PHYSICS 231 INTRODUCTORY PHYSICS I · Fictitious Force: Derivation Eq. of motion in fixed frame x=v...

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PHYSICS 231 INTRODUCTORY PHYSICS I Lecture 5

Transcript of PHYSICS 231 INTRODUCTORY PHYSICS I · Fictitious Force: Derivation Eq. of motion in fixed frame x=v...

PHYSICS 231

INTRODUCTORY PHYSICS I

Lecture 5

• Range (yi=yf):

• Relative velocity:

• Newton’s Laws:

1. If Σ F = 0, velocity doesn’t change.

2.3.

Main points of last lecture

R =vi2

gsin2!

vbe = vbr + vre

rF! = m

ra

rF12= !

rF21

Newton’s Third Law

• Single isolated force cannot exist• For every action there is an equal and opposite

reaction• Action and Reaction Forces act on different

objects

Force on “1” due to “2”

rF12= !

rF21

Free-body Diagram

• Draw arrows for allforces acting ON object

• If the object isin “equilibrium” (at restor constant v)

• Otherwise, findacceleration by

Ignore rotational motion for now.Treat object as a particle.

rF! = 0

rF! = m

ra

Example 4.1

Two blocks sit on a frictionless table. The masses areM1=2 kg and M2=3 Kg. A horizontal force F=5 N is applied to Block 1.

1. What is the acceleration of the blocks?

2. What is the force of block 1 on block 2?

1. a = 1 m/s2

2. F21= 3 N

M1M2

F

Mechanical Forces

• Gravity: w=mg (down)• Normal forces• Strings, ropes and Pulleys• Friction• Springs (later)

Rules for Ropes and Pulleys

• Force from rope points AWAY fromobject• (Rope can only pull)

• Magnitude of the force is Tension• Tension is same everywhere in the rope

• Tension does not change when goingover pulley

Approximations: Neglect mass of rope and pulley, neglect friction in pulley

Example 4.2

I pull a 5 kg mass up with a rope, so that itaccelerates 2 m/s2. What is the tension in the rope?

T = 59 N

Example 4.3 - Atwood Machine

a) Find accelerationb) Find T, the tension in thestringc) Find force ceiling must exerton pulley

a) a = g/6 = 1.635 m/s2

b) T = 57.2 Nc) Fpulley=2T = 114.5 N

5 kg7 kg

Example 4.4a

Which statements are correct?Assume the objects are in staticequilibrium.

T1 is _____ T2

cos(10o)=0.985 sin(10o)=0.173

A) Less thanB) Equal toC) Greater than

Example 4.4b

Which statements are correct?Assume the objects are static.

T2 is ______ T3

cos(10o)=0.985 sin(10o)=0.173

A) Less thanB) Equal toC) Greater than

Example 4.4c

Which statements are correct?Assume the objects are static.

T3 is ______ Mg

cos(10o)=0.985 sin(10o)=0.173

A) Less thanB) Equal toC) Greater than

Example 4.4d

Which statements are correct?Assume the objects are static.

T1+T2 is ______ Mg

cos(10o)=0.985 sin(10o)=0.173

A) Less thanB) Equal toC) Greater than

Example 4.4

Given that Mlight = 25 kg, find all three tensions

T3 = 245.3 N, T1 = 147.4 N, T2 = 195.7 N

Cable Pull Demo

Inclined Planes

• Choose x along theincline and yperpendicular to incline

• Replace force of gravitywith its components

Fg,x = mgsin!

Fg,y = mgcos!30.0°

Example 4.5

Find the acceleration and the tension

a = 4.43 m/s2, T= 53.7 N

Example 4.6

Find M such that the box slides at constant v

M=15.6 kg

M

Frictional Forces

• RESISTIVE force between object and neighborsor the medium

• Examples:• Sliding a box• Air resistance• Rolling resistance

Sliding Friction

• Parallel to surface,opposing direction of motion

• ~ independent ofthe area of contact

• Depends on the surfaces in contact• Object at rest: Static friction• Object in motion: Kinetic friction

Static Friction, ƒs

• Just enough force tokeep object at rest.

