Unformatted text preview: Physics 211
Lecture 17 Today's Concepts:
Torque due to gravity
Static Equilibrium Physics 211 Lecture 17, Slide 1
Physics I would like to know how torque and statics are going to be combined with
would
rotation problems on the future exams.
How does friction provide torque but do no work on rolling objects? Also
why is the shortest distance to the force applied the lever arm in torque.
can the preflights and prelectures go up earlier?
and
go
Can we please discuss everything static equilibrium? I am so lost and scared
for exam 3. (seriously).
If I am in space and traveling at the speed of light, so the light from me does not
travel faster than my body, and I hold a mirror in front of me, am I invisible?
Maybe it's just because none of the concepts are really new, but I prefer to
think it's the masterful prelecture. Either way, these concepts seem very
clear.
Can we just do a lot of these? They don't seem too bad, but we need to get
used to using them.
I really like torque. It's my physics happy place.
If the Yankees win we should have a pizza party…
is the person who does the prelectures from Cal Poly? when i refreshed that
from
page, it asked me to log onto a Cal Poly account. I don't have any questions, but I'd like to point out that
don't
physics is the one thing stopping me from pursuing my
dream to be a ninja. New Topic, Old Physics:
New Statics:
Statics: As the name implies, “statics” is the study of systems that don’t
move. Ladders, signposts, balanced beams, buildings, bridges...
The key equations are familiar to us: ∑ F = ma ∑ τ = Iα If an object doesn’t move: a = 0 τ = 0
If ∑F =0
∑τ=0 The net force on the object is zero
The
The net torque on the object is zero
(for any axis)
Physics 211 Lecture 17, Slide 4 Statics:
Statics:
Example:
What are all of the forces acting on a car parked on a hill? y x N
f θ mg
Physics 211 Lecture 17, Slide 5
Physics Car on Hill:
Car
• Use Newton’s 2nd Law: FNET = MACM = 0
Use
• Resolve this into x and y components:
Resolve
and components: x: f  mg sin θ = 0
f = mg sin θ y: N  mg cos θ = 0 y x N
f N = mg cos θ θ mg
Physics 211 Lecture 17, Slide 6
Physics Torque due to Gravity
Torque
Magnitude:
R⊥ RCM τ = R cm M g sin(θ )
= M gR ⊥ Lever arm Physics 211 Lecture 17, Slide 7 Example:
Example:
Now consider a plank of mass M suspended by two strings as
suspended
shown. We want to find the tension in each string: T1 T2
M x cm
L/4 L/2 Mg y
x Physics 211 Lecture 17, Slide 8 Balance forces:
Balance ∑F =0
T1 + T2 = Mg
T1 T2
M x cm
L/4 L/2 Mg y
x Physics 211 Lecture 17, Slide 9
Physics Balance torques ∑τ=0 Choose the rotation axis to be out of the
Choose
page through the CM: The torque due to the string
on the right about this axis is: L
τ 2 = T2
4
The torque due to the string on
the left about this axis is: L
τ1 = −T1
2 T1 T2
M x cm
L/4 L/2 Mg
Gravity exerts no
torque about the CM y
x Physics 211 Lecture 17, Slide 10 Finish the problem
The sum of all torques must be 0:
The L
L
−T1 + T2 = 0
2
4 τ1 + τ 2 = 0
T 2 = 2T1 T1 M x cm We already found that
T1 + T2 = Mg
T1 = T2 L/4 L/2
1
Mg
3 2
T 2 = Mg
3 Mg y
x Physics 211 Lecture 17, Slide 11 What if you choose a different axis? ∑τ=0 Choose the rotation axis to be out of the
Choose
page at the left end of the beam: The torque due to the string
on the right about this axis is: 3L
τ 2 = T2
4 T1 The torque due to the string on
the left is zero τ1 = 0
Torque due to gravity: L
τ g = − Mg
2 T2
M x cm
L/4 L/2 Mg y
x Physics 211 Lecture 17, Slide 12 You end up with the same answer!
The sum of all torques must be 0:
The τ1 + τ 2 + τ g = 0
T2 = T2 3L
L
− Mg = 0
4
2 2
Mg
3 T1 T2 We already found that
T1 + T2 = Mg L/4 L/2
T1 = M x cm 1
Mg
3 2
T 2 = Mg
3 Mg y
x Physics 211 Lecture 17, Slide 13 Approach to Statics: Summary
Approach
In general, we can use the two equations ∑F =0 ∑τ=0 to solve any statics problem.
