Equilibrium and Torque
Equilibrium
An object is in “Equilibrium” when:
1. There is no net force acting on the object
2. There is no net Torque (we’ll get to this later)
In other words, the object is NOT experiencing
linear acceleration or rotational acceleration.
v
0
t


0
t
a
We’ll get to this later
Static Equilibrium
An object is in “Static Equilibrium” when it is
NOT MOVING.
x
=0
t
v
a
0
t
v

Dynamic Equilibrium
An object is in “Dynamic Equilibrium” when it is
MOVING with constant linear velocity
and/or rotating with constant angular velocity.
v
0
t


0
t
a

Equilibrium
Let’s focus on condition 1: net force = 0
F  0
The x components of force cancel
Fx  0
The y components of force cancel

 Fy  0


Condition 1: No net Force
We have already looked at situations where the net force = zero.
Determine the magnitude of the forces acting on each of the
2 kg masses at rest below.
60°
30°
30°
30°
Condition 1: No net Force
∑Fx = 0
N = 20 N
mg = 20 N
and
∑Fy = 0
∑Fy = 0
N - mg = 0
N = mg = 20 N
Condition 1: No net Force
∑Fx = 0
T1 = 10 N
T2 = 10 N
20 N
and
∑Fy = 0
∑Fy = 0
T1 + T2 - mg = 0
T1 = T2 = T
T + T = mg
2T = 20 N
T = 10 N
Condition 1: No net Force
N = mgcos 60
N = 10 N
Fx - f = 0
f = Fx = mgsin 60
f = 17.4 N
f = 17.4 N
60°
mg =20 N
N = 10 N
Condition 1: No net Force
∑Fx = 0
30°
and
∑Fy = 0
30°
T2 = 20 N
T1 = 20 N
T2y = 10 N
T2 = 20 N
30°
T2x
mg = 20 N
∑Fx = 0
T2x - T1x = 0
T1x = T2x
∑Fy = 0
T1y + T2y - mg = 0
2Ty = mg = 20 N
Ty/T = sin 30
T = Ty/sin 30
T = (10 N)/sin30
Equal angles ==> T1 = T2
Ty = mg/2 = 10 N
T = 20 N
Note: unequal angles ==> T1 ≠ T2
Condition 1: No net Force
∑Fx = 0
30°
T1 = 20 N
and
30°
T2 = 20 N
mg = 20 N
Note:
The y-components cancel, so
T1y and T2y both equal 10 N
∑Fy = 0
Condition 1: No net Force
∑Fx = 0
and
∑Fy = 0
30°
T1 = 40 N
T2 = 35 N
20 N
∑Fy = 0
T1y - mg = 0
T1y = mg = 20 N
T1y/T1 = sin 30
T1 = Ty/sin 30 = 40 N
∑Fx = 0
T2 - T1x = 0
T2 = T1x= T1cos30
T2 = (40 N)cos30
T2 = 35 N
Condition 1: No net Force
∑Fx = 0
and
30°
T1 = 40 N
T2 = 35 N
20 N
Note:
The x-components cancel
The y-components cancel
∑Fy = 0
Condition 1: No net Force
A Harder Problem!
a. Which string has the greater tension?
b. What is the tension in each string?
60°
30°
a. Which string has the greater tension?
∑Fx = 0 so T1x = T2x
60°
30°
T1
T2
T1 must be greater in order to have the same x-component as T2.
What is the tension in each string?
60°
30°
T1
∑Fx = 0
T2x-T1x = 0
T1x = T2x
T1cos60 = T2cos30
T2
∑Fy = 0
T1y + T2y - mg = 0
T1sin60 + T2sin30 - mg = 0
T1sin60 + T2sin30 = 20 N
Note: unequal angles ==> T1 ≠ T2
Solve
simultaneous
equations!
Equilibrium
An object is in “Equilibrium” when:
1. There is no net force acting on the object
2. There is no net Torque
In other words, the object is NOT experiencing
linear acceleration or rotational acceleration.
v
0
t


0
t
a
What is Torque?
Torque is like “twisting force”
The more torque you apply to a wheel the
more quickly its rate of spin changes
Math Review:
1. Definition of angle in “radians”
  s /r
arc length
 
radius
s

r

 2. One revolution = 360° = 2π radians
ex: π radians = 180°
ex: π/2 radians = 90°
Linear vs. Rotational Motion
• Linear Definitions
x
x
v
t
v
a
t
• Rotational Definitions
 in radians

 
in radians/sec or rev/min
t

 
in radians/sec/sec
t
f
x

xi
xf




i
Linear vs. Rotational Velocity
• A car drives 400 m in
20 seconds:
a. Find the avg linear velocity
• A wheel spins thru an angle of
400π radians in 20 seconds:
a. Find the avg angular velocity
 400 radians
 
=
t
20s
  20 rad/sec
 10 rev/sec
 600 rev/min
x 400m
v

 20m /s
t
20s
x



xi
xf

Linear vs. Rotational
Net Force ==> linear acceleration
The linear velocity changes
Net Torque ==> angular acceleration
The angular velocity changes
(the rate of spin changes)
   net
a Fnet

