Physics 2111 Unit 19 - College of DuPage Lecture 17, Slide 1 Physics 2111 Unit 19 Today’s Concepts: a) Static Equilibrium b) Young’s Modulus

Mechanics Lecture 17, Slide 1

Physics 2111

Unit 19

Today’s Concepts:

a) Static Equilibrium

b) Young’s Modulus


New Topic, Old Physics:

As the name implies, “statics” is the study of systems that don’t move. Ladders, sign-posts, balanced beams, buildings, bridges...

The key equations are familiar to us:

If an object doesn’t move:

The net force on the object is zero

The net torque on the object is zero (for any axis)

Statics:

F ma I

0a 0

0F

0


Statics:

Example: What are all of the forces acting on a car parked on a hill?

x y

q

N f

mg


x y

q

N f

mg

Car on Hill:

Use Newton’s 2nd Law: FNET MACM 0

Resolve this into x and y components:

x: f - mg sinq 0

f mg sinq

y: N - mg cosq 0

N = mg cosq


Magnitude:

RCMMg sin (q )

Torque Due to Gravity

MgR

Lever arm

R

CMR


Prelecture question

Method 1: S = 0 = Ften*L*sin(q) - mgL/2

Ften = mgL/(2*L*sin(q))

Two rods of equal length and mass are supported by

cables as shown below. In which case has a greater

tension in the cable greater?

q q

Method 2: S = 0 = Ften*l - mgL/2

Ften = mgL/(2l) (l is “lever arm”)

l l


Example:

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

M x cm

T1 T2

Mg y

x


Balance forces:

T1 + T2 Mg

0F

L/2 L/4

M x cm

T1 T2

Mg y

x


L/2 L/4

M x cm

T1 T2

Mg y

x

Balance Torques

Choose the rotation axis to be out of the page through the CM:

The torque due to the string on the right about this axis is:

The torque due to the string on the left about this axis is:

Gravity acts at CM, so it exerts no torque about the CM

2 T2 L/4

1 T1 L/2

0


Finish the problem

The sum of all torques must be 0:

We already found that

T1 + T2 Mg

1 2 0 +

12 2TT

1 2 02 4

L LT T- +

MgT3

11

MgT3

22

L/2 L/4

M x cm

T1 T2

Mg y

x


L/2 L/4

M x cm

T1 T2

Mg y

x

What if you choose a different axis?

Choose the rotation axis to be out of the page at the left end of the beam:

The torque due to the string on the right about this axis is:

The torque due to the string on the left is zero

Torque due to gravity:

0

2 2

3

4

LT

1 0

2g

LMg -


You end up with the same answer!

The sum of all torques must be 0:

We already found that

T1 + T2 Mg

Same as before

1 2 0g + + 2

30

4 2

L LT Mg-

MgT3

22

MgT3

11

MgT3

22

L/2 L/4

M x cm

T1 T2

Mg y

x


Example 19.2 (beam on cables)

A uniform beam of mass M=200kg and length L=2.0m is supported by two cables. One of the cables is attached at the far left end and the other is attached 0.5m from the right end. What is the tension in each of the cables?

M

T1 T2

y

x


Approach to Statics: Same as before

When choosing axes about which to calculate torque, we can sometimes be clever and make the problem easier....

1)Draw FBD

0F

2)Apply

03) Apply


In Case 1 one end of a horizontal massless rod of length L is attached to a vertical wall by a hinge, and the other end holds a ball of mass M. In Case 2 the massless rod holds the same ball but is twice as long and makes an angle of 30o with the wall as shown. In which case is the total torque about the hinge biggest? A) Case 1 B) Case 2 C) Both are the same

gravity

CheckPoint

Case 1 Case 2

L

90o

M

30o

M

CheckPoint


An object is made by hanging a ball of mass M from one end of a plank having the same mass and length L. The object is then pivoted at a point a distance L/4 from the end of the plank supporting the ball, as shown below. Is the object balanced? A) Yes B) No, it will fall left C) No, it will fall right

M

M

L L/4 gravity

Question


How far to the right of the pivot is the center of mass of just the plank. A) L/4 B) L/2 C) 3L/4

M x

CheckPoint


L/2

M

In Case 1, one end of a horizontal plank of mass M and length L is attached to a wall by a hinge and the other end is held up by a wire attached to the wall. In Case 2 the plank is half the length but has the same mass as in Case 1, and the wire makes the same angle with the plank. In which case is the tension in the wire biggest? A) Case 1 B) Case 2 C) Both are the same

gravity

Case 1 Case 2

L

M


Usefulness of a material in construction can depend on:

What isYoung’s Modulus?

