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8/22/2019 AP Equilibrium Torque
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Equilibrium and Torque
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EquilibriumAn object is in Equilibrium when:
1. There is no net force acting on the object
2. There is no net Torque (well get to this later)
In other words, the object is NOT experiencing
linearacceleration or rotational acceleration.
a v
t 0
t 0 Well get to this later
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Static Equilibrium
An object is in Static Equilibrium when it is
NOT MOVING.
v x
t= 0
a v
t 0
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Dynamic Equilibrium
An object is in Dynamic Equilibrium when it is
MOVING with constant linear velocity
and/or rotating with constant angular velocity.
a v
t 0
t 0
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EquilibriumLets focus on condition 1: net force = 0
The x components of force cancel
The y components of force cancel
F 0
Fx 0
Fy 0
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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
3030 30
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Condition 1: No net Force
Fx = 0 and Fy = 0
mg = 20 N
N = 20 NFy = 0
N - mg = 0N = mg = 20 N
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Condition 1: No net Force
Fx = 0 and Fy = 0
Fy = 0
T1 + T2 - mg = 0
T1 = T2 = T
T + T = mg
2T = 20 N
T = 10 N
T1 = 10 N
20 N
T2 = 10 N
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Condition 1: No net Force
60
mg =20 N
f = 17.4 N N = 10 N
Fx - f = 0
f = Fx = mgsin 60
f =17.4 N
N = mgcos 60
N = 10 N
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Condition 1: No net Force
Fx = 0 and Fy = 0
3030
mg = 20 N
T1 = 20 N T2 = 20 N
Fy = 0
T1y + T2y - mg = 02Ty = mg = 20 N
Ty = mg/2 = 10 N
T2 = 20 N
T2x
T2y = 10 N
30
Ty/T = sin 30
T = Ty/sin 30T = (10 N)/sin30
T = 20 N
Fx = 0
T2x - T1x = 0
T1x = T2x
Equal angles ==> T1 = T2
Note: unequal angles ==> T1 T2
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Condition 1: No net Force
Fx = 0 and Fy = 0
Note:
The y-components cancel, so
T1y and T2y both equal 10 N
3030
mg = 20 N
T1 = 20 N T2 = 20 N
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Condition 1: No net Force
Fx = 0 and Fy = 0
30
20 N
T1 = 40 N
T2 = 35 N
Fy = 0
T1y - mg = 0T1y = mg = 20 N
T1y/T1 = sin 30
T1 = Ty/sin 30 = 40 N
Fx = 0
T2- T1x = 0T2= T1x= T1cos30
T2 = (40 N)cos30
T2 = 35 N
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Condition 1: No net Force
Fx = 0 and Fy = 0
Note:
The x-components cancel
The y-components cancel
30
20 N
T1 = 40 N
T2 = 35 N
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3060
Condition 1: No net Force
A Harder Problem!
a. Which string has the greater tension?
b. What is the tension in each string?
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a. Which string has the greater tension?
Fx = 0 so T1x = T2x
3060
T1 T2
T1 must be greater in order to have the same x-component as T2.
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Fy = 0
T1y + T2y - mg = 0
T1sin60 + T2sin30 - mg = 0
T1sin60 + T2sin30 = 20 N
Solve
simultaneous
equations!
Fx = 0
T2x-T1x = 0
T1x = T2xT1cos60 = T2cos30
3060
T1 T2
Note: unequal angles ==> T1 T2
What is the tension in each string?
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EquilibriumAn 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 accelerationorrotational acceleration.
a v
t 0
t 0
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What is Torque?
Torque is like twisting force
The more torque you apply to a wheel the
more quickly its rate of spin changes
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Math Review:
1. Definition of angle in radians
2. One revolution = 360 = 2 radians
ex: radians = 180ex: /2 radians = 90
s /r
arc length
radius
s
r
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Linear vs. RotationalMotion
Linear Definitions Rotational Definitions
x
v x
t
a v
t
in radians
t
in radians/sec or rev/min
t in radians/sec/sec
x
xi xf
i
f
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Linear vs. RotationalVelocity
A car drives 400 m in20 seconds:
a. Find the avg linear velocity
A wheel spins thru an angle of400 radians in 20 seconds:
a. Find the avg angular velocity
v x
t
400m
20s 20m /s
t=
400 radians
20s
20 rad/sec
10 rev/sec
600 rev/min
x
xi xf
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Linear vs. Rotational
Net Force ==> linear acceleration
The linear velocity changes
aFnet
Net Torque ==> angular acceleration
The angular velocity changes
(the rate of spin changes)
net
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Torque
Torque is like twisting force
The more torque you apply to a
wheel, the more quickly its rateof spin changes
Torque = Frsin
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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 producesa more effective twisting force?
