Download Physics 170 Week 5, Lecture 2

Physics 170 Week 5, Lecture 2
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Physics 170 Week 5 Lecture 2
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Textbook Chapter 5:Section 5.5-5.7
Physics 170 Week 5 Lecture 2
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Learning Goals:
• Review the condition for equilibrium of a rigid body
• Consider the special case of two-force and three-force members
• Learn how to recognize and use what we learn about two-force
and three-force members in the context of an example.
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Review: Equilibrium of a rigid body
Consider a static rigid body with a number of forces F⃗1 , ..., F⃗k
acting on it at points ⃗r1 , ..., ⃗rk and a number of couple moments
⃗ ◦1 , ...M
⃗ ◦q .
M
The conditions for equilibrium are
k
∑
F⃗i = 0
(1)
i=1
q
∑
j=1
M◦j +
k
∑
i=1
⃗ Oi =
M
q
∑
j=1
M◦j +
k
∑
(⃗ri − ⃗rO ) × F⃗i = 0
(2)
i=1
⃗ Oi are moments due to forces, taken about a point O. The sum
M
over moments are independent of the position of the point O.
Furthermore, forces are “sliding vectors” – they can be moved
along their lines of action.
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Two-force members
Theorem: Consider a subsystem of a larger system which is in
equilibrium and which has only two forces and no couple moments
acting on it. If that subsystem is to be in equilibrium, the forces
must have equal magnitude, opposite directions and they
must have the same line of action.
∑ ⃗
Proof: Since we must have i Fi = 0 and there are only two
forces,
F⃗1 + F⃗2 = 0 → F⃗2 = −F⃗1
Furthermore, the sum of moments must vanish. If F⃗1 acts at ⃗r1 and
F⃗2 acts at ⃗r2 , the sum of moments about O located at ⃗rO is
(⃗r1 − ⃗rO ) × F⃗1 + (⃗r2 − ⃗rO ) × F⃗2 = (⃗r1 − ⃗rO ) × F⃗1 − (⃗r2 − ⃗rO ) × F⃗1
= (⃗r1 − ⃗rO − ⃗r2 + ⃗rO ) × F⃗1 = (⃗r1 − ⃗r2 ) × F⃗1 = 0
which can only be true if ⃗r1 − ⃗r2 || F⃗1 . QED
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Example:
The force on the pins at A and B must have equal magnitudes and
opposite directions and they must have a common line of action.
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Three Force Members
Theorem: If three non-parallel forces and no couple moments act
on a body in equilibrium, the forces are concurrent, that is, their
lines of action must have a common point of intersection.
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Proof:
• Consider the resultant force F⃗1+2 = F⃗1 + F⃗2 . Together with F⃗3 ,
they form a two-force system, i.e.
F⃗1+2 = −F⃗3
and (by the previous theorem) they have the same line of
action, so can be taken as acting at the same point.
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F1
r1
ro
F3
r2
F2
• Moments computed about any point must vanish, so let us
compute the moments about a point which is on the line of
action of F⃗3 and therefore on the line of action of F⃗1 + F⃗1 .
(⃗r1 −rO )×F⃗1 +(⃗r2 −rO )×F⃗2 = 0 → (⃗r1 −rO )×F⃗1 = −(⃗r2 −rO )×F⃗2
and, in particular, F⃗1 , F⃗2 ,(⃗r1 − rO ) and (⃗r2 − rO ) are all in the
same plane. Thus, in particular, the lines of action of F⃗1 and
F⃗2 must intersect (nonparallel lines in a plane must intersect).
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F1
r1
ro
F3
r2
F2
• It is a simple fact of geometry (parallelogram law of addition)
that F⃗1 , F⃗2 and F⃗1+2 lie in the same plane and that their lines
of action intersect at the same point.
• Since F⃗3 = −F⃗1+2 its line of action intersects the lines of action
of F⃗1 and F⃗2 at that point.
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Example
Consider the lever assembly. Find the direction of the reaction
force at A.
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The short link is a two-force member
The short link BD is a two-force member, so the resultant forces at
pins D and B must be equal, opposite, and collinear. Although the
magnitude of the force is unknown, the line of action must pass
through B and D.
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The lever ABC is a three-force member
The three nonparallel forces acting on it must be concurrent at O.
The distance CO must be 0.5 m
tan θ = 0.7m/0.4m → θ = 60.3
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Find the magnitude of the force at A:
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Describe the three forces:
F⃗C
F⃗
F⃗A
Physics 170 Week 5 Lecture 2
= 400N î
)
F (
= − √ î + ĵ
2
(
)
= FA cos θî + sin θĵ
15
F⃗C = 400N î
(
)
F
F⃗ = − √ î + ĵ
2
(
)
F⃗A = FA cos θî + sin θĵ
Decompose into components and apply equilibrium condition:
√
∑
Fxi = 0: FA cos θ − F/ 2 + 400N = 0
√
∑i
i Fyi = 0: FA sin θ − F/ 2 = 0
The second equation implies
FA = √
F
2 sin θ
Plug this into the first equation:
F cos θ
F
√
− √ + 400N = 0
2 sin θ
2
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The equation which remains to be solved is
F cos θ
F
√
− √ + 400N = 0
2 sin θ
2
which we can re-write as
√
√
√
( 2)(400) N
( 2)(400) N
(.7)( 2)(400) N
F =
=
=
1 − cot θ
1 − (.4)/(.7)
(.3)
where we recalled the result of the previous computation that
tan θ = .7/.4. Finally,
F = 1320 N
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Example:
Find the angle of the ladder θ at equilibrium (so that the ladder
will stay there without friction).
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Strategy for finding a solution:
• We recognize that the ladder is a three-force member.
• This means that the three forces acting on it must be
concurrent.
• We can then solve the problem by finding the angle θ such that
the three forces are indeed concurrent.
• We will assume that the reaction forces at the roof and wall are
orthogonal to the roof and the wall, respectively.
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Free body diagram
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Geometry
The vertical distance from the bottom to top of ladder is 18 sin θ.
The horizontal distance from the bottom of the ladder to its center
of gravity is 9 cos θ.
The tangent of the angle between the normal to the roof and the
horizontal, i.e. (90-40)=50 degrees, is equal to the ratio of these
two distances,
18 sin θ
1
tan 50 =
= 2 tan θ → θ = arctan tan 50
9 cos θ
2
θ = 30.8 degrees
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Enrichment problem:
Assume that the ladder is in equilibrium, that is θ = 30.8 degrees.
Is this configuration stable?
That is, if you change the angle a little bit, does the ladder tend to
move back to its equilibrium position or does it tend to move away
even farther from equilibrium?
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For the next lecture, please read
Textbook Chapter 8:Section 8.1-8.2
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