newton’s law(s) and order chapter 5.6-5.7 important vocabulary: normal force contact force...
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Newton’s Law(s) and Order
Chapter 5.6-5.7
Important Vocabulary:Normal ForceContact Force
“It’s not just an equation, it’s the Law!”
Getting hammered?
1. Left side picture is better.2. Right side picture is better.3. Neither one has an advantage, the
forces are the same.
Which picture describes the better approach for tightening a loose hammer-head?
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?
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.
m
aTTVx
NsmkgMaF
smT
VVa
3.15
)2.1)(42.5(2
120.1*0.16
2
1
5150)/42.5)(950(
/42.5
221
2
22.10.165.912
M=950kg, V1, V2 and T=1.2s are given.What is the average Force?How far did the car travel?
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.
• Givens: M1 and a
• Asks: Mp
F1
Flift Flift
F2
a = -0.56 m/s2
M1=1220kgM2=M1+Mp
Prob. 63: Solution
M1g
Flift Flift
M2g
a = -0.56 m/s2
M1=1220kgM2=M1+mp
Initially, acceleration is zero, so force of gravity is equal to force of lift.
LIFTFgM 1
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.
ag
aMm
aMgMgM
aMFgM
p
LIFT
1
212
22
Mr. Ed the talking horse.
?
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.
F-F
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.”
F=-Mg
N
Forces are vectors!From problem 5-23.
Fx1
Fx2
Fy1
Fy2
Fxtot=Fx1 + Fx1
Fytot=Fy1-Fy2
(NOTE: adding magnitudes)