This one!
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Torque is like twisting force
Imagine a bicycle wheel that can only spin about its axle.
What affects the torque?1. The place where the force is applied: the distance r
2. The strength of the force
3. The angle of the force
r
F
F// to r
F to r
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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 rthat the force is applied
2. The component of force perpendicular to the r-vector
r
F
F to r
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Imagine a bicycle wheel that can only spin about its axle.
Torque = (the component of force perpendicular to r)(r)
r
F
F Fsin
(F
)(r)
(Fsin)(r)
Frsin
Torque = Frsin
Torque (F)(r)
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Torque is like twisting force
Imagine a bicycle wheel that can only spin about its axle.
r
F
F Fsin
(F
)(r)
(Fsin)(r)
Frsin
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(F )(r) Frsin
r
F
F Fsin
F
r
Since and are supplementary angle
(ie : + = 180)sin = sin
Cross r with F and choose any angle
to plug into the equation for torque
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Torque = (Fsin)(r)
Two different ways of looking at torque
r
F
F Fsin
Torque = (F)(rsin)
r
F
r
r
F Fsin
r
F
Torque (F)(r)
Torque (F)(r)
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Imagine a bicycle wheel that can only spin about its axle.
Torque = (F)(rsin)
r
F
r
r
F
r is called the "moment arm" or "moment"
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EquilibriumAn 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 accelerationorrotational acceleration.
a vt
0
t 0
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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
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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 ??
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Condition 2: No net Torque
Weights are attached to an 8 meter long leverat rest.Determine the unknown weight below
20 N ??
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Condition 2: No net Torque
F1 = 20 N
r1 = 4 m r2 = 4 m
F2 =??
Ts = 0
T2 - T1 = 0
T2 = T1F2r2sin2 = F1r1sin1
(F2)(4)(sin90) = (20)(4)(sin90)
F2= 20 N same as F1
Upward force
from the fulcrum
produces no torque
(since r = 0)
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Condition 2: No net Torque
20 N 20 N
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Condition 2: No net Torque
20 N ??
Weights are attached to an 8 meter long leverat rest.Determine the unknown weight below
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Condition 2: No net Torque
Ts = 0
T2 - T1 = 0T2 = T1
F2r2sin2 = F1r1sin1
(F2)(2)(sin90) = (20)(4)(sin90)
F2 = 40 NF2 =??
F1 = 20 N
r1 = 4 m r2 = 2 m
(force at the fulcrum is not shown)
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Condition 2: No net Torque
20 N 40 N
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Condition 2: No net Torque
20 N ??
Weights are attached to an 8 meter long leverat rest.Determine the unknown weight below
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Condition 2: No net Torque
F2 =??
F1 = 20 N
r1 = 3 m r2 = 2 m
Ts = 0
T2 - T1 = 0
T2 = T1F2r2sin2 = F1r1sin1
(F2)(2)(sin90) = (20)(3)(sin90)
F2 = 30 N
(force at the fulcrum is not shown)
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Condition 2: No net Torque
20 N 30 N
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In this special case where
- the pivot point is in the middle of the lever,
- and 1 = 2
F1R1sin1 = F2R2sin2
F1R1= F2R2
20 N 20 N
20 N 40 N
20 N 30 N
(20)(4) = (20)(4)
(20)(4) = (40)(2)
(20)(3) = (30)(2)
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More interesting problems
(the pivot is not at the center of mass)
Masses are attached to an 8 meter long leverat rest.
The lever has a mass of 10 kg.
Determine the unknown weight below.
20 N ??
CM
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Trick: gravity applies a torque equivalent to(the weight of the lever)(Rcm)
Tcm =(mg)(rcm) = (100 N)(2 m) = 200 Nm
More interesting problems
(the pivot is not at the center of mass)
20 N??
CM
Weight of lever
Masses are attached to an 8 meter long leverat rest.
