matter in a body.
- Measured with a balance
- Scalar quantity
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First Law
Also known as the law of inertia
An object will remain at rest or continue to move at a constant speed unless acted upon by a net force.
Net force – vector sum of all forces acting on an object.
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Second Law
When a net force acts on a body, the acceleration is directly proportional to the force and indirectly proportional to the mass.
Equation: ∑F = ma or Fnet = ma Net force on an object = mass x acceleration Net = sum of the forces
Units of measurement for force metric system - Newton (N)
metric system - Dyne (dyn)english system - Pound (lb)
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Third Law
For every action, there is an equal and opposite reaction.
Forces occur in pairs.
Net force and acceleration will always be in the same direction.
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Applied Force (Fa)
Pushing, pulling or lifting an object.
Applied force is a vector quantity.
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Normal Force (FN)
Normal force, FN , is a component of the force the surface exerts on an object.
Normal force is always perpendicular to the surface.
Normal force is a vector quantity.
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Weight (W)
Weight – depends on the acceleration due to gravity. - Measured with a spring scale. - Vector quantity - Is a force
- Equation: W = mg - Units of measurement
metric system - Newtonmetric system - Dyneenglish system - Pound
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Weight (W) and Mass (m)
Weight and mass are not the same. Weight changes with a change in acceleration due to gravity. Mass does not change with a change in acceleration due to gravity.Example: Your weight will change if you go from the earth to the moon. Your mass will remain the same.
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Friction
Occurs when objects are in contact Acts opposite to the direction of motion Two kinds of frictional forces
Static (object at rest) = fs
Kinetic (object is moving) = fk
Equations
Static fsmax =µsFN
Kinetic fk = µkFN
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Coefficient of Friction (µ)
Ratio of the frictional force to the normal force. That is, you are comparing one force to another.
Static Kinetic (sliding)
µs = fs µk = fk
FN FN
No unit of measurement because you are dividing a force by a force.
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Kinetic Frictional Force (fk)
The frictional force always act opposite to the applied force.
Think of these two forces as being x-components. One is positive and the other is negative.
Fafk
Direction of motion
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Kinetic Frictional Force (fk)
Example 3: Object sliding down an
inclined plane No applied force
fk
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Static Frictional Force (fs)
- Friction between atoms or
molecules
- No applied force
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Fnet = ma
Applications of this equation
in the x direction
Applied Force (Fa) and Frictional Force (fk)
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Fnet = ma
Applications of this equation
in the y direction
Normal Force (FN) and Weight (W)
Applied Force (FN) and Weight (W)
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Solving Force Problems
1. Label all forces
2. Write an expression for the forces acting in the x direction.
3. Write an expression for the forces acting in the y direction.
4. Substitute the known values into the expressions.
5. Solve for the unknown.
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Example 1
What is the magnitude and direction of FN?
Mass = 20.0 kg
FN
W = mg = 196 N
FN - W = ma
a. Label forces(No forces in x direction)
b. Write expression
c. FN - W = ma FN = W + ma =196 N + (20.0 kg)(0 m/s2) FN = 196 N, upward
ay = 0 m/s2
At rest
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Example 2
A 12.0 kg object is pulled upward by a massless rope with an acceleration of 3.00 m/s2. What is the tension (T) in the rope?
Fa = T
W = (12.0 kg)(9.80 m/s2) = 118 N
T – W = ma T = 118 N + (12.0 kg)(3.00 m/s2) T = W + ma T = 154 N
m = 12.0 kga = 3.00 m/s2
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Example 3
A 12.0 kg object is accelerating downward at 3.00 m/s2. What is the tension (T) in the rope?
Fa = T
W = (12.0 kg)(9.80 m/s2) = 118 N
T – W = ma T = 118 N + (12.0 kg)(-3.00 m/s2) T = W + ma T = 82.0 N
m = 12.0 kga = -3.00 m/s2
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Example 4
A 100. N crate is pulled across the floor. The tension in the rope is 75.0 N. Calculate the maximum frictional force.
(µk = 0.5)
fkFa = T
FN
W = 100 n
a. Label forces
b. Write expressions
a
Fk = µk FN
= (0.5)(100 N)Fk = 50.0 N
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Example 4
Calculate the resultant force (Fnet) in Example 4.