When choosing axes about which to calculate torque, we
can sometimes be clever and make the problem easier.... Physics 211 Lecture 17, Slide 14
Physics Hanging Lamp Again
Hanging
A lamp of mass m hangs from the end of plank of mass M and
length L. One end of the plank is held to a wall by a hinge, and
the other end is supported by a massless string that makes an
angle θ with the plank.
a) What is the tension in the string?
b) What are the forces supplied by the hinge on the plank? θ
M hinge
L m
Physics 211 Lecture 17, Slide 15
Physics First use the fact that
x:
y: ∑ F = 0 in both x and y directions: T cos θ  Fx = 0
T sin θ + Fy  Mg  mg = 0 y
x Now use ∑ τ = 0
Just to be different, choose the rotation
axis to be through the end of the rod. T
Fy θ Now mg and T and Fx will not
enter into the torque equation:
L
FY L − M g = 0
2 M
m L/2 L/2 Fx Mg mg
Physics 211 Lecture 17, Slide 16 So we have three equations and
So
three unknowns:
T cos θ  Fx = 0
T sin θ + Fy  Mg  mg = 0 y L
FY L − Mg = 0
2 x which we can solve to find:
T= T ( M / 2 + m )g
sin(θ ) ( M / 2 + m )g
Fx =
tan(θ )
1
F y = mg
2 Fy θ
M
m L/2 L/2 Fx Mg mg
Physics 211 Lecture 17, Slide 17
Physics Same as in the prelecture T
m Fy θ L/2 M Fx L/2
mg y Mg
x
( M / 2 + m )g
sin(θ )
( M / 2 + m )g
Fx =
tan(θ ) T= Fy = 1
mg
2 Physics 211 Lecture 17, Slide 18
Physics Preflight
In case 1 one end of a horizontal massless rod of length L
In
one
is attached to a vertical wall by a hinge, and the other end holds a
is
ball of mass M.
In case 2 the massless rod holds the same ball but is twice as
the
long and makes an angle of 30o with the wall as shown.
with
In which case is the total torque about the hinge biggest?
A) Case 1 B) Case 2
C) Both are the same
B)
C) gravity 90o M 30o 2L M L
Case 1
67% got this right Case 2
Physics 211 Lecture 17, Slide 19 In which case is the total torque about the hinge biggest?
In
A) Case 1 B) Case 2
C) Both are the same
B)
C)
M
Case 2
90o M 30o 2L Case 1 L A) the force of gravity is perpendicular to the rod, which will cause more
torque.
B) f is the same in both cases, and r is 2L in case 2, so the torque is bigger
in case 2.
C) The torque is force times its lever arm. In this question, since gravity
always point down, the lever will be the horizontal distance between the
ball and the wall, which in both case is equal to L. Physics 211 Lecture 17, Slide 20 Preflight
An object is made by hanging a ball of mass M from one end
An
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 L/4 gravity C) No, it will fall right L
M M 62% got this right Physics 211 Lecture 17, Slide 21 Act
How far to the right of the pivot is the center of mass of just
How
the plank.
A) L/4 B) L/2 C) 3L/4 x M Physics 211 Lecture 17, Slide 22 Is the object balanced?
Is
A) Yes
Yes
B) B) No, it will fall left
B)
C) C) No, it will fall right
C) L/4
M L/4 x
M A) Ball of mass M balanced at L/4 left of pivot; beam of mass M balanced
L/4 right of pivot. So system balanced.
B) because part of the plank is on the left side of the pivot so there is more
there
mass on the left side
C) The torque is greater on the right because the distance from the center
of mass is greater on the right. . Physics 211 Lecture 17, Slide 23 Preflight
In case 1, one end of a horizontal plank of mass M and
In
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
as
wire makes the same angle with the plank.
with
In which case is the tension in the wire biggest?
A) Case 1 B) Case 2
C) Both are the same
B)
C) gravity
M
L
42% got this right M
L/2 Physics 211 Lecture 17, Slide 24 Act
Which equation correctly expresses the fact that the total
Which
torque about the hinge is zero? L
=0
2 L
B) TL − M g = 0
2
C) TL − M gL sin(θ ) = 0 The L cancels out:
cancels T L si
n(θ
) A) TL sin(θ ) − M g θ
Mg
L Mg
T sin(θ ) =
2
Physics 211 Lecture 17, Slide 25 In which case is the tension in the wire biggest?
In
A) Case 1
B) Case 2
C) Same
B)
C) M M L L/2 A) the force is at a point that is twice the length away so the torque is twice
as big
B) they both have a net force and torque of zero case two has a shorter
lever arm so it needs a bigger force to produce the same torque to cancel
out their equal weights.
C) The tension will be the same in both cases because T=Mg/2sin(theta),
and theta and M are equal in both cases.
Physics 211 Lecture 17, Slide 26 ...
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This note was uploaded on 09/22/2011 for the course PHYSICS 211 taught by Professor Selig during the Fall '10 term at University of Illinois, Urbana Champaign.
 Fall '10
 Selig
 Physics, Gravity, Static Equilibrium

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