The lever has a mass of 10 kg.
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Masses are attached to an 8 meter long leverat rest.
The lever has a mass of 10 kg.
Determine the unknown weight below.
F1 = 20 N
CM
R1 = 6 m R2 = 2 m
F2 = ??
Rcm = 2 m
Fcm = 100 N
Ts = 0
T2 - T1 - Tcm = 0T2 = T1 + Tcm
F2r2sin2 = F1r1sin1 + FcmRcmsincm(F2)(2)(sin90)=(20)(6)(sin90)+(100)(2)(sin90)
F2 = 160 N
(force at the fulcrum is not shown)
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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
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Sign on a wall #1
Eat at Joes
30
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.
30
Pivot point
T Ty = mg = 200N
mg = 200N
Ty/T = sin30
T = Ty/sin30 = 400N
We dont need to use torque
if the rod is massless!
(force at the pivot point is not shown)
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Sign on a wall #2
Eat at Joes
30
A 20 kg sign hangs from a 2 meter long rod that
has a mass of 10 kgand is supported by a cable at an
angle of 30 as shown. Determine the tensionin the cable FT
30
Pivot point
FT
mg = 200N
Fcm= 100N
(force at the pivot point is not shown)
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Sign on a wall #2
A 20 kg sign hangs from a 2 meter long rod that
has a mass of 10 kgand is supported by a cable at an
angle of 30 as shown. Determine the tension in the cable.
30
Pivot point
FT
mg = 200N
Fcm = 100N
T = 0
TFT = Tcm + Tmg
FT(2)sin30 =100(1)sin90 + (200)(2)sin90
FT = 500 N
(force at the pivot point is not shown)
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Diving board
bolt
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.
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Diving board
A 4 meter long diving board with a mass of 40 kg.
a. Determine the downward force of the bolt.(Balance Torques)
bolt
Rcm = 1R1 = 1
Fcm = 400 NFbolt = 400 N
T = 0
(force at the fulcrum is not shown)
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Diving boardA 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)
bolt
Fcm = 400 NFbolt = 400 N
F = 800 N F = 0
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Remember:
An object is in Equilibrium when:
a. There is no net Torque
b. There is no net force acting on the object
F 0
0
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Push-ups #1
A 100 kg man does push-ups as shown
Find the force on his hands and his feet
1 m
0.5 m
30
Fhands
Ffeet
CM
Answer:
Fhands = 667 N
Ffeet = 333 N
A 100 kg man does push ups as shown
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A 100 kg man does push-ups as shown
Find the force on his hands and his feet
T = 0
TH = Tcm
FH(1.5)sin60 =1000(1)sin60
FH = 667 N
F = 0
Ffeet + Fhands = mg = 1,000 N
Ffeet = 1,000 N - Fhands = 1000 N - 667 N
FFeet = 333 N
1 m
0.5 m
30
Fhands
Ffeet
CM
mg = 1000 N
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Push-ups #2
A 100 kg man does push-ups as shown
Find the force acting on his hands
1 m
0.5 m
30
Fhands
CM90
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Push-ups #2A 100 kg man does push-ups as shown
T = 0
TH = TcmFH(1.5)sin90 =1000(1)sin60
FH = 577 N
1 m
0.5 m
30
Fhands
CM
90
mg = 1000 N
Force on hands:
(force at the feet is not shown)
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Sign on a wall, again
A 20 kg sign hangs from a 2 meter long rod that
has a mass of 10 kgand is supported by a cable at an
angle of 30 as shown
Find the force exerted by the wall on the rod
Eat at Joes
30 30
FT = 500N
mg = 200N
Fcm = 100N
FW = ?
(forces and angles NOT drawn to scale!
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Find the force exerted by the wall on the rod
FWx = FTx = 500N(cos30)
FWx= 433N
FWy + FTy = Fcm +mg
FWy = Fcm + mg - FTy
FWy= 300N - 250
FW 50N (forces and angles NOT drawn to scale!)
Eat at Joes
30 30
FT = 500N
mg = 200NFcm = 100N
FW
FWy= 50NFWx= 433N
FW= 436N
FW= 436N