Mr. Boroskyphysics Section 4.3 Notespage 1 of 2

Mr. BoroskyPhysics Section 4.3 NotesPage 1 of 2

Section 4.3 Interaction Forces

Objectives

Define Newton’s third law.
Explain the tension in ropes and strings in terms of Newton’s third law.
Define the normal force.
Determine the value of the normal force by applying Newton’s second law.

Read intro paragraph

IDENTIFYING INTERACTION FORCES

Read section.

A Force is the result of an interaction between 2 objects.

What happens if you are on rollerblades and you push a friend that is also on rollerblades? Your friend will go in the direction in which you pushed him and you will go backwards.

Forces always come in pairs.

Interaction Pair – are 2 forces that are equal in magnitude but opposite in direction. They are also called the Action-Reaction Forces.

NEWTON’S THIRD LAW

Read Section.

Newton’s Third Law – also called the Law of Interaction or the Action Reaction Law. This law states “For every action there is an equal and opposite reaction.” Or in terms of the old book “When one object exerts a force on a second object, the second object exerts a force on the first object that is equal in magnitude but opposite in direction.” Or in terms of the new book “Forces come in pairs that are equal in strength but opposite in direction.

If you put a small force on a ball then it puts a small force on you. If you put a large force on a ball then it puts a large force on you.

These 2 forces are called the Action Reaction Forces.

It is important to keep in mind that an interaction pair must consist of 2 forces equal in magnitude but pointing in opposite directions.

Read p.103 for the example

Do Example 3 p. 104

FEarthonBall = maFBallOnEarth = maF = ma

FEarthonBall = .18(-9.8) FBallOnEarth = .18(9.8)1.764 = 6*1024 a

FEarthonBall = -1.764 N FBallOnEarth = 1.764 N.294 * 10-24 m/s2 = a

2.94 * 10-25 m/s2 = a

Do Practice Problems p. 104 # 28-31

FORCES OF ROPES AND STRINGS

Tension – the specific name for the force exerted by a rope or string.

Go over example with tug of war bottom of p. 105

Do Example 4 p. 106

First find accelerationFnet = FTension – Fg

vf2 = vi2 + 2adFTension = Fnet + Fg

32 = 0 + 2a(3)FTension = ma + mg

9 = 6aFTension = 50(1.5) + 50(9.8)

1.5 m/s2 = aFTension = 75 + 490

FTension = 565 N

Thus the rope is in danger of breaking.

Do Practice Problems p. 106 # 32-33

THE NORMAL FORCE

Read Section.

Normal Force – the perpendicular contact force exerted by a surface on another object. It is equal to the weight of the object. This will be key in the next chapter when dealing with resistance.

If a stone is hung from a rope with no mass, at which place on the rope will there be more tension?

The tension will be same throughout the rope

In a tug-of-war event, both teams A and B exert an equal tension of 200 N on the rope. What is the tension in the rope? In which direction will the rope move? Explain with the help of Newton’s third law.

200 N; the rope will not move since net force = 0

Do 4.3 Section Review p. 107 # 34-39