Fk =50.0 N Fa = T = 75.0 N
FN=100. N
W = 100 n
a. Label forces
b. Write expressions
a
Fy = FN – W Fx = T – fk FN = W = 75.0 N – 50.0 NFy = 100. N – 100. N Fx = 25.0 NFy = 0 N
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Example 4
Calculate the acceleration of the object in Example 4.
fk= 50.0 NFa = T= 75.0 N
FN
W = 100. N m = w = 100. N = 10.2 kg g 9.80 m/s2
a. Label forces
b. Write expressionsfor direction ofmotion only.
T – fk = ma a = 75.0 N – 50.0 N a = T – fk 10.2 kg m a = 2.45 m/s2
a
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Example 5
A 20.0 kg object is at rest on an inclined plane
(µs = 0.10) What is the value of fs?
Ө = 10.0º fs = µsFN
fs
FN
WӨ
FN ≠ W
Reason: They aren’t opposite each other
“W” is a vector quantity and is acting at
an angle. Calculate x- and y-
components,Next
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Example 5
W = mg
= 196 N
fs
FN
WxӨ
Wy = W cos Ө
Wx = W sin Ө
Note: Ө is the same for both the large and small right triangles.
Wx will always act “down” the incline.. ( -Wx)
Next
Wy
W Ө
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Example 5
FN = Wy
fs
FN
WxӨ
Fs = µsFN
= µsWy
= µs ( W cos Ө)
= (0.10)(196 N)(cos 10.0º)
fs = 19.3 N
Wy
W Ө
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Example 6
A 20.0 kg object is sliding down an incline at a constant velocity under the influence of gravity. Find the value of fK.
fk
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Example 6
Label Forces and write expressions
fkFN
WxӨ
Object isn’t moving in the y direction. So…..don’t use the y expression: FN - Wy = ma
Object is moving along the incline. Use…. ∑Fx fk - Wx = ma a = 0. Object is moving at a constant velocity.
So….. Fk = Wx
Next
Wy
W Ө
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Example 7
A 20.0 kg object is pulled up a 10.0º incline and µk = 0.05. The object is accelerated at a rate of 1.25 m/s2. What is the value of the applied force?
W = 196 N
fk
FN
WxӨNext
Wy
W Ө
Fa
(i) Fa - fk - Wx = ma Fa = fk + Wx + ma
*’a” is positive. The object is moving up the incline.
fk = µkFN
= µkWy
fk = µkW cos Ө
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Example 7
W = 196 N
fk
FN
WxӨ
Wy
W Ө
Fa
(i) Fa - fk - Wx = ma Fa = fk + Wx + ma
(ii) Fa = (µkW cos Ө) +W sin Ө + ma
= (0.05)(196 N)(cos 10.0º) + (196 N)(sin 10.0º) + (20.0 kg)(1.25 m/s2)
= 9.65 N + 34.0 N + 25.0 N Fa = 68.7 N
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Reviewing Concepts
1. A push or pull is a(n) ______
2. Quantities that require magnitude, direction, and point of origin for their description are ____ quantities.
3. A physical quantity that can affect the motion of an object is a(n) ____.
4. When no net force acts on a body, either no force acts on the body or the ___ of all force acting on the body is ____.
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Reviewing Concepts
5. If there is no net force acting on a body, it will continue in its state of __ or will continue moving along a(n) __line with ___speed.
6. The property of matter that is the concern of the first law of motion is__.
7. A common unbalanced force that makes it difficult to prove Newton’s first law of motion experimentally is ___
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Things to Remember
Weight will not change when you are near the surface of the earth.
The normal force, FN, is changing. It is the scale reading.
Forces occur in pairs. The person exerts a force on the scale and the scale exerts a force on the person.
The inertia of the person wants to remain at rest. The elevator floor and scale will either push up on the person causing an increase in scale reading or drop out from underneath the person causing a decrease in scale reading.
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Think About Your ride on an Elevator
The elevator door closes and You are moving upward Slowing down as it comes to a stop You are moving downward Slowing down as it comes to a stop
Do you feel heavier, lighter or nothing?
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Elevator at Rest
Person standing on scale in elevator.
FN – W = ma
a = 0 m/s2
FN = WW
FN
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Example 8. Elevator at Rest
A 70.0 kg person stands on a scale in an elevator that is at rest. What is the scale reading?
FN = W = mg
= (70.0 kg)(9.80 m/s2)
FN = 686 N
W
FN
3, please
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Elevator Moving Upward and Speeding Up
FN – W = ma
FN = W + ma
W
FN
a
Any brakes on thistub?
