Transcript Lecture 08: Equilibrium II

Physics 106: Mechanics
Lecture 08
Wenda Cao
NJIT Physics Department
Static Equilibrium
Important announcement
 Equilibrium and static
equilibrium
 Static equilibrium
conditions




Net external force must
equal zero
Net external torque must
equal zero
Solving static equilibrium
problems
March 10, 2011
Static and Dynamic Equilibrium
Equilibrium implies the object is at rest (static)
or its center of mass moves with a constant
velocity (dynamic)
 106 deals only with the special case in which
linear and angular velocities are equal to zero,
called “static equilibrium” : vCM = 0 and w = 0
 Examples






Book on table
Puck sliding on ice in a constant velocity
Ceiling fan – off
Ceiling fan – on
Ladder leaning against wall (foot in groove)
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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
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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
This must be true for any axis
of rotation
March 10, 2011
Conditions for Equilibrium
 The

F  0
If the object is modeled as a particle, then
this is the only condition that must be
satisfied
 The

net force equals zero
net torque equals zero

0
This is needed if the object cannot be
modeled as a particle
 These
conditions describe the rigid objects
in equilibrium analysis model
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Equilibrium Equations


Fnet   Fext  0 : Fnet , x  0 Fnet , y  0 Fnet , z  0


 net   ext  0 :  net , x  0  net , y  0  net , z  0
Equation 1:
 Equation 2:
 We will restrict the applications to situations
in which all the forces lie in the xy plane
 There are three resulting equations

Fnet , x   Fext , x  0
Fnet , y   Fext , y  0
 net , z    ext , z  0
March 10, 2011
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.
Fnet , y  n  mg  Mg  m pl g  0
n  mg  Mg  m pl g
 net , z   d   f   pl   n
 mgd  Mgx  0  0  0
mgd  Mgx
m
x   d  d
M 
Fnet , x   Fext , x  0
Fnet , y   Fext , y  0
 net , z    ext , z  0
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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

March 10, 2011
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
 mgd  Mgx  0  0  0
mgd  Mgx
 0  Mg (d  x)  m pl gd  nd  0
m
x
M
mgd  Mgx
P
 Mgd  Mgx  m pl gd  ( Mg  mg  m pl g )d  0

d

m
x
M

d

O
Fnet , x   Fext , x  0
Fnet , y   Fext , y  0
 net , z    ext , z  0
March 10, 2011
Center of Gravity
The torque due to the gravitational force on an
object of mass M is the force Mg acting at the
center of gravity of the object
 If g is uniform over the object, then the center
of gravity of the object coincides with its center
of mass
 If the object is homogeneous and symmetrical,
the center of gravity coincides with its geometric
center

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Problem-Solving Strategy 1








Draw sketch, decide what is in or out the system
Draw a free body diagram (FBD)
Show and label all external forces acting on the object
Indicate the locations of all the forces
Establish a convenient coordinate system
Find the components of the forces along the two axes
Apply the first condition for equilibrium
Be careful of signs
F  F 0
net , x
ext , x
Fnet , y   Fext , y  0
March 10, 2011
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.
March 10, 2011
Horizontal Beam Example, 2

Analyze


Draw a free body
diagram
Use the pivot in the
problem (at the wall)
as the pivot


This will generally be
easiest
Note there are three
unknowns (T, R, q)
March 10, 2011
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

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Problem-Solving Strategy 2

Choose a convenient axis for calculating the net torque
on the object


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




Remember the choice of the axis is arbitrary
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
March 10, 2011
Horizontal Beam Example, 3
l
 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 cos q  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 sin 
 W p  Wb  T sin  
  71.7 
T sin 


q  tan 1 
T cos  (313 N ) cos 53
R

 581N
cos q
cos 71.7 
March 10, 2011
Problem-Solving Strategy 3





The two conditions of equilibrium will give a system of
equations
Solve the equations simultaneously
Make sure your results are consistent with your free
body diagram
If the solution gives a negative for a force, it is in the
opposite direction to what you drew in the free body
diagram
Fnet , x   Fext , x  0
Check your results to confirm
Fnet , y   Fext , y  0
 net , z    ext , z  0
March 10, 2011
Example 3:
A uniform beam, of length L and mass m = 1.8 kg, is at rest with its
ends on two scales (see figure). A uniform block, with mass M =
2.7 kg, is at rest on the beam, with its center a distance L / 4 from
the beam's left end.
What do the scales read?
Solve as equilibrium system
Fnet , x   Fext , x  0
Fnet , y   Fext , y  0
 net , z    ext , z  0
March 10, 2011
1:
Fnet , x   Fext , x  0
2:
Fnet , y   Fext , y  Fl  Fr  Mg  mg  0
3: Choose axis at O
 net , z    ext , z   l   Mg   mg   r
L
L
 mg   Fr  L  0
4
2
L
L
Fr L  Mg  mg
4
2
Mg mg
Fr 

 15 N
4
2
 Fl  0  Mg 
From 2:
Fl   Fr  Mg  mg  Mg  mg 
O
Mg mg

 29 N
4
2
March 10, 2011
Example 4:
A safe whose mass is M = 430 kg is
hanging by a rope from a boom with
dimensions a = 1.9 m and b = 2.5 m.
The boom consists of a hinged
beam and a horizontal cable that
connects the beam to a wall. The
uniform beam has a mass m of 85
kg; the mass of the cable and rope
Fv
are negligible.
Tc
Tr
mg
Fh
(a) What is the tension Tc in the
cable; i. e., what is the magnitude of
the force Tc on the beam from the
horizontal cable?
(b) What is the force at the hinge?
March 10, 2011
Fnet , x   Fext , x  Fh  Tc  0
Fh  Tc
Fnet , y   Fext , y  Fv  Mg  mg  0
Tr  Mg
Choose axis at hinge
 net , z    ext , z   h   v   mg   Tc   Tr
b
 0  0  mg  Tc a  Tr b  0
2
b
 mg  Tc a  Mgb  0
2
b
b
m b
Tc  Mg  mg
 ( M  ) g  6093 N
a
2a
2 a
Fh  Tc  6093 N
Fv  Mg  mg  5047 N
F  Fh  Fv  7912 N
2
2
March 10, 2011
Example 7: Rock Climber
In the figure, a rock climber with mass
m = 55 kg rests during a “chimney
climb,” pressing only with her
shoulders and feet against the walls of
a fissure of width w = 1.0 m. Her center
of mass is a horizontal distance d =
0.20 m from the wall against which her
shoulders are pressed. The coefficient
of static friction between her shoes
and the wall is m1 = 1.1, and between
her shoulders and the wall it is m2 =
0.70. To rest, the climber wants to
minimize her horizontal push on the
walls. The minimum occurs when her
feet and her shoulders are both on the
verge of sliding.
m2
m1
(a) What is the minimum horizontal push on the walls?
(b) What should the vertical distance h be between the shoulders and
feet, in order for her to remain stationary?
March 10, 2011
(a) What is the minimum horizontal push on the walls?
(b) What should the vertical distance h be between the shoulders and
feet, in order for her to remain stationary?
Fnet , x   Fext , x  N1  N 2  0
N1  N 2
Fnet , y   Fext , y  f1  f 2  mg  0
f1  m1 N1
f 2  m2 N 2
O
m1 N1  m 2 N 2  mg  0
mg
N
 300 N
m1  m 2
Choose axis at “O”
 net ,O    ext ,O   f 2   N 2   mg   f 1   N 1
 0  0  mgd  f1w  N1h  0
w = 1.0 m
f1w  mgd m1 N1w  mgd
h

 0.74m
N1
N1
March 10, 2011