Newton’s Law(s) and Order
Chapter 5.6-5.7
Important Vocabulary:
Normal Force
Contact Force
“It’s not just an equation, it’s the Law!”
Which string breaks first?
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.
Problem 5-9: force and acceleration.
•
•
•
M=950kg. dT=1.20s. V1=16.0m/s
V2=9.5m/s
What is the average Force?
How far did the car travel?
M=950kg, V1, V2 and T=1.2s are given.
What is the average Force?
How far did the car travel?
F=Ma The mass is given, so need to find a.
a is equal to change in velocity over time (definition).
Once have “a”, get distance from master equation.
V2  V1 9.516.0
 1.2  5.42m / s 2
T
F  Ma  (950kg)( 5.42m / s 2 )  5150 N
1
1
x  V1T  aT 2  16.0 *1.20  (5.42)(1.2) 2
2
2
 15.3m
a
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
M1g
M2g
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.
M1 g  FLIFT
M 2 g  FLIFT  M 2 a
M 2 g  M1g  M 2a
a
mp  M1
g a
Mr. Ed the talking horse.
-F
F
?
Paradox? According to Newtons’ laws, Mr Ed says, I can’t
move the cart. If I pull with force F, the wagon pulls back with
an equal and opposite force F. The net force is ZERO, so the
cart doesn’t move.
Mr. Ed Paradox.
1. Trust the horse, the cart doesn’t move.
2. From the perspective of the cart, there is a
net force.
3. From the perspective of the horse, there is
a net force.
4. Both 3 and 4 are correct.
The Normal Force
• “Normal” refers to the direction of the force
• The Normal Force is the contact force due to gravity, acting in the
direction opposite to gravity.
• When an object is moving with constant speed under the influence
of gravity, the normal force equals the force of gravity—the “weight.”
N
F=-Mg
Normal Force is not always equal to the weight
Equal
Not equal
Forces are vectors!
From problem 5-23.
Fx1
Fy1
Fx2
Fy2
Fxtot=Fx1 + Fx1
Fytot=Fy1-Fy2
(NOTE: adding magnitudes)
Forces are vectors, Prob. 5-25.
Given the mass of the
skier, and neglecting
friction, what are the
forces on the skier
(direction and magnitude)?
HINTS ON FINDING EQUAL
ANGLES:
N
Fnet
Mg
1. Parallel line rule.
2. Perpendicular line rule.