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Example 9. Elevator Moving Upward and Speeding Up
Mass of person = 70.0 kgWeight = 686 N
The elevator is accelerating upward at 1.10 m/s2. What is the scale reading?
FN – W = ma
FN = W + ma
= 686 N + (70.0 kg)(1.10 m/s2)
FN = 763 NW
FN
a
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Elevator Moving Upward and Slowing Down
FN – W = m(-a)
FN = W - ma
W
FNa
Whoa. I’m Getting dizzy.
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Example 10. Elevator Moving Upward and Slowing Down
Mass of person = 70.0 kg
Weight = 686 N
The elevator is accelerating downward at 1.10 m/s2. What is the apparent weight?
FN – W = m(-a)
FN = W – ma
= 686 N – (70.0 kg)(1.10 m/s2)
FN = 609 N
W
FNa
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Elevator Moving Downward and Speeding Up
FN – W = m(-a)
FN = W - ma
W
FNa
My goodness. Please stop.
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Example 11. Elevator Moving Downward and Speeding Up
Mass of person = 70.0 kg
Weight = 686 N
A = 1.10 m/s2
FN – W = m(-a) FN = W – ma
= 686 N – (70.0 kg)(1.10 m/s2)
FN = 609 N
W
FNa
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Elevator Moving Downward and Slowing Down
FN – W = ma
FN = W + ma
W
FNa
Yea…fresh air soon
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Example 12. Elevator Moving Downward and Slowing Down
Mass of person = 70.0 kg
Weight = 686 N
FN – W = ma
FN = W + ma
= 686 N + (70.0 kg)(1.10 m/s2)
FN = 763 NW
FNa
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Example 13. Elevator in Free Fall
FN – W = ma
FN = 0 N (No contact)
- W = ma
-mg = ma
-g = a
-9.80 m/s2 = aW
FNa
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Example 14. Applied Force (T) on Elevator
A woman stands on a scale in a moving elevator. Her mass is 70.0 kg and the combined mass of the elevator and scale is an additional 790. kg. Starting from rest, the elevator accelerates upward. During the acceleration, there is a tension (T) of 8500. N in the hoisting cable. What is the reading on the scale during the acceleration?
T –We - Ww = mta a = 8500. N – 7742 N - 686 N (70.0 + 790.)kg a = 0.0837m/s2
FN = W + ma = mg + ma = m(g + a) = 70.0 kg(9.80 + 0.0837)m/s2
FN = 692 N
T
FN
Ww =(70.0kg)(9.80 m/s2) =686 NWe= (790kg)(9.80 m/s2)=7742 N
a
scale
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Reviewing Concepts
8. If an object is stationary, its inertia tends to keep it ___; if an object is in motion, its inertia tends to keep it ___.
9. The acceleration of a body is __ proportional to the mass of the body.
10. The acceleration of a body is in the __ direction as the applied force.
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Reviewing Concepts
11. The acceleration of a body is ___ proportional to the force exerted.
12. If a body is moving in a straight line and a force is applied in the direction of its motion, the body will __ in speed as long as the force continues.
13. If a body is moving in a straight line and a force is applied in the direction __ to its motion, the velocity is changing, the body may stop, and the body may move in the opposite direction.
Applied Force Acting At An Angle with the Horizontal
FN ≠ WAlways calculate and use the x- and y-
components of a vector acting at an angle
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Vector Acting at an Angle
Fa(x) = Fa cos Ө
= 20.0 N cos 20.0º
Fa(x) = 18.8 N
Fa(y) = Fa sin Ө
= 20.0 N sin 20.0 º
Fa(y) = 6.84 N
Ө = 20.0 º
Fa = 20.0 N
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∑Fy
Object Moving to the right FN ≠ W
FN = W – Fa(y)
Why? ∑Fy = FN + Fa(y) – W
FN + Fa(y) – W = ma a = 0 m/s2 in the y direction
So… FN + Fa(y) – W = 0Or FN = W – Fa(y)
Fa(y) = F sin Ө
FN Fa
W
fk
Ө
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∑Fx
Object Moving to the rightFa(x) = F cos Ө
∑Fx = Fa(x) - fk
So….Fa(x) - fk = ma
fk = µkFN
= µk(W – Fa(y))
or = µk(mg - Fasin Ө)
FN Fa
W
fk
Ө
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Example 16
A box weighing 450. N is pulled along a level floor at constant speed by a rope that makes an angle of 30.0º with the floor. If the force on the rope is 300. N, what is the coefficient of sliding friction?
