free-body diagram

Physics 211: Lecture 6, Pg 1

Physics 211: Lecture 6Physics 211: Lecture 6

Today’s AgendaToday’s Agenda

RecapRecap Problems...problems...problems!!Problems...problems...problems!!

Accelerometer Inclined plane Motion in a circle


Something from Opportunity Set 2Something from Opportunity Set 2


Total acceleration = tangential plus radialTotal acceleration = tangential plus radial

Tangentialdeceleration

blue arrowis velocity Look at this in polar

coordinates

Tangential motionTangential accel

Radial motion?Radial accel

2 2TANG RADIALa a a




blue arrowis velocity Tang accel is given

Radial accel depends on v and R

aR=v2/R

Figure out v at bottom, to figure out a there.

2 2TANG RADIALa a a



2 20 2 TANGv v a path length


blue arrowis velocity Tang accel is given

Radial accel depends on v and R

aR=v2/R

Figure out v at bottom, to figure out a there.

2 2TANG RADIALa a a

given

Centripetal =v2

R

path length = R

2


ReviewReview

Discussion of dynamics.

Review Newton’s 3 Laws

The Free Body DiagramFree Body Diagram

The tools we have for making & solving problems:» Ropes & Pulleys (tension)» Hooke’s Law (springs)


Review: Pegs & PulleysReview: Pegs & Pulleys

Used to change the direction of forces An ideal massless pulley or ideal smooth peg will

change the direction of an applied force without altering the magnitude: The tension is the same on both sides The tension is the same on both sides for a massless rope!for a massless rope!

FF1 = -T ii ideal peg or pulley

FF2 = T jj

| FF1 | = | FF2 |

massless rope


Review: SpringsReview: Springs

Hooke’s Law: Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position.

FX = -kx Where x is the displacement from the equilibrium and k is the constant of proportionality.

relaxed position

FX = 0

x

Spring demo


Lecture 6, Lecture 6, Act 1Act 1Springs Springs

A spring with spring constant 40 N/m has a relaxed length of 1 m. When the spring is stretched so that it is 1.5 m long, what force is exerted on a block attached to the end of the spring?

x = 0

M

k kM

x = 0 x = 1.5

(a) -20 N (b) 60 N (c) -60 N

x = 1


Lecture 6, Lecture 6, Act 1Act 1Solution Solution

Recall Hooke’s law:

FX = -kx Where x is the displacement from equilibrium.

FX = - (40) ( .5)

FX = - 20 N

(a) -20 N (b) 60 N (c) -60 N


Problem: Problem: Pendulum AccelerometerPendulum Accelerometer

A weight of mass m is hung from the ceiling of a car with a massless string. The car travels on a horizontal road (duh, it’s Illinois), and has an acceleration a in the x direction. The string makes an angle with respect to the vertical (y) axis. Solve for in terms of a and g.

tells you what a is, then. The pendulum is an accelerometer

a

i i


Accelerometer...Accelerometer...

Draw a free body diagram for the mass: What are all of the forces acting?

m

TT (string tension)

mgg (gravitational force)

i i

RememberFnet=ma



Using components Using components (recommended) you decompose or resolve (same thing) F=ma into components:

ii: FX = TX = T sin = ma

jj: FY = TY mg = T cos mg = 0

TT

mgg

mmaa

jj

ii

TX

TY



Using components Using components :

ii: T sin = ma

jj: T cos - mg = 0

Eliminate T :

mgg

mmaaT sin = ma

T cos = mgtan

ag

TX

TY jj

ii

TT

It’s an accelerometer!For every there is a particular a



Let’s put in some numbers:

Say the car goes from 0 to 60 mph in 10 seconds: 60 mph = 60 x 0.45 m/s = 27 m/s. Acceleration a = Δv/Δt = 2.7 m/s2. So a/g = 2.7 / 9.8 = 0.28 (no units).

= arctan (a/g) = 15.6 deg

tan ag

a

Cart w/ accelerometer

helium accelerometer


Another Problem: Inclined planeAnother Problem: Inclined plane

A block of mass m slides down a frictionless ramp that makes angle with respect to the horizontal. What is its acceleration a ?

ma


Inclined plane...Inclined plane...

Define convenient axes parallel and perpendicular to plane: Acceleration a is in x direction only.

ma

ii

jj

What’s the “physics” here? Gravity provides downward force.Ramp constrains the motionWhere’s block going to go?

g


Inclined plane...Inclined plane...Identify the forces in the problemIdentify the forces in the problem

free body diagramfree body diagram

Force components: consider x and y components separately: just draw forces acting on box

ii: mg sin =ma. a = g sin

jj: N - mg cos = 0. N = mg cos

mgg

NN

maa

ii

jj

mg cos

mg sin

Incline


Angles of an Inclined planeAngles of an Inclined plane

ma = mg sin

mgN

The triangles are similar, so the angles are the same!

Prove the triangles are similar


Lecture 6, Lecture 6, Act 2Act 2Forces and MotionForces and Motion

A block of mass M = 5.1 kg is supported on a frictionless ramp by a spring having constant k = 125 N/m. When the ramp is horizontal the equilibrium position of the mass is at x = 0. When the angle of the ramp is changed to 30o what is the new equilibrium position of the block x1?

