WE WILL PRACTICE APPLYING NEWTON’S SECOND LAW TO …

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WE WILL PRACTICE APPLYING NEWTON’S SECOND LAW TO SYSTEMS OF FORCES. I WILL SOLVE PROBLEMS INVOLVING TENSION FORCES Newton’s Laws:

Transcript of WE WILL PRACTICE APPLYING NEWTON’S SECOND LAW TO …

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W E W I L L P R A C T I C E A P P L Y I N G N E W T O N ’ S S E C O N D L A W T O S Y S T E M S O F F O R C E S .

I W I L L S O L V E P R O B L E M S I N V O L V I N G

T E N S I O N F O R C E S

Newton’s Laws:

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Warm up -

1. Pick up the warm up worksheet next to the bin. There are 2 sides. Ties go on the same line.

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Review of the formulas we have learned so far:

��⃑�𝐹 = 𝑚𝑚�⃑�𝑎

𝐹𝐹𝑔𝑔 = 𝑚𝑚𝑚𝑚

𝐹𝐹𝐺𝐺 =𝐺𝐺𝑚𝑚1𝑚𝑚2𝑟𝑟2

𝐹𝐹𝑓𝑓𝑓𝑓 ≤ 𝜇𝜇𝑓𝑓𝐹𝐹𝑁𝑁 𝐹𝐹𝑓𝑓𝑓𝑓 = 𝜇𝜇𝑓𝑓𝐹𝐹𝑁𝑁

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Types of Forces

FA – applied force Fg – weight / force of gravity between an object

and the Earth FG – force of gravity between any 2 objects FN – Normal Force Ff – Force of Friction FT – Tension

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Static Equilibrium example 1

1. A 2 kg block is hanging from a string attached to the ceiling. Draw a FBD and determine the tension in the string

2. A 1.25 kg block is attached below the 2 kg block from question (1). Determine the Tension in both strings. (it will be different from question 1)

Presenter
Presentation Notes
19.6 N Top string: 31.85 N bottom string: 12.25 N
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Static equilibrium example 2

Determine the tension in all three of the strings in the diagram to the left.

4.00 kg

1 kg

FT1 FT2

FT3

Presenter
Presentation Notes
FT1= 24.5 N FT2= 24.5 N FT3= 9.8 N
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Static Equilibrium Example 3

1. Draw FBDs for each box.

2. If m1= 4 kg; m2= 2 kg and m3=1.5 kg, what is the normal force exerted by

a) Box 1 on box 2? b) The ground on box 1?

Presenter
Presentation Notes
2. a) 4.9 N b) 44.1 N
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Systems of Forces

When objects are attached by ropes/strings – they accelerate at the same rate.

You can apply ∑𝑭𝑭 = 𝒎𝒎𝒎𝒎 to the entire system or just one piece/object of the system.

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Pulley Example 1

A 3.0 kg mass (m1) and a 1.0 kg mass (m2) are connected by a string. The string is stretched over a pulley. Determine the acceleration of the masses and the tension in the string.

Presenter
Presentation Notes
a= 3.92 m/s/s FT= 13.72 N
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Systems of Forces 1

500 N 500 N 800 N

800 N

How is (a) different from (b)?

Presenter
Presentation Notes
Same forces – different mass of the system a=2.26 m/s/s a= 5.88 m/s/s
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Pulley Example 2

A 1.50 kg mass (m2) is attached to a 3 kg mass on a frictionless surface. Determine the acceleration of the system and the tension in the string.

Presenter
Presentation Notes
a=3.267 m/s/s FT= 9.8 N
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Pulley Example 3

A 1.50 kg mass (m2) is attached to a 3 kg mass on a surface with a coefficient of friction, µ=.225. Determine the acceleration of the system and the tension in the string.

Presenter
Presentation Notes
a=1.797 m/s/s FT=12.006 N
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Homework

Newton’s Laws HW set 2 – quiz on 12/4 for A-days; 12/5 for B-days Answers are online One correction: the homework quiz will require you to draw Free Body Diagrams