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Equilibrium and Torque
EquilibriumAn object is in “Equilibrium” when: 1. There is no net force acting on the object2. There is no net Torque (we’ll get to this later)
In other words, the object is NOT experiencing linear acceleration or rotational acceleration.
†
a =DvDt
= 0
a =DwDt
= 0 We’ll get to this later
Static EquilibriumAn object is in “Static Equilibrium” when it is
NOT MOVING.
†
v =DxDt
= 0
a =DvDt
= 0
Dynamic EquilibriumAn object is in “Dynamic Equilibrium” when it isMOVING with constant linear velocityand/or rotating with constant angular velocity.
†
a =DvDt
= 0
a =DwDt
= 0
EquilibriumLet’s focus on condition 1: net force = 0
The x components of force cancel
The y components of force cancel
†
Âr F = 0
†
Âr F x = 0
†
Âr F y = 0
Condition 1: No net ForceWe have already looked at situations where the net force = zero.Determine the magnitude of the forces acting on each of the2 kg masses at rest below.
60°
30°30° 30°
Condition 1: No net Force
∑Fx = 0 and ∑Fy = 0
mg = 20 N
N = 20 N∑Fy = 0N - mg = 0N = mg = 20 N
Condition 1: No net Force
∑Fx = 0 and ∑Fy = 0
∑Fy = 0T1 + T2 - mg = 0
T1 = T2 = T
T + T = mg2T = 20 NT = 10 N
T1 = 10 N
20 N
T2 = 10 N
Condition 1: No net Force
60°
mg =20 N
f = 17.4 N N = 10 N
Fx - f = 0f = Fx = mgsin 60f = 17.4 N
N = mgcos 60 N = 10 N
Condition 1: No net Force
∑Fx = 0 and ∑Fy = 0
30°30°
mg = 20 N
T1 = 20 N T2 = 20 N
∑Fy = 0T1y + T2y - mg = 02Ty = mg = 20 N
Ty = mg/2 = 10 N
T2 = 20 N
T2x
T2y = 10 N30°
Ty/T = sin 30T = Ty/sin 30 T = (10 N)/sin30 T = 20 N
∑Fx = 0T2x - T1x = 0T1x = T2x
Equal angles ==> T1 = T2
Note: unequal angles ==> T1 ≠ T2
Condition 1: No net Force
∑Fx = 0 and ∑Fy = 0
Note:The y-components cancel, soT1y and T2y both equal 10 N
30°30°
mg = 20 N
T1 = 20 N T2 = 20 N
Condition 1: No net Force
∑Fx = 0 and ∑Fy = 0
30°
20 N
T1 = 40 N
T2 = 35 N
∑Fy = 0T1y - mg = 0T1y = mg = 20 N
T1y/T1 = sin 30T1 = Ty/sin 30 = 40 N
∑Fx = 0T2 - T1x = 0T2 = T1x= T1cos30T2 = (40 N)cos30
T2 = 35 N
Condition 1: No net Force
∑Fx = 0 and ∑Fy = 0
Note:The x-components cancelThe y-components cancel
30°
20 N
T1 = 40 N
T2 = 35 N
30°60°
Condition 1: No net Force
A Harder Problem!
a. Which string has the greater tension?b. What is the tension in each string?
a. Which string has the greater tension?
∑Fx = 0 so T1x = T2x
30°60°
T1 T2
T1 must be greater in order to have the same x-component as T2.
∑Fy = 0T1y + T2y - mg = 0T1sin60 + T2sin30 - mg = 0T1sin60 + T2sin30 = 20 N
Solvesimultaneousequations!
∑Fx = 0T2x-T1x = 0T1x = T2xT1cos60 = T2cos30
30°60°
T1 T2
Note: unequal angles ==> T1 ≠ T2
What is the tension in each string?
EquilibriumAn object is in “Equilibrium” when: 1. There is no net force acting on the object2. There is no net Torque
In other words, the object is NOT experiencing linear acceleration or rotational acceleration.
†
a =DvDt
= 0
a =DwDt
= 0
What is Torque?
Torque is like “twisting force”
The more torque you apply to a wheel themore quickly its rate of spin changes
Math Review:
1. Definition of angle in “radians”
2. One revolution = 360° = 2π radiansex: π radians = 180°ex: π/2 radians = 90°
†
Dq = s /r
Dq =arc length
radius†
Dq
s
r
Linear vs. Rotational Motion
• Linear Definitions • Rotational Definitions
†
Dx
v = DxDt
a = DvDt
†
Dq in radians
w =DqDt
in radians/sec or rev/min
a =DwDt
in radians/sec/sec
†
Dx
†
xi
†
x f
†
Dq
†
qi†
q f
Linear vs. Rotational Velocity• A car drives 400 m in 20 seconds:a. Find the avg linear velocity
• A wheel spins thru an angle of400π radians in 20 seconds:
a. Find the avg angular velocity
†
v = DxDt
=400m20s
= 20m /s
†
w =DqDt
= 400p radians20s
w = 20p rad/sec =10 rev/sec = 600 rev/min
†
Dx
†
xi
†
x f
†
Dq
Linear vs. Rotational
Net Force ==> linear accelerationThe linear velocity changes
†
a µ Fnet
Net Torque ==> angular accelerationThe angular velocity changes(the rate of spin changes)
†
a µt net
Torque
Torque is like “twisting force”
The more torque you apply to awheel, 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. The place where the force is applied: the distance “r”2. The strength of the force3. The angle of the force
rø
F
†
F// to r
†
F^ to r
ø
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 applied2. The component of force perpendicular to the r-vector
rø
F
†
F^ to r
ø
Imagine a bicycle wheel that can only spin about its axle.
