newton’s laws of motion chapter 4. force force newton stated that the change in velocity of an...
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Newton’s Laws of Motion
Chapter 4
Force
FORCE Newton stated that the change
in velocity of an object is caused by FORCES.
When the velocity of an object is constant, or if the object is at rest, it is said to be in equilibrium.
FORCE Contact forces: forces that result
from physical contact between two objects. Examples: friction, a push or a pull
Field forces: forces that can act at a distance. Examples: magnetism, gravity
FORCE DIAGRAMS Force is a vector.
Force diagrams – show forces vectors as arrows
Free body diagrams – shows only the forces acting on a single object
Newton’s First Law
Newton’s First Law (Law of Inertia)
A body at rest will remain at rest, a body in motion will remain in motion, traveling with a constant velocity in a straight line, unless an unbalanced force acts on it.
INERTIA = a measure of a body’s ability to resist changes in velocity. INERTIA ≈ MASS (the greater the mass of a body, the
less it will accelerate under the action of an applied force)
It’s easier to push a Volkswagen than a Mack Truck
But, once it’s moving, it’s harder to stop the truck
That’s because the Mack Truck has more
mass, and thereby more inertia.
Warm Up 2/25• If you are standing on a scale
in an elevator and the elevator is accelerating upwards, would the scale show your true weight?
• What if it was travelling at a constant velocity upwards?
• What if it was accelerating downwards?
• Explain your reasoning.
Newton’s Second and Third Laws
Newton’s Second LawF = ma
The acceleration of an object is directly proportional to the resultant force acting on it and inversely proportional to its mass.
The direction of the acceleration is in the direction of the resultant force.
The SI unit of force is the NEWTON. 1N = 1 kg•m/s2
Weight: w = mg
Any acceleration requires a force.
Q
Acceleration can be…
Newton’s Third Law
If two bodies interact, the force exerted on a body 1 by body 2 is equal in magnitude, but opposite in direction to the force exerted on body 2 by 1.
OR…. For every action there is an equal and opposite reaction.
Action Reaction Pairs
Ball being thrown
Car hitting wall
Elevator pulled up by cable
http://ed.ted.com/lessons/joshua-manley-newton-s-3-laws-with-a-bicycle
Implicit in these laws are the following ideas.
• If no unbalanced force acts on a body, its acceleration must be zero.
• An unbalanced force is one whose vector sum does not equal zero.
Implicit in these laws are the following ideas.
If an unbalanced force acts on a body, it must accelerate. It will continue to accelerate for as long as the force(s) are unbalanced.
Implicit in these laws are the following ideas.
If a body has no acceleration, the vector sum of all the forces acting on it must be zero.
What is the vector sum of all the
forces acting on this point?
Zero
A 1 kg rock is thrown at 10 m/s straight upward. Neglecting air resistance, what is the net force that acts
on it when it is half way to the top of its path?
1. -9.8 N 2. -9.8/2 N 3. 0 N 4. -9.8/4 N
Everyday Forces
Weight vs Mass
The weight of an object is the force of gravity acting on the mass of that object. Measured in pounds in the USA Measured in Newtons elsewhere (and
in this class) Weight = mg or (mass) x (9.8m/s2)
On Earth
Normal Force
Normal force: the upward force acting on an object on a surface; the normal force is perpendicular to the surface on which the object is sitting. W = mg
N
In this case, N = mg;But not always
Tension
Tension (T) is a force acting on a rope. When a rope is taught there is tension in the rope.
Friction
Friction (Ff) is a resistive force that opposes motion.
How to draw a FBD
All the forces on one object are drawn as vectors (arrows)
The tail of the arrow starts on the object and point off in the direction of the force.
If the force is at an angle, include the angle.
Example: A car hits a wall. Draw the FBD of the car
Normal Force (N)
Weight (mg)
Force from wall (Fwall)
Multi-Object Force Diagrams
Draw all objects as boxes. Create a table of interactions
within the system. Using the table of interactions,
draw the force pairs acting on all objects.
Example 1: Car hits wall
Wall Car
Earth
Car Wall Earth
Car 0
Wall 0
Earth
0
+
+ +
+
++
Example 2: Elephant vs. Man
Elephant Rope
Earth
Man
Eleph
Rope
Man Earth
Eleph
0
Rope
0
Man 0
Earth
0
++ -++
-+++
+ + +
Example 3: Hanging Box
Rope
Box
Earth
Ceiling Ceiling
Rope Box Earth
Ceiling
0
Rope 0
Box 0
Earth 0
++ -
++-
+++
+ ++
How to solve problems with Newton’s Laws
Draw the picture or Free Body Diagram
Label ALL forces (in x and y direction)
Pick +X (or +y) to be in the direction of acceleration
Sum the forces in each direction Solve
Example 1: A car engine will push a 2500 kg car forward with 800. N of force on flat ground. If the air resistance is 300. N, what is the acceleration of the car? X Y
Fcar
Fair
w
N
800.N300.N
N
mg
Fx = max
800. - 300. = (2500kg)ax
500. = (2500kg)ax
ax = 0.20m/s2
800. 0
-300. 0
0 -mg
0 N
m = W/g = 8800/9.8
= 898 kg
Example: What is the tension in the cable of an elevator with a weight of 8800 N that ascends with an acceleration of 1.30 m/s2?
