Welcome back to Physics 211

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# Welcome back to Physics 211 - PowerPoint PPT Presentation

Welcome back to Physics 211. Today’s agenda: Applying Newton’s Laws with motion Weight, elevators, and normal forces Static and kinetic friction. Current homework assignments. WHW5: In blue Tutorials in Physics homework book HW-32 #3, HW-33 #4, HW-34 #5, HW-35 #6

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Presentation Transcript

### Welcome back to Physics 211

Today’s agenda:

Applying Newton’s Laws with motion

Weight, elevators, and normal forces

Static and kinetic friction

Current homework assignments
• WHW5:
• In blue Tutorials in Physics homework book
• HW-32 #3, HW-33 #4, HW-34 #5, HW-35 #6
• due Wednesday, Oct. 3rd in recitation
• FHW3:
• From end of chapters 4 & 5 in University Physics
• 4.40, 5.61, 5.88, and #4 from PHY211 website
• due Wednesday, Oct. 10th in recitation

The string that holds the block breaks, so there is no more tension force exerted on the block.

Will the magnitudes of either of the other two forces on the block (i.e., the weight and the normal force) change?

1. Both weight and normal force will change.

2. Only the weight force will change.

3. Only the normal force will change.

4. Neither one of the two forces will change.

Discussion
• Say normal does not change. What will be the new net force? Same as old, but “minus” the tension, i.e., straight to the left (and not zero).
• So what would the block do? Accelerate. OK -- Straight to the left? NO, can’t be correct answer.
• Acceleration is down the incline, so normal gets smaller.
Geometry (2)

*component of WBE

at 900 to plane

is WBEcos(q)

*component along

plane is WBEsin(q)

N

q

q

W

q

Force components: Block on frictionless incline
• The string that holds the block breaks.
• What does happen?
• Take components now along and perpendicular to incline
Summary
• To solve problems in mechanics, identify all forces and draw free body diagrams for all objects
• If more than one object, use Newton’s Third law to reduce number of independent forces
• Use Newton’s Second law for all components of net force on each object
• Choose component directions to simplify equations
Weight, mass, and acceleration
• What does a weighing scale ``weigh’’?
• Does it depend on your frame of reference?
• Consider elevators….

### Cinema Classics

Person on scale in elevator

A person is standing on a bathroom scale while riding an express elevator in a tall office building. When the elevator is at rest, the scale reads about 160 lbs.

While the elevator is moving, the reading is frequently changing, with values ranging anywhere from about 120 lbs to about 200 lbs.

At a moment when the scale shows the maximum reading (i.e. 200 lbs) the elevator

1. must be going up

2. must be going down

3. could be going up or going down

4. I’m not sure.

Motion of elevator

(if a )

Moving upward and slowing down,

OR

Moving downward and speeding up.

Motion of elevator

(if a )

Moving upward and speeding up,

OR

Moving downward and slowing down.

A person is standing on a bathroom scale while riding an express elevator in a tall office building. When the elevator is at rest, the scale reads about 160 lbs.

While the elevator is accelerating, a different reading is observed, with values ranging anywhere from about 120 lbs to about 200 lbs.

At a moment when the scale shows the maximum reading (i.e. 200 lbs) the acceleration of the elevator is approximately

1. 1 m/s2

2. 2.5 m/s2

3. 5 m/s2

4. 12.5 m/s2

Conclusions
• Scale reads magnitude of normal force |NPS|
• Reading on scale does not depend on velocity (principle of relativity again!)
• Depends on acceleration only

* a > 0  normal force bigger

* a < 0  normal force smaller

Reminder of free-fall experiment
• Objects fall even when there is no atmosphere (i.e., weight force is not due to air pressure).
• When there is no “air drag” things fall “equally fast.” i.e. same acceleration
• From Newton’s 2nd law, a = W/m is independent of m -- means W = mg
• The weight (i.e., the force that makes objects in free fall accelerate downward) is proportional to mass.
Inertial and gravitational mass
• Newton’s second law:

F = mIa

• For an object in “free fall”

W = mG g

• If a independent of mI, must have

mI = mG

Principle of equivalence

Forces of friction
• There are two types of situations in which frictional forces occur:
• Two objects “stick to each other” while at rest relative to one another (static friction).
• Two objects “rub against each other” while moving relative to each other (kinetic friction).
• We will use a macroscopic description of friction which was obtained by experiment.
Friction demo
• Static friction: depends on surface and normal force for pulled block
• Kinetic friction: generally less than maximal static friction
• depends on the type of surfaces of the objects
• depends on the normal force that the objects exert on each other
• does not depend on the surface area where the two objects are touching

The actual magnitude of the force of static friction is generally less than the maximum value.

A 2.4-kg block of wood is at rest on a concrete floor. (Using g = 10 m/s2, its weight force is about 24 N.)

No other object is in contact with the block. If the coefficient of static friction is ms = 0.5, the frictional force on the block is:

1. 0 N 3. 12 N

2. 8 N 4. 24 N

A 2.4-kg block of wood is at rest on a concrete floor. (Using g = 10 m/s2, its weight force is about 24 N.)

Somebody is pulling on a rope that is attached to the block, such that the rope is exerting a horizontal force of 8 N on the block. If the coefficient of static friction is ms = 0.5, the frictional force on the block is:

1. 0 N 3. 12 N

2. 8 N 4. 24 N

Having no choice, you have parked your old heavy car on an icy hill, but you are worried that it will start to slide down the hill.

Would a lighter car be less likely to slide when you park it on that icy hill?

1. No, the lighter car would start sliding at a less steep incline.

2. It doesn’t matter. The lighter car would start sliding at an incline of the same angle.

3. Yes, you could park a lighter car on a steeper hill without sliding.

Initially at rest
• What is the largest angle before the block slips?
• Resolve perpendicular to plane

 N = Wcosq

• Resolve parallel F = Wsinq
• Since F ≤ msN, we have

Wsinq ≤ msWcosq i.e.

tanq ≤ ms

What if  > tan-1ms ?

• depends on the type of surfaces of the objects
• depends on the normal force that the objects exert on each other
• does not depend on the surface area where the two objects are touching
• does not depend on the speed with which one object is moving relative to the other
What if  > tan-1ms ?
• Block begins to slide
• Resolve along plane:

Wsinq- mKWcosq= ma

• Or:

a = g(sinq- mKcosq)

Summary of friction
• 2 laws of friction: static and kinetic
• Static friction tends to oppose motion and is governed by inequality

Fs ≤ msN

• Kinetic friction is given by equality FK = mKN