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Force is part of an interaction

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Presentation on theme: "Force is part of an interaction"— Presentation transcript:

1 Force is part of an interaction
Force is push or pull It is an interaction between one thing and another. Example; An apple exacts a force on someone’s palm (force of gravity). The person’s palm exacts an upward force of reaction. There are therefore two forces acting in an interaction.

2 Newton’s 3rd law of motion
Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first object. One force is called Action Force The other is called Reaction Force Neither force exist without the other. Newton’s 3rd law can be summarized; To every action there is an equal opposite reaction.

3 Action = reaction If these two individuals are not in motion, which of the two is pulling with a bigger force?_______

4 Pairs of Action and Reaction
Event Pairs of objects in an interaction Object exacting the Action force Object exacting the Reaction force Moving car Car tires and the Road Tires push road backwards Road push tires forward Firing a rocket Rocket and Gas Falling boulder Earth and the Boulder Firing a cannon Cannon and cannon ball Hitting a nail with a hammer Nail and hammer Hammer on nail Nail on hammer

5 Action –reaction pairs do cause motion
If action equal reaction, how can motion be caused by such pairs of forces? Explain the question above by showing how Michelangelo's assistant is able to accelerate. (pg80) How the sled is able to move forward.

6 Action and reaction Vs acceleration?
How do equal forces (Action and reaction) produce unequal acceleration? Because force is applied on objects with unequal masses. The cannon has a large mass m, The cannon ball has a small mass m The action force F is equal to reaction force F. Therefore; acceleration of cannon is smaller than that of the cannon ball.

7 4-6: Weight-The force of Gravity and The Normal Force
Consider the forces acting on this box FN= Since the box is at rest, the Net force is equal to zero. Therefore, ∑F= FN-mg= 0 Which means FN=mg Here, Net force is equal to weight. FG=mg

8 Action reaction forces acts on different objects
Why are weight and normal force not considered action reaction pairs. (pg 81)

9 When is Normal force not equal to weight?
Consider a box that is at rest on a table and another force is acting on it from the top as shown. If M1=10 kg, and M2= 20 kg, calculate FN1 and FN2. ∑F= FN1-m1g= 0 ∑F= FN2-m2g-m1g= 0 FN2 FN1 m1g m2g

10 class work and Homework
Textbook page 97 #

11 Elevator problem Dave stands on a bathroom scale in an elevator . What is Dave's mass if the scale reads 720 N while the elevator accelerates upward at 2.2 m/s2 Use the formula Net force = apparent weight – actual weight

12 answer ∑F= FN- mg Ma = 720 - mg 720 = ma + mg Ma +mg = 720N
m (a +g) = 720N m ( ) = 720N m (12) = 720 m = 720/12 m= 60 kg Net Force (ma) Normal force (Action) Apparent Weight (reaction) Actual weight (mg)

13 Apparent weight loss A 65 kg woman descends in an elevator that briefly accelerates at 0.20 g downward. If she’s standing on a scale, what is it reading? since ∑F= ma then ∑F= mg-FN What does the scale read if she descends at a constant speed of 2 m/s

14 Example A 10 kg bucket is lowered vertically by a rope in which there is 150 N tension. What is the acceleration of the bucket and is it going up or down? Solution: Choose up to be the positive direction. Write Newton’s 2nd law for the vertical direction, and solve for the acceleration. ∑F= FN- mg

15 Class work Textbook pg 98 # 7-12, 15-18

16 Friction (textbook pg 91)
What are the two types of friction?

17 Coefficient of Kinetic friction.(μk)

18 Example (4-16) A box of mass 10 kg rests on a horizontal force. The coefficient of static friction is 0.4, and the coefficient of sliding friction is 0.3. determine the force of static friction Ffr if an external force of 10 N is applied, will the box move? How much force is required to cause the box to move?

19 Example 4-19 A 10 kg box is pulled along a horizontal surface by a force Fp of 40 N applied at an angle of 300. Calculate the acceleration f the box if the surface has a coefficient of kinetic friction of 0.3. 40 N 300 10 kg

20 Class work Workbook pg # 7-11 Workbook pg 47, # A6-A8

21 Class work Textbook pg 99 # 21-25,


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