Download Static Equilibrium

Statics
Physics 116 2017
Tues. 2/7 and Thurs. 2/9
Conditions for Static Equilibrium
• Equilibrium and static
equilibrium
• Static equilibrium
conditions
– Net external force must
equal zero
– Net external torque must
equal zero
• Center of gravity or
center of mass
• Solving static equilibrium
problems
Conditions for Equilibrium
• The first condition of
equilibrium is a statement of
translational equilibrium
• The net external force on the
object must equal zero
• It states that the translational
acceleration of the object’s
center of mass must be zero
Conditions for Equilibrium
• The first condition of
equilibrium is a statement of
translational equilibrium
• The net external force on the
object must equal zero



Fnet   Fext  ma  0
• It states that the translational
acceleration of the object’s
center of mass must be zero
Conditions for Equilibrium
• The first condition of
equilibrium is a statement of
translational equilibrium
• The net external force on the
object must equal zero



Fnet   Fext  ma  0
• It states that the translational
acceleration of the object’s
center of mass must be zero
Conditions for Equilibrium
• The second condition of
equilibrium is a statement of
rotational equilibrium
• The net external torque on
the object must equal zero
• It states the angular
acceleration of the object to
be zero
• This must be true for any axis
of rotation
Conditions for Equilibrium
• The second condition of
equilibrium is a statement of
rotational equilibrium
• The net external torque on
the object must equal zero



 net   ext  I  0
• It states the angular
acceleration of the object to
be zero for any axis of
rotation
Conditions for Equilibrium
• The second condition of
equilibrium is a statement of
rotational equilibrium
• The net external torque on
the object must equal zero



 net   ext  I  0
• It states the angular
acceleration of the object to
be zero for any axis of
rotation
A seesaw consisting of a uniform board of mass mpl and
length L supports at rest a father and daughter with masses
M and m, respectively. The support is under the center of
gravity of the board, the father is a distance x from the
center, and the daughter is a distance 2.00 m from the
center.
A) Find the magnitude of the upward force n exerted by the
support on the board.
B) Find where the father should sit to balance the system at
rest.
A seesaw consisting of a uniform board of mass mpl and
length L supports at rest a father and daughter with masses
M and m, respectively. The support is under the center of
gravity of the board, the father is a distance x from the
center, and the daughter is a distance 2.00 m from the
center.
A) Find the magnitude of the upward force n exerted by the
support on the board.
B) Find where the father should sit to balance the system at
rest.
A) Find the magnitude of the upward force n exerted by the support on
the board.
A) Find the magnitude of the upward force n exerted by the support on
the board.


Fnet , x   Fext , x  0


Fnet , y   Fext , y  0


 net , z   ext , z  0
A) Find the magnitude of the upward force n exerted by the support on
the board.


Fnet , x   Fext , x  0






Fnet , y   Fext , y  n  mg  m pl g  Mg  0




n  mg  m pl g  Mg


Fnet , x   Fext , x  0


Fnet , y   Fext , y  0


 net , z   ext , z  0
A) Find the magnitude of the upward
force n exerted by the support on the
board.
B) Find where the father should sit
to balance the system at rest.