T
W=mgW=8800N
0 T
0 -8800
X Y
Fy = may
T - 8800 = (m)(1.30)
T - 8800 = (898kg)(1.30)
T = 9,967 NT = 1.0 x 104 N
A coach pulls a 35 kg bag of soccer balls across the field with a force of 84 N directed at an angle of 35o above the horizontal. If the bag travels at a constant velocity, (a) how strong are the resistive forces from the field on the bag and (b) how strong is the normal force on the bag?
N
mg
Fr = ?
84 N
35o
Fx = max
68.81 – Fr = 0
Fr = 68.81 N
Fr = 69 N
84cos35o
= 68.8184sin35o
= 48.81
0 N
0 -mg
- Fr 0
X Y
84cos35o
= 69.6384sin35o
= 48.18
0 N
0 -mg
- Fr 0
X Y
A coach pulls a 35 kg bag of soccer balls across the field with a force of 84 N directed at an angle of 35o above
the horizontal. If the bag travels at a constant velocity, (a) how strong are the resistive forces from the field on the bag and (b) how strong is the normal force on
the bag?Fy = may
48.75 + N - mg = 0
N = 294.82 N
48.18 + N – (35)(9.8) = 0
48.18 + N - 343 = 0
N = 290 N
N
mg
Fr = ?
84 N
35o
Example: A child holds a sled at rest on a frictionless snow-covered hill. If the sled weighs 100. N, find the force the child must exert on the rope and the force the hill exerts on the sled.
40
40
W=100N
Tn
40
W=100N
TnX Y
0 N
T 0
- 100sin40o
= - 64.28- 100cos40o
= - 76.60
Example: A child holds a sled at rest on a frictionless snow-covered hill. If the sled weighs 100. N, find the force the child must exert on the rope and the force the hill exerts on the sled.
Fx = max
T – 64.28 = 0
T = 64.28 = 64.3 N
Fy = may
N – 76.60 = 0
N = 76.60 = 76.6 N
Example: A traffic light weighing 100. N hangs from a cable tied to two other cables fastened to a support. The upper cables make angles of 37 and 53 with the horizontal. Find the tension in each of the three cables.
53T2
T3
37T1
T3
W
T1 T2
T3=W
37 53
Example: A traffic light weighing 100. N hangs from a cable tied to two other cables fastened to a support. The upper cables make angles of 37 and 53 with the horizontal. Find the tension in each of the three cables.
T3
W
T1 T2
T3=W
37 53
X y
T1
T2
T3
-T1cos37
T1sin37
T2cos53 T2sin53
0 -100 N
Example: A traffic light weighing 100. N hangs from a cable tied to two other cables fastened to a support. The upper cables make angles of 37 and 53 with the horizontal. Find the tension in each of the three cables.
x y
T1
T2
T3
-T1cos37 T1sin37
T2cos53T2sin53
0 -100 N
Fx = 0 = T2cos53-T1cos37T2cos53 = T1cos37
T2 = T1(cos37/cos53)
Fy = 0 = T1sin37+T2sin53-100
100 = T1sin37+1.33T1sin53
100 = T1(.602)+T1(1.06)
T1= 60.2 N
T2 = 1.33T1
Example: A traffic light weighing 100. N hangs from a cable tied to two other cables fastened to a support. The upper cables make angles of 37 and 53 with the horizontal. Find the tension in each of the three cables.
T3 = W = 100N
T1= 60.2 N
T2 = 1.33T1
T2 = 1.33(60 N)T2= 80.1 N
T3= 100 N = 1.00 x 102 N
Forces of Friction
The term friction refers to the resistive forces that arise to oppose the motion of a body past another with which it is in contact.
Forces of Friction
Kinetic friction (sliding friction) is the frictional resistance a body in motion experiences.
Static friction is the frictional resistance a stationary body must overcome in order to be set in motion.
Ff=μN
Where μ = coefficient of friction N = normal force
The magnitude of μ depends on the nature of the surfaces in contactNOTE: friction does NOT depend on the area of contact!
Example: You need to move a box of books into your dormitory room. To do so, you attach a rope to the box and pull on it with a force of 90.0 N at an angle of 30.0.
The box of books has a mass of 20.0 kg, and the coefficient of friction between the bottom of the box and the hallway surface is 0.5. Find the acceleration of the box.
