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Whenever one object exerts a force on a second object the second object exerts a force equal in strength, BUT opposite in direction back on the first object.

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Presentation on theme: "Whenever one object exerts a force on a second object the second object exerts a force equal in strength, BUT opposite in direction back on the first object."— Presentation transcript:

1 Whenever one object exerts a force on a second object the second object exerts a force equal in strength, BUT opposite in direction back on the first object.

2 Action/Reaction? Action and Reaction forces act simultaneously. It is often hard to identify the pair of action and reaction forces. Remember that Action: Object A exerts a force on object B. Reaction: Object B exerts a force (same strength, and type) on object A in the opposite direction. Identify the action-reaction pairs in following situations: Driving a car swimming falling boulder.

3 Driving A car’s wheels spin so that they rub the road’s surface creating a friction force pointing to the rear of the car. The road “reacts” by exerting a friction force of equal magnitude on the car wheels in the forward direction. Friction Force from the car pushing on the road Friction Force from the road pushing on the car

4 Swimming A fish’s tail moves so that it exerts a contact force on the water, pushing the water backwards. The water “reacts” by pushing on the fish’s tail with a contact force of equal magnitude, but opposite direction making the fish move forward. Contact force of the fish on the water Contact force of the water on the fish

5 Falling Boulder The force of gravity from the planet pulls downward on the boulder causing it to fall. The falling boulder reacts by pulling upward on the Earth (trying to “lift it”) with a gravitational force. F g of planet on boulder F g of boulder on planet

6 Visualizing Newton's 3 rd law Whenever something places a force on another object, that object being pushed exerts a force back onto the thing that is pushing it. The Box exerts a force of gravity pushing down on the floor The floor pushes back on the box just as hard

7 A person pushes a block with a force The block pushes back on the person just as hard The Box exerts a Force of gravity Pushing down on the floor The floor pushes back on the box just as hard Smooth floor The Person exerts a Force of gravity Pushing down on the floor The floor pushes back on the person just as hard X X X X

8 Newton’s 3 rd law and accelerations Newton’s third law states that the forces between two objects pushing on each other are the same in size, however it does not state anything about acceleration. The accelerations of the these objects is governed by Newton’s 2 nd Law, not the 3 rd law.

9 Smooth floor A 50kg person pushes on a 20 kg box with a force of 10 Newtons. If both are standing on a smooth (frictionless) surface. What are the accelerations of the two objects? A person pushes a block with a force of +10 N The block pushes back on the person just as hard with -10 N The Box exerts a Force of gravity Pushing down on the floor Fg = (20kg)*(-9.8m/s 2 ) The floor pushes back on the box just as hard F floor = +196 N The Person exerts a Force of gravity Pushing down on the floor Fg = (50kg)*(-9.8m/s 2 ) The floor pushes back on the person just as hard F floor = +490 N Fg = -196 NFg = -490 N F floor = +490 N F floor = +196 N +10 N -10 N Step 1: Draw a Free Body Diagram

10 Smooth floor A 50kg person pushes on a 20 kg box with a force of 10 Newtons, both standing on a smooth (frictionless) surface. Fg = -196 NFg = -490 N F floor = +490 N F floor = +196 N +10 N -10 N Step 2: Look at each object separately

11 Smooth floor A 50kg person pushes on a 20 kg box with a force of 10 Newtons, both standing on a smooth (frictionless) surface. Fg = -490 N F floor = +490 N -10 N Step 2: Look at each object separately The forces for the Y axis cancel F net Y = F floor + F g = ma Y F net Y = +490 N + (-490 N) = (50 kg)a Y F net Y = 0 N= (50 kg)a Y This leaves only one force along the X axis. a Y = 0 m/s 2 F net X = F right + F Left = ma X F net X = 0N + -(10N) = (50 kg)a X a X = -(1/5) m/s 2

12 Smooth floor A 50kg person pushes on a 20 kg box with a force of 10 Newtons, both standing on a smooth (frictionless) surface. Fg = -196 NFg = -490 N F floor = +490 N F floor = +196 N +10 N -10 N Step 2: Look at each object separately

13 Smooth floor A 50kg person pushes on a 20 kg box with a force of 10 Newtons, both standing on a smooth (frictionless) surface. Fg = -196 N F floor = +196 N +10 N Step 2: Look at each object separately The forces for the Y axis cancel F net Y = F floor + F g = ma Y F net Y = +196 N + (-196 N) = (20 kg)a Y F net Y = 0 N= (20 kg)a Y This leaves only one force along the X axis. a Y = 0 m/s 2 F net X = F right + F Left = ma X F net X = 10N + (0N) = (20 kg)a X a X = (+1/2) m/s 2

14 Smooth floor +10 N -10 N 60 Kg 20 Kg a person = -(1/6) m/s 2 a Box = (+1/2) m/s 2 Net force on block = - net force on person Mass block * Acceleration Block = - Mass person * Acceleration person

15 In Short When dealing with two objects pulling or pushing on each other, and there are no other forces being applied we can always say Mass 1 * Acceleration 1 = -Mass 2 * Acceleration 2 This means that the acceleration of the block is based on the ratio of the masses of both the person and the block itself Acceleration 1 = - (Mass 2 / Mass 1 )* Acceleration 2


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