Torque
Torque is like “twisting force”
The more torque you apply to a
wheel, the more quickly its rate
of spin changes
Torque = Frsinø
Torque is like “twisting force”
Imagine a bicycle wheel that can only spin about its axle.
If the force is the same in each case, which case produces
a more effective “twisting force”?
This one!
Torque is like “twisting force”
Imagine a bicycle wheel that can only spin about its axle.
What affects the torque?
1.
2.
3.
The place where the force is applied: the distance “r”
The strength of the force
The angle of the force
F to r
ø

F
ø
F// to r
r

Torque is like “twisting force”
Imagine a bicycle wheel that can only spin about its axle.
What affects the torque?
1. The distance from the axis rotation “r” that the force is applied
2. The component of force perpendicular to the r-vector
F to r
ø

ø
r
F
Torque = Frsinø
Imagine a bicycle wheel that can only spin about its axle.
Torque = (the component of force perpendicular to r)(r)
Torque  (F )(r)
F  F sin ø
ø
  (F )(r)
  (F sin ø)(r)
  Frsin ø

ø
r
F
Torque is like “twisting force”
Imagine a bicycle wheel that can only spin about its axle.
F  F sin ø
  (F )(r)
  (F sin ø)(r)
  Frsin ø
ø

ø
r
F
Cross “r” with “F” and choose any angle
to plug into the equation for torque
  (F )(r)
  Frsin ø
F

ø
r
Since  and ø are supplementary angles
(ie :  + ø = 180)
 sin  = sinø

F  F sin ø

ø
r
ø
F
Two different ways of looking at torque
Torque = (Fsinø)(r)
Torque = (F)(rsinø)
Torque  (F )(r)
Torque  (F)(r )
F  F sin ø

F

r
F  F sin ø
ø

F
r
F

ø
r
r
r

ø
ø
Imagine a bicycle wheel that can only spin about its axle.
Torque = (F)(rsinø)
r is called the " moment arm" or "moment"
F
r
r

F
ø
r
ø

Equilibrium
An object is in “Equilibrium” when:
1. There is no net force acting on the object
2. There is no net Torque
In other words, the object is NOT experiencing
linear acceleration or rotational acceleration.
v
0
t


0
t
a
Condition 2: net torque = 0
Torque that makes a wheel want to rotate clockwise is +
Torque that makes a wheel want to rotate counterclockwise is -