Tensile or compressive strength

• How much force it takes to break it.

• How much force it takes to deform it.

Young’s Modulus, Y

Ductility

• How much it can deform before breaking


Define:

Load/unit area Stress

e.g Newtons/square meter (N/m2)

Pounds/square inch (psi)

Deformation/unit length Strain

e.g meters/meter

inch/inch

Stress-Strain Diagram for Steel

0 0.05 0.10 0.15 0.20 0.25

Strain (inch/inch)

80

60

40

20

0

Str

ess (

ksi)

Yield Stress

Ultimate Stress

Rupture Point

Stress-Strain Diagram for Steel

0 0.05 0.10 0.15 0.20 0.25

Strain (inch/inch)

80

60

40

20

0

Str

ess (

ksi)

Elastic Region

Plastic Region

Strain Hardening

Necking

Steel Compared to other materials

Cast iron in compression

Cast iron in tension

Steel Compared to other materials

Concrete in compression

Concrete in tension

How did they do it?


Hagia Sophia (532AD)

Open Span 250ft

Pantheon (126AD)

Open Span 142ft Santa Maria Cathedral

(1436AD)

Open Span 152ft

Flying Buttresses


Notre Dame de Paris

(construction started 1163)


F = -kDx

Remember Hooke’s Law?

F/A = -YDL/L Stress = Y * Strain

Young’s Modulus Like k with cross

sectional area and

length of the spring

removed.

Young’s Modulus / Modulus of Elastiscity

Slope of this part of the

graph



Example 19.1 (elevator cable)

A 1000kg elevator car is supported by a steel cable 3cm in diameter and 300m long.

How much will the cable stretch if the maximum acceleration is 1.5m/sec2?


Shear Stress

Is a small shear

Modulus good?

Lubricating Oil

Fluids are

extremely weak in

shear.


Example 20.3 (Hanging Beam)

A beams is hinged to a wall to

hold up a sign for new fancy

coffee shop. The beam has a

mass of Mb=6kg and the sign

has a mass of ms=17kg. The

length of the beam is 2.81m.

The sign is attached at the

very end of the beam and the

horizontal wire holding up the

beam is attached 2/3 of way

up the beam. The angle the

wire makes with the beam is

34.1o.

Karter’s

Koffee

a) What is the tension in the wire?

b) What is force the hinge places on the wall?


c o sm g L q

0s ig n b e a m w ir e + +

01

cos2

Mg L q+2

sin3

T L q-

3 3cot

2 4

m MT g q

+

Use hinge as axis

c o sL q

1cos

2L q

Mg 2

sin3

L q

T

Hx

Hy

mg


0xF

mg

Mg

T

0xT H-

Hx

Hy

0yF

xH T

y

x

0yH m g M g- -

( )yH m M g +

2 2

x yH H H +


Example 20.3 (Hanging Beam)

A beams is hinged to a wall to

hold up a sign for new fancy

coffee shop. The beam has a

mass of Mb=6kg and the sign

has a mass of ms=17kg. The

length of the beam is 2.81m.

The sign is attached at the

very end of the beam and the

horizontal wire holding up the

beam is attached 2/3 of way

up the beam. The angle the

wire makes with the beam is

34.1o.

Karter’s

Koffee

a) What is the tension in the wire?

b) What is force the hinge places on the wall?


Example 20.4 (Block and Strut)

The system shown above is in equilibrium. The steel block has a mass

m1 = 249.0 kg and the uniform rigid aluminum strut has a mass

m2 = 48.0 kg. The strut is hinged so that it can pivot freely about point A.

The angle between the left wire and the ground is Θ = 33.0o and the angle

between the strut and the ground is Φ = 45.0o

What is the tension in the left wire?

A

B

C

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Physics 2111 Unit 19 - College of DuPage Lecture 17, Slide 1 Physics 2111 Unit 19 Today’s Concepts: a) Static Equilibrium b) Young’s Modulus