Physics 2111 Unit 19 - College of DuPage Lecture 17, Slide 1 Physics 2111 Unit 19 Today’s...
Transcript of Physics 2111 Unit 19 - College of DuPage Lecture 17, Slide 1 Physics 2111 Unit 19 Today’s...
Mechanics Lecture 17, Slide 1
Physics 2111
Unit 19
Today’s Concepts:
a) Static Equilibrium
b) Young’s Modulus
Mechanics Lecture 17, Slide 2
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
Mechanics Lecture 17, Slide 3
Statics:
Example: What are all of the forces acting on a car parked on a hill?
x y
q
N f
mg
Mechanics Lecture 17, Slide 4
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
Mechanics Lecture 17, Slide 6
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
Mechanics Lecture 17, Slide 7
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
Mechanics Lecture 17, Slide 9
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
Mechanics Lecture 17, Slide 10
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
Mechanics Lecture 17, Slide 11
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 -
Mechanics Lecture 17, Slide 12
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
Mechanics Lecture 17, Slide 13
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
Mechanics Lecture 17, Slide 14
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
Mechanics Lecture 17, Slide 15
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
Mechanics Lecture 17, Slide 16
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
Mechanics Lecture 17, Slide 17
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
Mechanics Lecture 17, Slide 18
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
Mechanics Lecture 17, Slide 19
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
Mechanics Lecture 17, Slide 20
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
How did they do it?
Mechanics Lecture 17, Slide 25
Hagia Sophia (532AD)
Open Span 250ft
Pantheon (126AD)
Open Span 142ft Santa Maria Cathedral
(1436AD)
Open Span 152ft
Mechanics Lecture 17, Slide 27
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
Mechanics Lecture 17, Slide 29
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?
Mechanics Lecture 17, Slide 30
Shear Stress
Is a small shear
Modulus good?
Lubricating Oil
Fluids are
extremely weak in
shear.
Mechanics Lecture 17, Slide 31
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?
Mechanics Lecture 17, Slide 32
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
Mechanics Lecture 17, Slide 33
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 +
Mechanics Lecture 17, Slide 34
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?
Mechanics Lecture 17, Slide 35
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