Physics 170 Week 5, Lecture 3
http://www.phas.ubc.ca/∼gordonws/170
Physics 170 203 Week 5 Lecture 3
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Textbook Chapter 5:Section 5.5-5.7
Physics 170 203 Week 5 Lecture 3
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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 recoginize 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 F~1 , ..., F~k
acting on it at points ~r1 , ..., ~rk and a number of couple moments
~ ◦1 , ...M
~ ◦q .
M
The conditions for equilibrium are
k
X
F~i = 0
(1)
i=1
q
X
j=1
M◦j +
k
X
~ Oi =
M
i=1
q
X
j=1
M◦j +
k
X
(~ri − ~rO ) × F~i = 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.
P ~
Proof: Since we must have i Fi = 0 and there are only two
forces,
F~1 + F~2 = 0 → F~2 = −F~1
Furthermore, the sum of moments must vanish. If F~1 acts at ~r1 and
F~2 acts at ~r2 , the sum of moments about O located at ~rO is
(~r1 − ~rO ) × F~1 + (~r2 − ~rO ) × F~2 = (~r1 − ~rO ) × F~1 − (~r2 − ~rO ) × F~1
= (~r1 − ~rO − ~r2 + ~rO ) × F~1 = (~r1 − ~r2 ) × F~1 = 0
which can only be true if ~r1 − ~r2 || F~1 . 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:
• Take two of the vectors, say F~1 and F~2 .
• Being non-parallel, F~1 and F~2 define a plane.
• In that common plane, their lines of action must intersect
(since two non-parallel lines on a plane must intersect at
exactly one point).
• Now, consider the equivalent system where we move F~1 and F~2
along their lines of action to the point ~ri where they intersect.
³
´
• This equivalent system has two forces, F~1 + F~2 acting at ~ri
and F~3 acting at ~r3 .
• This equivalent system is a two-force member and we know
that the force vectors must
´ magnitude and opposite
³ have equal
direction, that is, F~3 = − F~1 + F~2 , which is just a condition
for equilibrium.
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• We also know that the lines of action of the two forces must be
~
the
³ same,´that is F3 has a common line of action with
F~1 + F~2 and which must therefore intersect ~ri .
• Thus we see that the three forces are concurrent with common
intersection point ~ri .
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Example
Cosider 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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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 latter is 18 sin θ.
The horizontal distance from the bottom of the ladder to its center
of gravity is 9 cos θ.
By geometry, the tangent of (90-40)=50 degrees is equal to the
ratio of these two distances,
18 tan θ
1
tan 50 =
→ θ = arctan tan 50
9 cos θ
2
θ = 30.8 degrees
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For the next lecture, please read
Textbook Chapter 8:Section 8.1-8.2
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