Transcript Potential Energy - McMaster University

Static Equilibrium
(Serway 12.1-12.3)
Physics 1D03
Equilibrium of a Rigid Body
For a particle, “Equilibrium” means Fnet = 0, and then v = constant.
For extended 2-D or 3-D objects, this is not enough!
F1
Here
F  0 , so aCM = 0.
But the object is not in equilibrium:
It will spin (angular acceleration).
-F1
A “force couple”
Physics 1D03
Equilibrium of a Rigid Body
Net external forces and net external torque must be zero
for a body in equilibrium.
1) a = 0 (no translational acceleration), so
F  0
(no net force)
2) a = 0 (no angular acceleration), so
τ  0
(no net torque)
Physics 1D03
Example:
Uniform beam AC:
weight w1 = 100 N
length L = 3.00 m
Hinge at A, cable from B to C
w2 = 200 N
Find: tension in BC, force at A.
B
f = 30°
A
5/
6
C
1/
6
L
w2
Steps:
1) Free-body diagram for the beam. The forces will be the
weights w1 and w2, and the forces from the supports at A and C.
2) Net force = 0 (2 equations), net torque =0 (1 equation).
3) Solve for up to 3 unknowns.
Physics 1D03
L
P
Forces on beam:
P
T
θ
y
30°
A
Px
w1
C
w2
Note the force P from the wall. The hinge at A prevents translational
motion in any direction, so it may exert a force P in any direction.
The unknowns are: tension T, force |P|, and angle q ;
or, T, Px and Py .
Physics 1D03
To find the tension, consider torques about A:
+
½ L w1 + 5/6 L w2 –LT sin30o = 0
so (after a little work),
5/
T = w1 + 3 w2
= 433 N
T
P
y
T sin30o
30°
A
Px
w1
½L
5/
6
C
w2
L
To find the forces at A, consider the net force on the beam:
x components,
y components,
Px - T cos30o = 0, so Px = T cos30o = 375 N
Py + T sin30o - w1 - w2 = 0, so Py = 83 N
Physics 1D03
P
P
T
θ
y
30°
A
Question:
Px
w1
C
w2
How can you generate equations for Px and Py
which don’t involve the tension T?
Physics 1D03
Answers:
1) torques about C: find Py (without using the value of tension!)
2) to find Px :
B
Take torques about B!
Py and T produce zero torque
about B, so we get an equation
in one unknown (Px ).
T
P
y
30°
A
Px
w1
C
w2
Physics 1D03
Equations of Equilibrium
In a 2-D problem, we can generate, at most, three independent
equations from the requirements of static equilibrium of a single
object. Possible combinations:
1) Two force equations,
one torque equation
2) One force equation (components along some axis)
two torque equations (torques about two different “pivots”).
3) Three torque equations (torques about 3 different “pivots”).
2) and 3) sometimes do not give 3 independent equations. You
shouldn’t pick 3 collinear pivot points; and if you use two pivot
points, taking force components along an axis perpendicular to the
line through these two points will not give a third independent
equation.
Physics 1D03
Static Friction Problems: “When does the ladder slip?”
Assume it is not slipping (so it is in equilibrium);
but is about to slip (so you can set fs = ms n, at each point that
has to slip for motion to occur).
We may want to know:
For what angles q will the ladder slip?
or
What coefficient of friction is required to
prevent slipping?
θ
Physics 1D03
Example: ladder
A uniform ladder of length L leans against a smooth (= frictionless)
wall at angle q to the ground. What is the smallest coefficient of
friction μs between ladder and ground which will prevent the ladder
from slipping?
Consider the free body diagram. Note the
force P from the wall is perpendicular (no
friction from the wall).
B
P
θ
C
Plan: treat the weight w, length L,
and angle q as known; use the
equations of equilibrium to find the
unknowns N and fs (and perhaps P).
final answer: minimum mS 
N
A
w
θ
fs
1
2 tan q
Physics 1D03
Take torques about A:
 A  w cosq  12 L  P sin q  L  0
P 
θ
w
2 tan q
y
x
C
N
Then, net force is zero:
B
P
A
Fx  f S  P  0
w
θ
fs
Fy  N  w  0
fS P
1
 minimummS 
 
N
w 2 t anq
Or the minimum angle:
1
tan(q ) 
2m s
Note that w, L cancelled.
Physics 1D03
What if someone is standing on the ladder? Does that change the
risk of slipping?
What if there is friction at both the floor and the wall?
What if the floor is smooth and the wall is rough (has friction)? Can
the ladder still stay in equilibrium for some range of angles?
Physics 1D03