Torque = (the component of force perpendicular to r)(r)
rø
F
†
F^ = F sinøø
†
t = (F^ )(r)t = (F sinø)(r)t = Frsinø
Torque = Frsinø
†
Torque = (F^ )(r)
Torque is like “twisting force”
Imagine a bicycle wheel that can only spin about its axle.
rø
F
†
F^ = F sinøø
†
t = (F^ )(r)t = (F sinø)(r)t = Frsinø
†
t = (F^ )(r)t = Frsinø
rø
F
†
F^ = F sinøø
F
rø
†
J
†
Since J and ø are supplementary angles(ie : J + ø = 180°)sinJ = sinø
Cross “r” with “F” and choose any angleto plug into the equation for torque
Torque = (Fsinø)(r)
Two different ways of looking at torque
rø
F
†
F^ = F sinøø
Torque = (F)(rsinø)
r ø
F
†
r̂ ø
r†
F^ = F sinø
†
r̂
F
†
Torque = (F^ )(r)
†
Torque = (F)(r̂ )
Imagine a bicycle wheel that can only spin about its axle.
Torque = (F)(rsinø)
r ø
F
†
r̂ ø
†
r̂
F†
r̂ is called the "moment arm" or "moment"
EquilibriumAn object is in “Equilibrium” when: 1. There is no net force acting on the object2. There is no net Torque
In other words, the object is NOT experiencing linear acceleration or rotational acceleration.
†
a =DvDt
= 0
a =DwDt
= 0
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 -
†
Dq
†
Dq
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
F1 = 20 N
r1 = 4 m r2 = 4 m
F2 =??
∑T’s = 0
T2 - T1 = 0T2 = T1F2r2sinø2 = F1r1sinø1(F2)(4)(sin90) = (20)(4)(sin90)
F2 = 20 N … same as F1
Upward force from the fulcrumproduces no torque(since r = 0)
Condition 2: No net Torque
20 N 20 N
Condition 2: No net Torque
20 N ??
Weights are attached to an 8 meter long lever at rest.Determine the unknown weight below
Condition 2: No net Torque
∑T’s = 0
T2 - T1 = 0T2 = T1F2r2sinø2 = F1r1sinø1(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)
Condition 2: No net Torque
20 N 40 N
Condition 2: No net Torque
20 N ??
Weights are attached to an 8 meter long lever at rest.Determine the unknown weight below
Condition 2: No net Torque
F2 =??
F1 = 20 N
r1 = 3 m r2 = 2 m∑T’s = 0
T2 - T1 = 0T2 = T1F2r2sinø2 = F1r1sinø1(F2)(2)(sin90) = (20)(3)(sin90)
F2 = 30 N
(force at the fulcrum is not shown)
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 = ø2F1R1sinø1 = F2R2sinø2F1R1= 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)
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.
20 N ??
CM
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 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.
F1 = 20 N
CM
R1 = 6 m R2 = 2 m
F2 = ??
Rcm = 2 m
Fcm = 100 N
∑T’s = 0
T2 - T1 - Tcm = 0T2 = T1 + TcmF2r2sinø2 = F1r1sinø1 + FcmRcmsinøcm(F2)(2)(sin90)=(20)(6)(sin90)+(100)(2)(sin90)
F2 = 160 N
(force at the fulcrum is not shown)
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
Eat at Joe’s
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
TTy = mg = 200N
mg = 200N
Ty/T = sin30T = Ty/sin30 = 400N
We don’t need to use torque if the rod is “massless”!
(force at the pivot point is not shown)
Sign on a wall #2
Eat at Joe’s
30°
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”
30°
Pivot point
FT
mg = 200NFcm = 100N
(force at the pivot point is not shown)
Sign on a wall #2A 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.
30°
Pivot point
FT
mg = 200NFcm = 100N
∑T = 0TFT = Tcm + TmgFT(2)sin30 =100(1)sin90 + (200)(2)sin90FT = 500 N
(force at the pivot point is not shown)
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.
Diving boardA 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)
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
Remember:
An object is in “Equilibrium” when:
a. There is no net Torque
b. There is no net force acting on the object
†
SF = 0†
St = 0
Push-ups #1A 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 NFfeet = 333 N
A 100 kg man does push-ups as shown
Find the force on his hands and his feet∑T = 0TH = TcmFH(1.5)sin60 =1000(1)sin60
FH = 667 N
∑F = 0Ffeet + Fhands = mg = 1,000 NFfeet = 1,000 N - Fhands = 1000 N - 667 N
FFeet = 333 N
1 m
0.5 m
30°
Fhands
Ffeet
CM
mg = 1000 N
Push-ups #2A 100 kg man does push-ups as shown
Find the force acting on his hands
1 m
0.5 m
30°
Fhands
CM90°
Push-ups #2A 100 kg man does push-ups as shown
∑T = 0TH = TcmFH(1.5)sin90 =1000(1)sin60
FH = 577 N
1 m
0.5 m
30°
Fhands
CM90°
mg = 1000 N
Force on hands:
(force at the feet is not shown)
Sign on a wall, againA 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
Eat at Joe’s
30° 30°
FT = 500N
mg = 200NFcm = 100N
FW = ?
(forces and angles NOT drawn to scale!
Find the force exerted by the wall on the rod
FWx = FTx = 500N(cos30) FWx= 433N
FWy + FTy = Fcm +mgFWy = Fcm + mg - FTy FWy= 300N - 250FWy= 50N (forces and angles NOT drawn to scale!)
Eat at Joe’s
30° 30°
FT = 500N
mg = 200NFcm = 100N
FW
FWy= 50N
FWx= 433N
FW= 436N
FW= 436N
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