Positive Torque


Negative Torque
Condition 2: No net Torque
Weights are attached to 8 meter long levers at rest.
Determine the unknown weights below
??
20 N
??
20 N
20 N
??
Condition 2: No net Torque
Weights are attached to an 8 meter long lever at rest.
Determine the unknown weight below
20 N
??
Condition 2: No net Torque
Upward force
from the fulcrum
produces no torque
(since r = 0)
r1 = 4 m
∑T’s = 0
T2 - T1 = 0
T2 = T1
F2r2sinø2 = F1r1sinø1
(F2)(4)(sin90) = (20)(4)(sin90)
r2 = 4 m
F2 = 20 N … same as F1
F1 = 20 N
F2 =??
Condition 2: No net Torque
20 N
20 N
Condition 2: No net Torque
Weights are attached to an 8 meter long lever at rest.
Determine the unknown weight below
20 N
??
Condition 2: No net Torque
r1 = 4 m
r2 = 2 m
∑T’s = 0
T2 - T1 = 0
T2 = T1
F2r2sinø2 = F1r1sinø1
(F2)(2)(sin90) = (20)(4)(sin90)
F1 = 20 N
F2 =??
(force at the fulcrum is not shown)
F2 = 40 N
Condition 2: No net Torque
20 N
40 N
Condition 2: No net Torque
Weights are attached to an 8 meter long lever at rest.
Determine the unknown weight below
20 N
??
Condition 2: No net Torque
r1 = 3 m
r2 = 2 m
∑T’s = 0
T2 - T1 = 0
T2 = T1
F2r2sinø2 = F1r1sinø1
(F2)(2)(sin90) = (20)(3)(sin90)
F1 = 20 N
F2 =??
(force at the fulcrum is not shown)
F2 = 30 N
Condition 2: No net Torque
20 N
30 N
In this special case where
- the pivot point is in the middle of the lever,
- and ø1 = ø2
F1R1sinø1 = F2R2sinø2
F1R1= F2R2
(20)(4) = (20)(4)
20 N
20 N
(20)(4) = (40)(2)
40 N
20 N
(20)(3) = (30)(2)
20 N
30 N
More interesting problems
(the pivot is not at the center of mass)
Masses are attached to an 8 meter long lever at rest.
The lever has a mass of 10 kg.
Determine the unknown weight below.
CM
20 N
??
More interesting problems
(the pivot is not at the center of mass)
Trick: gravity applies a torque “equivalent to”
(the weight of the lever)(Rcm)
Tcm =(mg)(rcm) = (100 N)(2 m) = 200 Nm
CM
??
20 N
Weight of lever
Masses are attached to an 8 meter long lever at rest.
The lever has a mass of 10 kg.
Masses are attached to an 8 meter long lever at rest.
The lever has a mass of 10 kg.
Determine the unknown weight below.
CM
R1 = 6 m
Rcm = 2 m
R2 = 2 m
∑T’s = 0
T2 - T1 - Tcm = 0
T2 = T1 + Tcm
F2r2sinø2 = F1r1sinø1 + FcmRcmsinøcm
(F2)(2)(sin90)=(20)(6)(sin90)+(100)(2)(sin90)
Fcm = 100 N
F1 = 20 N
F2 = ??
(force at the fulcrum is not shown)
F2 = 160 N
Other problems:
Sign on a wall#1 (massless rod)
Sign on a wall#2 (rod with mass)
Diving board (find ALL forces on the board)
Push ups (find force on hands and feet)
Sign on a wall, again
Sign on a wall #1
A 20 kg sign hangs from a 2 meter long massless rod
supported by a cable at an angle of 30° as shown.
Determine the tension in the cable.
(force at the pivot point is not shown)
Ty = mg = 200N
T
30°
Eat at Joe’s
We don’t need to use torque
if the rod is “massless”!
30°
Pivot point
mg = 200N
Ty/T = sin30
T = Ty/sin30 = 400N
Sign on a wall #2
A 20 kg sign hangs from a 2 meter long rod that
has a mass of 10 kg and is supported by a cable at an
angle of 30° as shown. Determine the tension in the cable “FT”
FT
30°
Eat at Joe’s
30°
Fcm = 100N
Pivot point
mg = 200N
(force at the pivot point is not shown)
Sign on a wall #2
A 20 kg sign hangs from a 2 meter long rod that
has a mass of 10 kg and is supported by a cable at an
angle of 30° as shown. Determine the tension in the cable.
∑T = 0
TFT = Tcm + Tmg
FT(2)sin30 =100(1)sin90 + (200)(2)sin90
FT = 500 N
FT
30°
Fcm = 100N
Pivot point
mg = 200N
(force at the pivot point is not shown)
Diving board
A 4 meter long diving board with a mass of 40 kg.
a. Determine the downward force of the bolt.
b. Determine the upward force applied by the fulcrum.
bolt
Diving board
A 4 meter long diving board with a mass of 40 kg.
a. Determine the downward force of the bolt.
(Balance Torques)
∑T = 0
bolt
R1 = 1
Fbolt = 400 N
Rcm = 1
Fcm = 400 N
(force at the fulcrum is not shown)
Diving board
A 4 meter long diving board with a mass of 40 kg.
a. Determine the downward force of the bolt.
(Balance Torques)
b. Determine the upward force applied by the fulcrum.
(Balance Forces)
F = 800 N
bolt
Fbolt = 400 N
Fcm = 400 N
∑F = 0
Remember:
An object is in “Equilibrium” when:
a. There is no net Torque
  0
b. There is no net force acting on the object


F  0
Push-ups #1
A 100 kg man does push-ups as shown
Fhands
Ffeet
0.5 m
1m
CM
30°
Find the force on his hands and his feet
Answer:
Fhands = 667 N
Ffeet = 333 N
A 100 kg man does push-ups as shown
Fhands
Ffeet
0.5 m
1m
CM
30°
mg = 1000 N
Find the force on his hands and his feet
∑T = 0
TH = Tcm
FH(1.5)sin60 =1000(1)sin60
∑F = 0
Ffeet + Fhands = mg = 1,000 N
Ffeet = 1,000 N - Fhands = 1000 N - 667 N
FH = 667 N
FFeet = 333 N
Push-ups #2
A 100 kg man does push-ups as shown
Fhands
0.5 m
1m
90°
CM
30°
Find the force acting on his hands
Push-ups #2
A 100 kg man does push-ups as shown
Fhands
(force at the feet is not shown)
0.5 m
1m
30°
Force on hands:
90°
CM
mg = 1000 N
∑T = 0
TH = Tcm
FH(1.5)sin90 =1000(1)sin60
FH = 577 N
Sign on a wall, again
A 20 kg sign hangs from a 2 meter long rod that
has a mass of 10 kg and is supported by a cable at an
angle of 30° as shown
Find the force exerted by the wall on the rod
FT = 500N
FW = ?
30°
Eat at Joe’s
30°
Fcm = 100N
mg = 200N
(forces and angles NOT drawn to scale!
Find the force exerted by the wall on the rod
FT = 500N
30°
Eat at Joe’s
FWx = FTx = 500N(cos30)
FWx= 433N
FWy + FTy = Fcm +mg
FWy = Fcm + mg - FTy
FWy= 300N - 250
FWy= 50N
FW
30°
Fcm = 100N
FW= 436N
mg = 200N
FWy= 50N
FWx= 433N
FW= 436N
(forces and angles NOT drawn to scale!)