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Newton’s Third Law Chapter 7 The 3 rd Law “For every force, there is an equal and opposite force.” The 3 rd Law Runner example: Does the runner push on the earth? Why does the runner move more? Does the earth move at all? Backward force for the earth Forward force for the runner

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the 3 rd law
The 3rd Law

“For every force, there is an equal and opposite force.”

the 3 rd law3
The 3rd Law

Runner example:

  • Does the runner push on the earth?
  • Why does the runner move more?
  • Does the earth move at all?

Backward force for the earth

Forward force for the runner

the 3 rd law4
The 3rd Law

Rocket Example:

  • Equal force up and down
  • Does the rocket move because it pushes against the earth?
  • If that is so, why does a rocket move in space?

Fforward

Fgases

the 3 rd law5
The 3rd Law

Sled of bricks on Ice:

  • Would the sled move?

ICE

the 3 rd law6
The 3rd Law

Why would Cyclops be in trouble?

tension
Tension
  • Flexible cord (can only pull)
  • Exerts a force Ft on the object it pulls
  • Usually neglect the mass of the cord
tension example 1
Tension: Example 1

Two boxes are connected by a cord as shown. They are then pulled by another short cord. Find the acceleration of each box and the tension in the cord between the boxes.

12.0 kg

10.0 kg

Fp= 40.0 N

slide10
First let’s find the acceleration (consider the boxes as one mass)

SF = Fp (this is the only horizontal force)

ma = Fp

a= Fp/m = 40.0 N/(12.0 kg + 10.0 kg)

a=1.82 m/s2

slide11
To find the tension, let’s deal with each box one at a time

SF = Fp – FT

m1a= Fp – FT

FT = m1a – Fp

FT= [(10 kg)(1.82 m/s2) – 40 N

FT= -21.8 N

10.0 kg

FT

FN

Fp= 40.0 N

m1g

slide12
The second box only has a horizontal pull from the tension.

SF = FT

m2a2 = FT

FT = (12.0 kg)(1.82 m/s2)

FT = 21.8 N

12.0 kg

FN

FT

m2g

Note that the sign of the tension varies depending on which box you consider.

tension example 2
Tension: Example 2

Calculate the acceleration of the elevator and the tension in the cable.

slide15
Set up the force equations:

SF = m1a1 = FT – m1g

SF = m2a2 = FT – m2g

Or

m1a1 = FT – m1g

m2a2 = FT – m2g

We have two equations, but three unknowns (FT, a1, and a2)

slide16
However, since the elevator will drop the counterweight will rise:

a1 = -a2

m1a1 = FT – m1g

m2a2 = FT – m2g

Three equations, three unknowns (FT, a1, and a2)

slide17
m1a1 = FT – m1g Substitute a1 = -a2

-m1a2 = FT – m1g Solve for FT

FT = m1g - m1a2

m2a2 = FT – m2g Substitute for FT

m2a2 = m1g - m1a2 – m2g

m2a2 + m1a2 = m1g– m2g

a2(m2+m1)=g(m1-m2)

a2 = g(m1-m2) = 9.8(1150-1000) = 0.68 m/s2

(m2+m1) (1000+1150)

slide18
Since we have solved for a2, it makes most sense to use this equation to find FT

m2a2 = FT – m2g

FT = m2a2 + m2g

FT = (1000 kg)(0.68m/s2) +(1000 kg)(9.8 m/s2)

FT =10500 N

tension example 3
Tension: Example 3

Mr. Fredericks uses a pulley to lift a 200 kg piano at a constant velocity. How much tension does he need to put on the rope?

slide21
SF = 2FT –mg = ma a = 0 (constant vel.)

0 = 2FT –mg Rearrange

FT = mg/2 Substitute

FT =(200kg)(9.8m/s2)/2

FT =980 N (Note how the pulley

doubles my effort force.)

tension example 4
Tension: Example 4

A physics student gets stuck in the mud. In order to get out, she ties a rope to a tree and pushes at the midpoint (Fpush=300 N). If the care begins to budge at an angle of 5o, calculate the force of the rope pulling on the car.

slide23
Note that the tension is always along the direction of the rope, and provided by the tree and the car. Since the car is just starting to budge, we will assume the sum of all the Forces is zero.
slide24
SFx = 0 = FT1x – FT2x

SFy = 0 = Fp - FT1y – FT2y

0 = FT1x – FT2x

0 = Fp - FT1y – FT2y Use trigonometry

0 = FT1cos5o – FT2cos 50

0 = 300N - FT1sin5o – FT2sin5o

slide25
FT1cos5o = FT2cos 5o Rearrange

FT1cos5o = FT2cos 5o

FT1 = FT2 Substitute

0 = 300N – FT2sin5o – FT2sin5o

300N = 2FT2sin5o

FT2= 300N/2sin5o = 1700 N

(Note that she magnified her force almost 6 times!!!!)

friction example 3
Friction: Example 3

In the following setup, the coefficient of kinetic friction between the box and the table is 0.20. What is the acceleration of the system?

m1=5.0 kg

m2=2.0 kg

slide27

FN

Ffr

FT

Box one does not move vertically, so we can just worry about the horizontal forces.

Ff r= (0.20)(5.0 kg)(9.8m/s2) = 9.8 N

SF=FT – Ffr

m1a = FT - 9.8 N

m1g

slide28

FT

In this case, the downward direction is considered positive since that is the direction of motion. That way a is the same for both boxes.

Box two only moves vertically.

SF=m2g - FT

m22 = (2.0 kg)(9.8 m/s2) - FT

m2a = 19.6 N - FT

m2g

slide29
m1a = FT - 9.8 N Two eqns, two

m2a = 19.6 N – FT unknowns

FT = m1a + 9.8 N

m2a = 19.6 N – m1a - 9.8 N

m2a + m1a = 9.8 N

a = 9.8 N/(m2+m1) = 9.8 N/7.0 kg = 1.4 m/s2

slide30
We can also calculate the tension:

FT = m1a + 9.8 N

FT = (5.0 kg)(1.4 m/s2) + 9.8 N

FT = 17 N

tension ex 4
Tension: Ex. 4

A 90.0 kg mountain climber climbs from the ropes as shown. The maximum tension that rope 3 can hold is 1500 N before it breaks. Calculate the maximum angle of q.

Rope 1

Rope 3

q

Rope 2

tension ex 5
Tension: Ex. 5

A 200 kg stage set is lifted down by a 100 kg stagehand as shown. Calculate the stagehand’s acceleration.

Ans: 3.27 m/s2

200 kg

100 kg