 net , z   ext   d   f   pl   n  0

 net , z  mgd  Mgx  0  0  0
mgd  Mgx
m
m
xd
 2  2.00m
M
M


Fnet , x   Fext , x  0


Fnet , y   Fext , y  0


 net , z   ext , z  0
Axis of Rotation
• The net torque is about an axis through any
point in the xy plane
• Does it matter which axis you choose for
calculating torques?
• NO. The choice of an axis is arbitrary
• If an object is in translational equilibrium and
the net torque is zero about one axis, then the
net torque must be zero about any other axis
• We should be smart to choose a rotation axis to
simplify problems
Axis of Rotation
• The net torque is about an axis through any
point in the xy plane
• Does it matter which axis you choose for
calculating torques?
• NO. The choice of an axis is arbitrary
• If an object is in translational equilibrium and
the net torque is zero about one axis, then the
net torque must be zero about any other axis
• We should be smart to choose a rotation axis to
simplify problems
Axis of Rotation
• The net torque is about an axis through any
point in the xy plane
• Does it matter which axis you choose for
calculating torques?
• NO. The choice of an axis is arbitrary
• If an object is in translational equilibrium and
the net torque is zero about one axis, then the
net torque must be zero about any other axis
• We should be smart to choose a rotation axis to
simplify problems
Axis of Rotation
• The net torque is about an axis through any
point in the xy plane
• Does it matter which axis you choose for
calculating torques?
• NO. The choice of an axis is arbitrary
• If an object is in translational equilibrium and
the net torque is zero about one axis, then the
net torque must be zero about any other axis
• We should be smart to choose a rotation axis to
simplify problems
Axis of Rotation
• The net torque is about an axis through any
point in the xy plane
• Does it matter which axis you choose for
calculating torques?
• NO. The choice of an axis is arbitrary
• If an object is in translational equilibrium and
the net torque is zero about one axis, then the
net torque must be zero about any other axis
• We should be smart to choose a rotation axis to
simplify problems
B) Find where the father should sit to balance the system at rest.
Rotation axis O

  

 net , z   d   f   pl   n
Rotation axis P





 net , z   d   f   pl   n


 net , z  0  Mg ( x  d )  m pl gd  nd  0
mgd  Mgx
m
m
xd
2
M
M
 net , z   Mgd  Mgx  m pl gd  ( Mg  mg  m pl g )d  0
 net , z  mgd  Mgx  0  0  0
P

mgd  Mgx
m
m
xd
2
M
M
O
i-Clicker!
Six identical 2-meter massless rods are supporting identical 12-Newton
weights. In each case, a vertical force F is holding the rods and the
weights at rest. The rods are marked at half-meter intervals.
Rank the magnitude of the vertical force F applied to the rod from the
strongest to weakest.



 net , z   M   F  0
A. F = E > C > D = A > B

 net , z  Mgd  Fr  0
B. F > E > C > D > A > B
C. F > C > E = A > D = B
Mgd  Fr
D. E > F > C > A > D > B
Mgd
F
E. Same for all cases
r
i-Clicker!
Six identical massless rods are all supported by a fulcrum and are
tilted at the same angle to the horizontal. Each rod is marked at 1meter intervals.
Rank the magnitude of the vertical force F applied to the end of the
rod from the strongest to the weakest.



 net , z   M   F  0
A. F > E > D > B > C > A

 net , z  Mgd  Fr  0
B. D > E > F > A > C > B
C. F > B > E > C > D > A
D. F > E > B > D > C > A
E. Same for all cases
i-Clicker!
Six identical massless rods are all supported by a fulcrum and are
tilted at the same angle to the horizontal. Each rod is marked at 1meter intervals.
Rank the magnitude of the vertical force F applied to the end of the
rod from the strongest to the weakest.



 net , z   M   F  0
A. F > E > D > B > C > A

 net , z  Mgd  Fr  0
B. D > E > F > A > C > B
C. F > B > E > C > D > A
Mgd  Fr
D. F > E > B > D > C > A
Mgd
F
E. Same for all cases
r
Horizontal Beam Example
A uniform horizontal beam with a
length of l = 8.00 m and a weight of
Wb = 200 N is attached to a wall by a
pin connection. Its far end is
supported by a cable that makes an
angle of  = 53 with the beam. A
person of weight Wp = 600 N stands
a distance d = 2.00 m from the wall.
Find the tension in the cable as well
as the magnitude and direction of the
force exerted by the wall on the
beam.
Horizontal Beam Example
• The beam is uniform
– So the center of gravity
is at the geometric
center of the beam
• The person is standing on
the beam
• What are the tension in
the cable and the force
exerted by the wall on
the beam?
Horizontal Beam Example, 2
• Analyze
– Draw a free body
diagram
– Use the pivot in the
problem (at the wall)
as the pivot point
• This will generally be
easiest
– Note there are three
unknowns (T, R, q)
Horizontal Beam Example, 3
• The forces can be
resolved into
components in the
free body diagram
• Apply the two
conditions of
equilibrium to obtain
three equations
• Solve for the
unknowns
Horizontal Beam Example, 3
l
Choose pivot point about hinge near wall
  z  (T sin  )(l )  W p d  Wb ( )  0
2
l
W p d  Wb ( )
2  (600 N )(2m)  (200 N )(4m)  313N
T
l sin 
(8m) sin 53
Horizontal Beam Example, 3
l
Choose pivot point about hinge near wall
  z  (T sin  )(l )  W p d  Wb ( )  0
2
l
W p d  Wb ( )
2  (600 N )(2m)  (200 N )(4m)  313N
T
l sin 
(8m) sin 53
 Fx  R cosq  T cos  0
 Fy  R sin q  T sin   W p  Wb  0
Horizontal Beam Example, 3
l
Choose pivot point about hinge near wall
  z  (T sin  )(l )  W p d  Wb ( )  0
2
l
W p d  Wb ( )
2  (600 N )(2m)  (200 N )(4m)  313N
T
l sin 
(8m) sin 53
 Fx  R cosq  T cos  0
 Fy  R sin q  T sin   W p  Wb  0
W p  Wb  T sin 
R sin q
 tan q 
R cos q
T cos 
 W p  Wb  T sin  
  71.1
q  tan 
T cos 