W=mg
n
30
T=90 N
Fkf
X Y
90.0sin30o
=90.0cos30o
=0
0
0
N
-mg
- Ff
Example: You need to move a box of books into your dormitory room. To do so, you attach a rope to the box and pull on it with a force of 90.0 N at an angle of 30.0. The box of books has a mass of 20.0 kg, and the coefficient of friction between the bottom of the box and the hallway surface is 0.500. Find the acceleration of the box.
W=mg
n
30
T=90 N
Fkf
Fy = 0 = n + 90sin30-(20)(9.8)
Fkf = μn = (0.5)(151 N)
n = 151 N
Fx = max = 90cos30-75.5
ax = +0.122 m/s2
Fkf = 75.5 N (to the left)
(20)(ax) = 90cos30-75.5
Example: A hockey puck is given an initial speed of 20.0 m/s on a frozen pond. The puck remains on the ice and slides 120. m before coming to rest. Determine the coeficient of friction between the puck and the ice.
W=mg
nd=120 m
Fkf
Example: A hockey puck is given an initial speed of 20.0 m/s on a frozen pond. The puck remains on the ice and slides 120. m before coming to rest. Determine the coeficient of friction between the puck and the ice.
W=mg
nd=120 m
Fkf
vf2 = vi
2 + 2ad
a = -1.67 m/s2
vi = 20 m/svf = 0 m/s
d = 120 m
0 = (20)2 + 2(a)(120)
Example: A hockey puck is given an initial speed of 20.0 m/s on a frozen pond. The puck remains on the ice and slides 120. m before coming to rest. Determine the coeficient of friction between the puck and the ice.
W=mg
n
d=120 m
Fkf
Fy = 0 = n - W
Thus, Fkf = μn =μmg
n = W = mgFx = max = -Fkf
μ = 0.170
Fkf = μmg = 9.8μm
(m)(-1.67) = -(9.8)(μ)(m)
Example: A 4.00-kg object is connected to a 7.00-kg object by a light string that passes over a frictionless pulley. The coefficient of sliding friction between the 4.00-kg object and the surface is 0.300. Find the acceleration of the two objects and the tension in the string.
4 kg
7 kg
m2g
N
m1
g
T
T
4.00 kg object:Fx = ma = T - Fkf
Fy = 0 = n – mg
4a = T - Fkf
n = 39.2 N
n = (4.00 kg)(9.8m/s2)
Fkf
Example: A 4.00-kg object is connected to a 7.00-kg object by a light string that passes over a frictionless pulley. The coefficient of sliding friction between the 4.00-kg object and the surface is 0.300. Find the acceleration of the two objects and the tension in the string.
4 kg
7 kg
m2g
N
m1
g
T
T
Fkfa = 5.16 m/s2
Example: A 4.00-kg object is connected to a 7.00-kg object by a light string that passes over a frictionless pulley. The coefficient of sliding friction between the 4.00-kg object and the surface is 0.300. Find the acceleration of the two objects and the tension in the string.
4 kg
7 kg
m2g
N
m1
g
T
T
FkfT = 32.5 N
a = 5.16 m/s2
Multiple Objects Hooked Together
• Treat it as a whole system together.
• Add all the masses up and use that as the total mass.
• Then just look at the forces on the outside.
Example: Three boxes are tied together by light strings as shown. They are then pulled along a horizontal floor with a force of 150 N directed parallel to the floor. If each box has a mass of 10.0 kg and experiences a frictional force of 25 N, what is the total acceleration of the system?
10.0 kg 10.0 kg 10.0 kg 150 N
25 N25 N25 N
F = mtotala
150 – 25 – 25 - 25 = (30.0)a
a = 2.5 m/s2
Air resistance & Terminal Velocity
Air resistance or Drag
Air resistance is similar to friction in that it opposes the direction of motion.
Air resistance is the resistance of the air that the object is falling through. When the object is moving slowly, it is not
moving through much air so the air resistance is small.
The faster the object moves, the more air it travels through and the larger the air resistance is.
Weight (mg)
Air Resistance
Weight (mg)
Air Resistance
When the forces are balanced, acceleration
stops.
Terminal velocity
weight
Air Resistance
The forces are balanced. That is, the vector sum is equal to zero.
Free Fall Motion
About Forces
Common misconceptions about forces
When a ball has been thrown, the force of the hand that threw it remains on it for awhile. NO! The force of the hand is a contact
force; therefore, once contact is broken, the force is no longer exerted.
Common misconceptions about forces
Even if no force acts on a moving object, it will eventually stop.
Weight and mass are two names for the same thing.
Air does not exert a force. NO. Air exerts a huge force, but because it
is balanced on all sides, it usually exerts no net force unless an object is moving.