• µs is coefficient ofstatic friction

• N is the normal force

f

F

fs ! µsN

KineticFriction, ƒk

• µk is coefficient ofkinetic friction

• Friction force opposesdirection of motion

• N is the normal forceF

f!

fk = µkN

Coefficientsof Friction

f ! µsN

f = µkN

µs > µk

Example 4.7

The man pushes/pulls with a force of 200 N. Thechild and sled combo has a mass of 30 kg and thecoefficient of kinetic friction is 0.15. For each case:What is the frictional force opposing his efforts?What is the acceleration of the child?

f=59 N, a=3.80 m/s2 / f=29.1 N, a=4.8 m/s2

Example 4.8

Given m1 = 10 kg and m2 = 5 kg:a) What value of µs would stop the block from sliding?b) If the box is sliding and µk = 0.2, what is theacceleration?c) What is the tension of the rope?

a) µs = 0.5 b) a=1.96 m/s2 c) 39.25 N

Example 4.9

What is the minimum µs required toprevent a sled from slipping down ahill of slope 30 degrees?

µs = 0.577

Other kinds of friction

• Air resistance, F ~ Area ⋅ v2

• Rolling resistance, F ~ v

Terminal velocity:

Fresistance = CAv2

= mg at terminal velocity

Coffee Filter Demo

Example 4.9

An elevator falls with acceleration a = 8.0 m/s2.If a 200-lb person stood on a bathroom scaleduring the fall, what would the scale read?

36.9 lbs

Accelerating Reference Frames

• Equivalent to “Fictitious” gravitational force

g fictitious = !a frame

Fictitious Force: Derivation

Eq. of motion in fixed framex = v

0t +1

2at2

= v0t +1

2

F

mt2

F-maf looks like force in new frame, maf acts like fake gravitational force!

x0 (t) =1

2a f t

2

x ! x0 (t) = v0t +1

2

(F ! maf )

mt2

Example 4.10

You are calibrating an accelerometer so that you canmeasure the steady horizontal acceleration of a car bymeasuring the angle a ball swings backwards.If M = 2.5 kg and the acceleration, a = 3.0 m/s2:a) At what angle does the ball swing backwards?b) What is the tension in the string?

θ = 17 degT= 25.6 N

θ

Example 4.11a

A fisherman catches a 20 lb trout (mass=9.072kg), and takes the trout in an elevator to the78th floor to impress his girl friend, who is theCEO of a large accounting firm. The fish ishanging on a scale, which reads 20 lb.s while thefisherman is stationary. Later, he returns via theelevator to the ground floor with the fish stillhanging from the scale.

In the instant just after the elevator begins tomove upward, the reading on the scale will be______________ 20 lbs.

a) Greater thanb) Less thanc) Equal to

Example 4.11b

A fisherman catches a 20 lb trout (mass=9.072 kg), andtakes the trout in an elevator to the 78th floor to impresshis girl friend, who is the CEO of a large accounting firm.The fish is hanging on a scale, which reads 20 lb.s whilethe fisherman is stationary. Later, he returns via theelevator to the ground floor with the fish still hangingfrom the scale.

On the way back down, while descending atconstant velocity, the reading on the scale willbe ________________ 20 lbs.

a) Greater thanb) Less thanc) Equal to

Example 4.11c

A fisherman catches a 20 lb trout (mass=9.072 kg), andtakes the trout in an elevator to the 78th floor to impresshis girl friend, who is the CEO of a large accounting firm.The fish is hanging on a scale, which reads 20 lb.s whilethe fisherman is stationary. Later, he returns via theelevator to the ground floor with the fish still hangingfrom the scale.

In the instant just before the elevator comesto a stop on the 78th floor, the mass of thefish will be ______________ 9.072 kg.

a) Greater thanb) Less thanc) Equal to