lecture 21 - university of oklahoma

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Physics 2514 Lecture 21 P. Gutierrez Department of Physics & Astronomy University of Oklahoma Physics 2514 – p. 1/12

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Page 1: Lecture 21 - University of Oklahoma

Physics 2514Lecture 21

P. Gutierrez

Department of Physics & AstronomyUniversity of Oklahoma

Physics 2514 – p. 1/12

Page 2: Lecture 21 - University of Oklahoma

Goals

Introduce and discuss Newton’s third law of motion.

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Page 3: Lecture 21 - University of Oklahoma

Newton’s Third Law

To every action there is always an opposite and equal reaction;or, the mutual actions of two bodies upon each other are alwaysequal and directed to contrary parts.

1. Newton’s first and second laws apply to a single object;Forces acting on an object cause it to accelerate;

2. Newton’s third law describes forces between two objects;If one object push on another, the other pushes back.

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Examples

A number of examples on the application of Newton’s third law

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Clicker

An object of mass 10 kg sits on a table. Gravity acts on theobject downward with a force of 98 N. What is the reaction tothis force?

1. The normal force acting upward with a force of 98 N;2. The object pushing down on the table with a force of 98 N;3. The object pulling upward on the Earth with a force of 98 N;4. The table pushing upward with a force of 98 N

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Clicker

An object of mass m is in free fall toward the Earth. Which of thefollowing is true as view from an inertial frame of reference?(Ignore the effect of other celestial objects on the Earth and theobject. In addition, neglect the effect of air resistance on theEarth and object.)

1. The Earth accelerates upward toward the object at the rateof a = (m/M)g, where M is the mass of the Earth;

2. The Earth does not accelerate toward the object;3. The object accelerates toward the Earth at a = (1/2)mg and

the Earth toward the object at a = (1/2)mg;4. This is too hard, give me a break.

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Example

Consider the system below. Calculate the forces on each box.Forces

Box A & B: (mA + mB)a = Fh

Box A: mAa = Fh − FAB

Box B: mBa = FAB

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Clicker

Recall that the tension in a “rope” is the force it exerted on anobject. For the figure below, the tension in rope 1 and rope 2 arerelated by:

1. 2T1 = T2

2. T1 = 2T2

3. T1 = T2

4. T1 = 4T2

5. 4T1 = T2

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Page 9: Lecture 21 - University of Oklahoma

Example

A 10 kg block on a table is connected by a string to a 5 kg massthat is hanging over the edge of the table. Assuming thatfrictional forces my be neglected, what is the magnitude ofacceleration of the 10 kg block when the other block is released?

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Page 10: Lecture 21 - University of Oklahoma

Example

Free Body Diagram:Constraint: a ≡ aA = aB

Forces on Upper Block

nA − mAg = 0

T = mAa

Forces on Lower Block

mBg − T = mBa

Solution:

a =mB

mA + mB

g

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Page 11: Lecture 21 - University of Oklahoma

Clicker

An object attached to a string is whirled at a constant speed in acircle of radius r in the horizontal plane. What is the cause of theoutward force acting on the string that keeps it stretched?

1. Object pulling outward;2. Inertial force;3. There is no outward force;4. Gravity;5. My hand whirling it.

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Assignment

Study for examWill cover chapters 5, 6, 7, 8

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