T cos  (313 N ) cos 53
R

 581N

cos q
cos 71.1
1
i-Clicker!
A beam is supported by a frictionless pin so that it is
horizontal. A mass is suspended from the right end of the
beam, and the left end is tethered by an anchored cable. Both
cases below are identical, except for the location of the
anchor point of the cable.
Is the magnitude of the torque due to the cable about the
pivot pin greater in case A, greater in case B, or the same in
both cases?
A. Greater in case A
B. Greater in case B
C. The same in both cases
i-Clicker!
A beam is supported by a frictionless pin so that it is
horizontal. A mass is suspended from the right end of the
beam, and the left end is tethered by an anchored cable. Both
cases below are identical, except for the location of the
anchor point of the cable.
Is the tension in the cable greater in case A, greater in case B,
or the same in both cases?
A. Greater in case A
B. Greater in case B
C. The same in both cases
Ladder Example
• A uniform ladder of length
l rests against a smooth,
vertical wall. The mass of
the ladder is m, and the
coefficient of static friction
between the ladder and
the ground is s = 0.40.
Find the minimum angle q
at which the ladder does
not slip.
A Problem-Solving Strategy
• Choose a convenient axis for calculating the net torque
on the object
– Remember the choice of the axis is arbitrary
• Choose an origin that simplifies the calculations as
much as possible
– A force that acts along a line passing through the origin
produces a zero torque
• Be careful of sign with respect to rotational axis
– I’ll choose positive if force tends to rotate object in CCW
– negative if force tends to rotate object in CW
– zero if force is on the rotational axis
• Apply the second condition for equilibrium