(a) x1 = 20cm (b) x1 = 25cm (c) x1 = 30cm

x = 0

M

k

x 1 = ?

Mk

= 30o


Lecture 6, Lecture 6, Act 2Act 2SolutionSolution

x 1

Mk

x

y

Choose the x-axis to be along downward direction of ramp. (smart)

Mg

FBD: The total force on the block is zero since it’s at rest. (aha!)

N

Fx,g = Mg sin

Force of gravity on block is Fx,g = Mg sin Consider x-direction:

Force of spring on block is Fx,s = -kx1

F x,s = -kx 1convenient coord system

0


Lecture 6, Lecture 6, Act 2Act 2SolutionSolution

x 1

Mk

x

y

Since the total force in the x-direction must be 0:

F x,g = Mg sin

F x,s = -kx 1

Mg sinkx1 κsin Μgx1

θ

m20mN125

50sm189kg15x2

1 ....


Problem: Two Blocks Problem: Two Blocks

Two blocks of masses m1 and m2 are placed in contact on a horizontal frictionless surface. If a force of magnitude F is applied to the box of mass m1, what is the force on the block of mass m2?

Intuition Technique (free body diagram or FBD)

mm11 mm22

F F2,1F1,2

Only force acting on 2Is F2,1


Problem: Two BlocksProblem: Two Blocks

Realize that F = (mm11+ mm22) a :

Draw FBD of block mm22 and apply FNET = ma:

F2,1F2,1 = mm22 a

F / (mm11+ mm22) = a

m22,1

m2m1

FF

Substitute for a :

mm22

(m1 + m2)m2F2,1 F


Problem: Two BlocksProblem: Two Blocks

What about the net force on mass 1?

F acts in +x-direction, but F1,2 acts in –x-direction

The net force on mass 1 matters

11

1 2NET onm

mF Fm m


Problem: Tension and AnglesProblem: Tension and Angles

A box is suspended from the ceiling by two ropes making an angle with the horizontal. What is the tension in each rope?

mm


Problem: Tension and AnglesProblem: Tension and Angles

Draw a FBD:

T1 T2

mg

Since the box isn’t going anywhere, Fx,NET = 0 and Fy,NET = 0

T1sin T2sin

T2cos T1cos

jj

ii

Fx,NET = T1cos - T2cos = 0 T1 = T2

2 sin mg

T1 = T2 =Fy,NET = T1sin + T2sin - mg = 0


Problem: Motion in a CircleProblem: Motion in a Circle

A boy ties a rock of mass m to the end of a string and twirls it in the vertical plane. The distance from his hand to the rock is R. The speed of the rock at the top of its trajectory is v.

What is the tension T in the string at the top of the rock’s trajectory?

R

v

TT

Tetherball


Motion in a Circle...Motion in a Circle...

Draw a Free Body Diagram (pick y-direction to be down): We will use FFNET = maa (surprise) First find FFNET in y direction:

FFNET = mg +T

TTmgg

y y


Motion in a Circle...Motion in a Circle...

FFNET = mg +T Acceleration in y direction is known: It’s centripetal…

maa = mv2 / R

mg + T = mv2 / R

T = mv2 / R - mg

R

TT

v

mgg

y y

F = ma


Motion in a Circle...Motion in a Circle... What is the minimum speed of the mass at the top of the trajectory such that the string does not go limp?

i.e. find v such that T = 0.

mv2 / R = mg + T

v2 / R = g

Notice that this doesnot depend on m.

R

mgg

v

T= 0

v Rg

Bucket


Lecture 6Lecture 6, , Act 3Act 3Motion in a CircleMotion in a Circle

A skier of mass m goes over a mogul having a radius of curvature R. How fast can she go without leaving the ground?

R

mgg NN

vv

(a) (b) (c) Rgv =mRgv =mRg

v =

Trackw/ bump


Lecture 6Lecture 6, , Act 3Act 3SolutionSolution

mv2 / R = mg - N

For N = 0:

R

v

mgg NN

v Rg


Recap of Today’s lecture:Recap of Today’s lecture:

Example ProblemsExample Problems

Accelerometer Inclined plane (Text: example 6-1) Motion in a circle (Text: 3-5, 5-2, 9-1)

Look at textbook problems Chapter 4: # 49 Chapter 5: # 65, 69

Next time: Friction…



Alternative solution using vectorsAlternative solution using vectors (elegant but not as (elegant but not as straightforward):straightforward):

Find the total vector force FFNET:

TTmgg

FFTOT

m

TT (string tension)




Alternative solution using vectorsAlternative solution using vectors (elegant but not as (elegant but not as systematic):systematic):

Find the total vector force FFNET: Recall that FFNET = ma:

So

maa

tan mamg

ag

TTmgg

tan ag

m

TT (string tension)



Inclined plane...Inclined plane...

Alternative solution using vectors:

m

mgN

a = g sin iiN = mg cos jj

ii

jj

How do we know what is?

free-body diagram

Documents

Transcript of free-body diagram