physics 201: lecture 21, pg 1 lecture 21 goals: discussion conditions for static equilibrium use...

Physics 201: Lecture 21, Pg 1

Lecture 21Goals:Goals:

• Discussion conditions for static equilibrium

• Use Free Body Diagrams prior to problem solving

• Introduce Young’s, Shear and Bulk modulus

Exam 3: Wednesday, April, 18th 7:15-8:45 PM


Statics

Equilibrium is established when

0 motion nalTranslatio F

0 motion Rotational

In 3D this implies SIX expressions (x, y & z)


Three force and three torque expressions

0

0

0

forces External

z

y

x

F

F

F

0

0

0

torquesExternal

z

y

x


Free Body Diagram Rules

For forces zero acceleration reflects the motion of the center of mass

For torques the position of the force vector matters

We are free to choose the axis of rotation (usually one that simplifies the algebra)


Statics: Using Torque

Now consider a plank of mass M suspended by two strings as shown.

We want to find the tension in each string:

L/2 L/4

Mx CM

T1 T2

Mg

y

x

0F

0


Statics: Using Torque Now consider a plank of mass M suspended by two

strings as shown. We want to find the tension in each string:

L/2 L/4

Mx CM

T1 T2

Mg y

x

0

0

y

x

F

F0 z

MgTTFy 210

00 2142 LL TTzMgTTFy 2220

MgT 32

2


Statics: Using Torque

Now consider a plank of mass M suspended by two strings as shown.

Using a different torque position

L/2 L/4

Mx CM

T1 T2

Mg y

x

0 z

243

20 LL MgTz

MgT 32

2


Another Example: See-saw

Two children (60 kg and 30 kg) sit on a 3.0 m long uniform horizontal teeter-totter of mass 30 kg. The larger child is 1.0 m from the pivot point while the smaller child is trying to figure out where to sit so that the teeter-totter remains motionless.

Position of the small boy?

0 mEquilibriuNet

30 kg60 kg

30 kg


Example: Soln.

Draw a Free Body diagram (assume g = 10 m/s2)

0 = 300 d + 300 x 0.5 + N x 0 – 600 x 1.0

0= 2d + 1 – 4

d = 1.5 m from pivot point

0 UseNet

30 kg

1 m

60 kg30 kg

0.5 m

300 N 300 N 600 N

N

d


Exercise: Statics A 1.0 kg ball is hung at the end of a uniform rod

exactly 1.0 m long. The system balances at a point on the rod 0.25m from the end holding the mass. What is the mass of the rod?

(a) 0.5 kg (b) 1 kg (c) 2 kg

1kg

1m


The classic ladder problem A uniform ladder of length L and mass m is propped up against

a frictionless wall. There is friction with the ground with static coefficient . The angle the ladder makes with the ground is .

What is the minimum value of to keep the latter from slipping?

N

N’

mg

f

=0 0F

0

'0 NfFx mgNFy 0

sin'cos0 2 LNmg L

z

Nf


The classic ladder problem A uniform ladder of length L and mass m is propped up

against a frictionless wall. There is friction with the ground with static coefficient . If the angle the ladder makes with the ground is .

What is the minimum value of that keeps the ladder from slipping?

N

N’

mg

f

=0

'0 NfFx mgNFy 0

sin'cos0 2 LNmg L

z

mgNf sin'cos2 Nmg

sincos2 mgmg

tan21


Example: Statics

A box is placed on a ramp in the configurations shown below. Friction prevents it from sliding. The center of mass of the box is indicated by a white dot in each case. In which case(s) does the box tip over ?

(a) all (b) 2 & 3 (c) 3 only

1 2 3


A stable equilibrium

Potential energy with tipping (left and right) x-axis corresponds to the x-position of the center-of mass

sU

x0

1


Balancing Acts

So, where is the center of mass for this construction?

Between the arrows!


Example problem The board in the picture has an average length of 1.4 m,

mass 2.0 kg and width 0.14 m. The angle of the board is 45°. The bottle is 1.5 m long, a center of mass 2/5 of the way from the bottom and is mounted 1/5 of way from the end. If the sketch represents the physics of the problem.

What are the minimum and maximum masses of the bottle that will allow this system to remain in balance?

-0.1 m

-0.5 m-0.8 m

0.1 m0.0 m

0.4 m


Example problem What are the minimum and maximum masses of the bottle

that will allow this system to remain in balance? X center of mass,

x, -0.1 m < x < 0.1 m

-0.1 m

-0.5 m-0.8 m

0.1 m0.0 m

0.4 m

2.0 kg

i

iiCM m

mxx CM

1.2

)(4.0)2(5.0

MM

xCM

12410

MM

kg 4M


Example problem What are the minimum and maximum masses of the bottle

that will allow this system to remain in balance? Check torque condition (g =10 m/s2) x, -0.1 m < x < 0.1 m

-0.1 m

-0.5 m-0.8 m

0.1 m0.0 m

0.4 m

2.0 kg

kg 04max .M

)6.0(20)3.0(40 z

Nm )1212( zNm 0 z

kg 61min .M


Real Physical Objects Representations of matter

Particles: No size, no shape can only translate along a path Extended objects are collections of point-like particles

They have size and shape so, to better reflect their motion, we introduce the center of mass

Rigid objects: Translation + Rotation

Deformable objects:

Regular solids:

Shape/size changes under stress (applied forces)

Reversible and irreversible deformation

Liquids:

They do not have fixed shape

Size (aka volume) will change under stress


For Thursday

All of chapter 12


Young’s modulus: measures the resistance of a solid to a change in its length.

Bulk modulus: measures the resistance of solids or liquids to changes in their volume.

Some definitions

L0 L

F

V0

V0 - V

F

Elastic properties of solids :

elasticity in length

volume elasticity

0

0

//

strain tensilestress tensileY

LLAF

0

0

//

BVVAF

physics 201: lecture 21, pg 1 lecture 21 goals: discussion conditions for static equilibrium use...

Documents

y x slide

n n d slide

algebra slide

pm slide

torque expressions slide

plank of mass

y z slide

expressions x