 net , z   ext , z  0
Choose an origin O that simplifies the
calculations as much as possible ?
• A uniform ladder of length l rests against a smooth, vertical
wall. The mass of the ladder is m, and the coefficient of
static friction between the ladder and the ground is s =
0.40. Find the minimum angle.
B)
A)
O
C)
D)
O
O
mg
O
mg
mg
mg
Choose an origin O that simplifies the
calculations as much as possible ?
• A uniform ladder of length l rests against a smooth, vertical
wall. The mass of the ladder is m, and the coefficient of
static friction between the ladder and the ground is s =
0.40. Find the minimum angle.
B)
A)
O
C)
D)
O
O
mg
mg
mg
mg
O
Want to choose an origin that
eliminates unknowns
Choose an origin O that simplifies the
calculations as much as possible ?
• A uniform ladder of length l rests against a smooth, vertical
wall. The mass of the ladder is m, and the coefficient of
static friction between the ladder and the ground is s =
0.40. Find the minimum angle.
B)
A)
O
C)
D)
O
O
mg
mg
mg
mg
O
Want to choose an origin
that keeps values we know
Want to choose an origin that
eliminates unknowns
Lever Arm
A lever arm is the shortest perpendicular distance between
the line of action of a force and the axis of rotation. It
provides an alternative way to determine the torque and is
especially useful when the forces do not lie along
coordinate axes.
lever arm=rsin
Torque=F*r*sin
Torque=F*lever arm
Sometimes it is simple to read the lever arm right off of the
diagram.
Three forces of equal
magnitude are applied to
a 3-m by 2-m rectangle.
Forces F1 and F2 act at 45°
angles to the vertical as
shown, while F3 acts
horizontally.
Is the motion due to the net torque about the point A
1. clockwise
2. counterclockwise
3. zero
What about point B? point C?
Another ladder example
A uniform ladder
weighing 60 N leans
against a frictionless
vertical wall. What is the
minimum coefficient of
friction necessary
between the ladder and
the floor if the ladder is
not to slip?
4m
3m
Another ladder example
Fwall(normal)
A uniform ladder
weighing 60 N leans
against a frictionless
vertical wall. What is the
minimum coefficient of
friction necessary
between the ladder and
the floor if the ladder is
not to slip?
Ffloor(friction)
Fearth(weight)
4m
3m
Ffloor(normal)
What do we know?
A uniform ladder
weighing 60 N leans
against a frictionless
vertical wall. What is the
minimum coefficient of
friction necessary
between the ladder and
the floor if the ladder is
not to slip?
Ffloor(friction)
Fwall(normal)
Fearth(weight)
4m
3m
Ffloor(normal)
Annotated solution
We are trying to find the coefficient We know the
of friction, 
forces have to
How is it related to what we know balance… SF=0
Fwall(normal)
(the weight)?
Ffloor(friction)=Ffloor(normal) 
Since SFvertical=0, by observation we
can see that Ffloor(normal) Fearth(weight)
 60 N
Fearth(weight)
4m
Since SFhorizontal=0, by observation
we can see that Fwall(normal)
Ffloor(friction) ??
Ffloor(friction)
There are two unknowns and only
one equation – we need another
3m
equation.
Ffloor(normal)
Sz=0 … to the rescue!
i-Clicker!
In this case, what are the best
choices for the axis of rotation?
Fwall(normal)
I. At the top of the ladder
II. At the middle of the ladder
III. At the bottom of the ladder
A.
B.
C.
D.
E.
I
II
III
I or II
II or III
Fearth(weight)
4m
Ffloor(friction)
3m
Ffloor(normal)
Annotated solution
We are trying to find the coefficient of
friction, 
How is the frictional torque related to the
weight torque about the base?
Ffloor(friction)=Ffloor(normal) 
and Ffloor(friction)=Fwall(normal)
Since Sbase=0, then
Fearth(weight)*(its lever arm) must balance
with Fwall(normal) *(its lever arm)
We know the
torques about any
axis have to
balance… S=0 Fwall(normal)
How do you find the lever arm?
It is the shortest distance from the line of
action of the force to the axis of rotation.
The lever arm for the weight=1.5 m and the
lever arm for the wall normal force is 4 m.
Fearth(weight)
4m
Ffloor(friction)
O
3m
Ffloor(normal)
Putting it all together
Sbase=0, so
Fearth(weight)*(its lever arm) must balance with
Fwall(normal) *(its lever arm)
Putting it all together
Sbase=0, so
Fearth(weight)*(1.5m) = Fwall(normal)*(4 m)
Fwall(normal) = Fearth(weight)*(1.5m)/(4 m)
Ffloor(friction)=Fwall(normal)  Fearth(weight)*0.375
  Fwall(normal)/Ffloor(normal)
Since in this case Fearth(weight) = Ffloor(normal)
  Fwall(normal)/Fearth(weight)
 = [Fearth(weight)*0.375] /Fearth(weight)
 = 0.375
How does this problem change if there
is a person of known weight standing
on the ladder at a known position?
Old exam problem
A 50 N monkey climbs up a
120 N ladder and stops 1/3 of
the way up. The upper and
lower ends of the ladder rest
on frictionless surfaces. The
lower end is fastened to the
wall by a massless horizontal
rope. Find the tension in the
rope when the monkey is 1/3
of the way up the ladder.
8m
6m
Old exam problem
A mass of M=225 kg hangs from the end of the
uniform strut whose mass is mstrut = 45.0 kg. Find:
(a) the tension T in the cable
(b) horizontal and vertical force components
exerted on the strut by the hinge.