constant speed --- a = 0 m/s2
F cos Ө - µk(W – Fa(y)) = ma
F cos Ө - µk(W – Fa(y)) = 0
F cos Ө - µk(W – Fa(y)) = - F cos Ө
- µk (W – Fa(y)) = - F cos Ө
- - µk = - F cos Ө - (W – Fa(y))
- = 300. N cos 30.0º- 450. N - 300. sin 30.0º
- µk = 0.866
FN Fa
W
fk
Ө
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Example 17
A 20.0 kg box is pulled along a horizontal surface by a force of 40.0 N, which is supplied at a 30.0º angle. The coefficient of kinetic friction is 0.30. Calculate the acceleration.
Fa(x) - fk = ma a = Fa(x) - fk
a = F cos Ө - µk(W – Fa(y)) m
= (40.0N cos 30.0º) –(0.30) [196 N – (40.0 sin 30.0º)]
20.0 kg
34.6 N – 52.8 N = 20.0 kg
a = -.907 m/s2
FN Fa
W
fk
Ө
W = (20.0 kg)(9.80 m/s2) = 196 N
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Reviewing Concepts
14. The maximum frictional force between stationary objects is known as ___ friction, and the frictional force between objects that are sliding with respect to one another is known as __ friction.
15. Starting friction my be (less than, greater than) sliding friction.
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Reviewing Concepts
16. The ratio of the force of sliding friction to the normal force pressing the surfaces together is called ___.
17. If an object moves along a horizontal surface, under the influence of a horizontal force, the normal force pressing the surfaces together is the __ of the object.
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Example 18. Solving for Tension (T) I
∑Fx= -T cos Ө + B = 0
T = B
cos Ө
∑Fy = T sin Ө - W = 0
T = W
sin Ө
T
Ө
WB
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Example 18. Solving for Tension (T) I
Calculate the magnitudes of “T” and “B”. ∑Fy = T sin Ө - W = 0 T = W sin Ө
T = (0.500 kg)(9.80 m/s2) sin 60º
T = 5.7 N
∑Fx= -T cos Ө + B = 0 B = T cos Ө
= 5.7 N cos 60º
B = 2.9 N
T
60º
m = 500 g
B
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Example 19. Solving For Tension (T) II
∑Fx = 0
-T1 cos Ө1 + T2 cos Ө = 0
Solve for T2 and substitute your answer in the statement below.
∑Fy = 0
T1 sin Ө1 + T2 sin Ө2 - W = 0
T1T2
Ө1 Ө2
W
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Example 19. Solving For Tension (T) II
∑Fx = 0
-T1 cos Ө1 + T2 cos Ө = 0
-T1 cos 10º + T2 cos 5º = 0
T2 = T1 cos 10º
cos 5º
T2 = 0.989T1
T1T2
Ө110º Ө2 = 5º
W = 90 N
Next
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Example 19. Solving For Tension (T) II
∑Fy = 0
T1 sin Ө1 + T2 sin Ө2 - W = 0
T1 sin 10º + T2 sin 5º - 90 N = 0
0.174T1+(0.989T1)(0.087) = 90 N
0.174T1 + 0.086T1 = 90 N
T1 = 346 N
T2 = 0.989T1
T2 = 342 N
T1T2
Ө110º Ө2 = 5º
W = 90 N
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Reviewing Concepts Select the correct answer(s)
18. The newton is
(a) the unit of force in the metric
system
(b) is the force required to accelerate
1 kg of mass at the rate of 1 m/s2
(c) is a derived unit,
(d) is 1 kg m/s2
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Reviewing Concepts Select the correct answer(s)19. The force required to accelerate an object of known weight is a. Directly proportional to the weight of the object. b. directly proportional to the acceleration desired. c. Directly proportional to the acceleration due to gravitation d. Not related in any way to the acceleration
due to gravitation.
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Reviewing Concepts Select the correct answer(s)
20. When an automobile is accelerated forward,
a. The tires exert a forward action force on
the road.
b. The road exerts a forward reaction force
on the tires.
c. The tires exert a rearward action force
on the road.
d. The road exerts a rearward reaction
force on the tires.
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