Newtons Third Law

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# Newtons Third Law - PowerPoint PPT Presentation

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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### Newton’s Third Law

Chapter 7

The 3rd Law

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

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 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 3rd Law

Sled of bricks on Ice:

• Would the sled move?

ICE

The 3rd Law

Why would Cyclops be in trouble?

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

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

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

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

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

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

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)

a1 = -a2

m1a1 = FT – m1g

m2a2 = FT – m2g

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

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)

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

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?

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

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.

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.
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

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

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

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

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

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

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

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

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