Transcript Chapter05_1-4_FA05

Newton’s Laws
The dark side of the force(s)……
• Objects change velocity only when they are
accelerated
• Acceleration happens when an object is acted on by
a force
• The acceleration is proportional to the force, and
inversely proportional to the mass (F=ma)
• Forces act in equal and opposite pairs
A book is held stationary in place. Which of
the following statements is most correct?
1.
2.
3.
4.
5.
The book is stationary, there are no
forces acting on the book.
The book exerts a downward force
on my hand.
My hand exerts an upward force on
the book.
The force of the book on my hand is
equal and opposite to the force of
my hand on the book.
Number 1 is false; 2,3 and 4 are all
true.
Forces on a Bicycle
• Bicycle accelerating to the right.
• Wheels are turning clockwise.
• What is the force of the road on the bicycle?
?
?
What is the force of the road on the
bicycle?
?
?
1. To the left.
2. To the right.
3. There is no left or right
force, only up.
Examples of Action-Reaction Force Pairs
FORCE PAIRS are also called CONTACT FORCES.
The total force is the same. Box 1 is twice as heavy as Box 2.
Does the contact force change?
?
?
Compare the contact forces in the two cases.
1. The total force is the same. By
symmetry, the contact forces
are the same.
2. The contact forces are higher
in the top diagram.
3. The contact forces are higher
in the bottom diagram.
The contact forces matter!
• Same force, different result!
• Bottom case breaks the egg, due to high contact forces.
Free body diagrams
• Two forces act on box 1 (neglecting friction)
• One force acts on box 2
What are the forces?
•
•
•
•
First step: Use F=Ma for TOTAL MASS to find acceleration, a.
Second step: Use F=Ma for MASS #1 to find Force on Box #1.
Third step: Use “equal and opposite” to find Force on Box #2.
Fourth step: Add up forces on Box #2.
Finding the forces.
•
•
•
•
First step: Use F=Ma for TOTAL MASS to
find acceleration, a.
Second step: Use F=Ma for MASS #1 to
find Force on Box #1.
Third step: Use “equal and opposite” to
find Force on Box #2.
Fourth step: Add up forces on Box #2.
M TOT  M 1  M 2  15kg
FTOT
20.0 N
a

 1.33m / s 2
M TOT 15.0kg
F1  M 1a  10.0kg 1.33m / s 2  13.3 N
F2  13.3 N
FBox2  FTOT  F2  20.0  13.3  6.67 N
Reaction Forces & Acceleration
Compare the ACCELERATION of
the Right Side boat to the Left.
1. They are equal because of
“equal and opposite forces”.
2. The right boat has higher
acceleration.
3. The right boat has lower
acceleration.
Force of gravity.
• Gravity (approximately) is a constant
acceleration, g, independent of mass.
• Therefore, the FORCE of gravity is proportional
to the mass (Galileo). Fgravity = -Mg
F = +mg
F = -mg
F = -mg
Tower of Babel
• All of the blocks have the same mass, M. They sit on top
of a table.
• What is the total force on the stack of blocks?
• What is the force of the table on the stack of blocks?
• What are the contact forces?
What is the total force on the Tower?
Each block has mass M.
1.
2.
3.
4.
0Mg
1Mg
2Mg
3Mg
What are the contact forces?
F_c
F_b
F_a
F_a, F_b, and F_c are equal to:
1. 1Mg, 2Mg, 3Mg
2. 3Mg, 2Mg, 1Mg
3. 2Mg, 1Mg, 0Mg
4. 0Mg, 1Mg, 2Mg
Problem 65: Addition of Forces
• When two people push in the same
direction on mass m, they cause an
acceleration a1. When they push in
opposite directions, the acceleration is a2.
• What is the force of each person in terms
of the given information?
•Step 1: DRAW A PICTURE!
•Step 2: “Givens” and “asks”
•Step 3: Relationships (F=ma)
Problem 65: solution
F1
M
F1  F2  Ma1
F2
a1
F1  F2  Ma2
F1
F2
Q: Find F1, F2.
M
a2
How do you complete the problem?
Hint: Add and subtract the equations.
Problem 65:
Given a1 and a2, the ratio F2/F1 is:
1.
2.
3.
4.
a2/a1
a1/a2
(a1-a2)/(a1+a2)
(a1+a2)/(a1-a2)
Problem 63: Hot air balloon
• A balloon with some passengers hovers
motionless at a total mass of 1220kg. A
last passenger climbs aboard, and the
balloon sinks at 0.56 m/s2.
• What was the mass of the last passenger?
•Step 1: DRAW A PICTURE!
•Step 2: “Givens” and “asks”
•Step 3: Relationships (F=ma)
Prob. 63: Setting it up.
Flift
Flift
a = -0.56 m/s2
F1
F2
M1=1220kg
• Givens: M1 and a
• Asks: Mp
M2=M1+Mp
Flift
Prob. 63: Solution
Flift
a = -0.56 m/s2
F1
F2
M1=1220kg
M2=M1+Mp
Initially, acceleration is zero, so
force of gravity is equal to force of
lift.
We know the force of gravity
initially, since the mass is given.
After the last passenger loads, the
difference between the force of gravity F2
and the lifting force leads to an
acceleration, a.
F1  FLIFT
F1  FLIFT  M1 g
F2  FLIFT  M 2 a
F2  M 2 g
M 2 g  M1g  M 2a
Getting hammered?
Which picture describes the
better approach for tightening
a loose hammer-head?
1. Left side picture is better.
2. Right side picture is
better.
3. Neither one has an
advantage, the forces are
the same.