Statics

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Statics
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Chapter 9 Bodies in Equilibrium Elasticity and
Fracture
We will only cover the sections on bodies in
equilibrium.
These sections are primarily applications of
physics you have already learned.
9.1 StaticsThe Study of Forces in Equilibrium
9.2 The Conditions for Equilibrium
9.3 Solving Statics Problems
These three sections are a single topic. I will
lecture on all at once.
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In order for an object to remain motionless (be
static), it must not translate or rotate
?Fx 0 ?Fy 0 ??z 0
Remember to choose your axes, including the
rotation axis, and explicitly indicate the
positive directions!
In this problem, the beam is the object of
interest
A massless beam of length L has its lower end
pivoted at P on the floor, making an angle ? with
the floor. A horizontal cable is attached from
its upper end E to a point A on a nearby wall. A
rope is attached at one-fourth of the way down
from the beams upper end, and hangs vertically
downward. A disgustingly cheery purple dinosaur
of mass M is attached motionless to the end of
the rope.
Choose any convenient point O about which to
calculate torques.
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Physics 23 Supplementary Problem Winter/Spring
2003
Test-Level Problem on Statics (from a Test given
in Winter 1996).
A stationary ladder of total length L13 m and
weight W216 N leans against a vertical smooth
wall, while its bottom legs rest on a rough
horizontal surface at a distance d5 m from the
base of the wall. Its top is located a distance
H12 m above the floor.
A rope is attached to the ladder at a distance
d2m from its base. A worker pulls horizontally
on the rope, producing a tension of T260 N.
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In the diagram to the right, sketch all the
forces acting on the ladder, and the location
where they act.
Weight.
Weight. At L/2 up the ladder.
Tension, of course.
Normal force exerted by wall.
The floor clearly exerts an upward normal force,
to balance W.
If T?N then a friction force acts on the ladder
due to the floor. We cant tell the direction of
this friction force from the data given.
The total floor force (call it P) is the sum of
the floor normal and friction forces (assumed in
directions).
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What are the vertical and horiz-ontal components
of the force applied on the ladder by the floor
if the ladder does not slip?
First think. We have been studying torques. Do
I need to use torques?
I will have two sum of force equations (x and y
components). I have three unknowns, Px, Py, and
N.
I need a third equation, so yes, I will have to
use the sum of torques equation.
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Think some more. Where is a good place to put
the rotation axis?
Answer at the point of application of one of
the forces.
I think force equations are easier than torque
equations (dont need to do R?F or RF? or RF
sin?).
Therefore, I choose to put the origin at the
floor contact point, where I have two unknown
torques.
Therefore, I choose to put the origin at the
floor contact point, where I have two unknown
torques. Ill pick clockwise for positive
rotations.
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OSE ?Fymay
Ny Wy Ty Py may
-W Py 0
Py W
OSE ?Fxmax
Nx Wx Tx Px max
-N T Px 0
Px N - T
Dont know N, dont know Px. Need another
equation.
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OSE ?tz I?z
tNz tWz tTz tPz I?z
-NH (D/2)W Td? 0
Dang! Forgot to show D/2 and d? on the diagram!
Better do it now!
Dang again! d? is not a system parameter!
Better figure it out now!
There are several ways to do it. I see two
similar triangles. Click twice to see them.
d?/d (H/2) / (L/2) ? d? d H/L
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Continuing
-NH (D/2)W Td? 0
-NH (D/2)W T d H/L 0
Two equations, two unknowns, solve for Px.
Px N - T
-NH (D/2)W T d H/L 0
NH (D/2)W T d H/L
N DW/2H T d /L
Px DW/2H T d / L - T
Numerically, Px -175 N, so it is directed in
the x direction. You are invited to check my
algebra.