Equilibrium and Torque

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Equilibrium and Torque
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Equilibrium

• An object is in Equilibrium when
• There is no net force acting on the object
• There is no net Torque (well get to this later)
• In other words, the object is NOT experiencing
• linear acceleration or rotational acceleration.

Well get to this later
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Static Equilibrium
An object is in Static Equilibrium when it
is NOT MOVING.
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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.
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Equilibrium

• Lets focus on condition 1 net force 0
• The x components of force cancel
• The y components of force cancel

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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.
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Condition 1 No net Force
?Fx 0 and ?Fy 0
?Fy 0 N - mg 0 N 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
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Condition 1 No net Force
N mgcos 60 N 10 N
Fx - f 0 f Fx mgsin 60 f 17.4 N
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Condition 1 No net Force
?Fx 0 and ?Fy 0
Ty/T sin 30 T Ty/sin 30 T (10 N)/sin30 T
20 N
?Fx 0 T2x - T1x 0 T1x T2x Equal angles gt
T1 T2
?Fy 0 T1y T2y - mg 0 2Ty mg 20 N Ty
mg/2 10 N
Note unequal angles gt 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
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Condition 1 No net Force
?Fx 0 and ?Fy 0
?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
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Condition 1 No net Force
?Fx 0 and ?Fy 0
Note The x-components cancel The y-components
cancel
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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
T1 must be greater in order to have the same
x-component as T2.
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What is the tension in each string?
?Fy 0 T1y T2y - mg 0 T1sin60 T2sin30 - mg
0 T1sin60 T2sin30 20 N
Solve simultaneous equations!
?Fx 0 T2x-T1x 0 T1x T2x T1cos60 T2cos30
Note unequal angles gt T1 ? T2
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Equilibrium

• An object is in Equilibrium when
• There is no net force acting on the object
• There is no net Torque
• In other words, the object is NOT experiencing
• linear acceleration or rotational acceleration.

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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
• Definition of angle in radians
• One revolution 360 2p radians
• ex p radians 180
• ex p/2 radians 90

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Linear vs. Rotational Motion

• Linear Definitions

• Rotational Definitions

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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 400p radians in 20
  seconds
• a. Find the avg angular velocity

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Linear vs. Rotational
Net Force gt linear acceleration The linear
velocity changes
Net Torque gt angular acceleration The angular
velocity changes (the rate of spin changes)
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Torque
Torque is like twisting force The more torque
you apply to a wheel, the more quickly its rate
of spin changes
Torque Frsinø
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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 produces a more effective twisting
force?
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Torque is like twisting force

• Imagine a bicycle wheel that can only spin about
  its axle.
• What affects the torque?
• The place where the force is applied the
  distance r
• The strength of the force
• The angle of the force

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Torque is like twisting force

• Imagine a bicycle wheel that can only spin about
  its axle.
• What affects the torque?
• The distance from the axis rotation r that the
  force is applied
• The component of force perpendicular to the
  r-vector

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Torque Frsinø
Imagine a bicycle wheel that can only spin about
its axle. Torque (the component of force
perpendicular to r)(r)
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Torque is like twisting force
Imagine a bicycle wheel that can only spin about
its axle.
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Cross r with F and choose any angle to plug
into the equation for torque
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Two different ways of looking at torque
Torque (Fsinø)(r)
Torque (F)(rsinø)
F
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Imagine a bicycle wheel that can only spin about
its axle.
Torque (F)(rsinø)
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Equilibrium

• An object is in Equilibrium when
• There is no net force acting on the object
• There is no net Torque
• In other words, the object is NOT experiencing
• linear acceleration or rotational acceleration.

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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
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Condition 2 No net Torque
Weights are attached to an 8 meter long lever at
rest. Determine the unknown weight below
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Condition 2 No net Torque
Upward force from the fulcrumproduces no torque
(since r 0)
?Ts 0 T2 - T1 0 T2 T1 F2r2sinø2
F1r1sinø1 (F2)(4)(sin90) (20)(4)(sin90) F2
20 N same as F1
r1 4 m
r2 4 m
F1 20 N
F2 ??
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Condition 2 No net Torque
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Condition 2 No net Torque
Weights are attached to an 8 meter long lever at
rest. Determine the unknown weight below
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Condition 2 No net Torque
?Ts 0 T2 - T1 0 T2 T1 F2r2sinø2
F1r1sinø1 (F2)(2)(sin90) (20)(4)(sin90) F2
40 N
(force at the fulcrum is not shown)
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Condition 2 No net Torque
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Condition 2 No net Torque
Weights are attached to an 8 meter long lever at
rest. Determine the unknown weight below
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Condition 2 No net Torque
?Ts 0 T2 - T1 0 T2 T1 F2r2sinø2
F1r1sinø1 (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
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• In this special case where
• - the pivot point is in the middle of the lever,
  
• and ø1 ø2
• F1R1sinø1 F2R2sinø2
• F1R1 F2R2

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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.
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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
Masses are attached to an 8 meter long lever at
rest. The lever has a mass of 10 kg.
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Masses are attached to an 8 meter long lever at
rest. The lever has a mass of 10 kg. Determine
the unknown weight below.
?Ts 0 T2 - T1 - Tcm 0 T2 T1
Tcm F2r2sinø2 F1r1sinø1 FcmRcmsinøcm (F2)(2)(s
in90)(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 wall1 (massless rod)
Sign on a wall2 (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
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)
We dont need to use torque if the rod is
massless!
Ty/T sin30 T Ty/sin30 400N
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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 kg and is supported by a cable
at an angle of 30 as shown. Determine the
tension in the cable FT
(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 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
(force at the pivot point is not shown)
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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.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.
• Determine the downward force of the
  bolt.(Balance Torques)

?T 0
(force at the fulcrum is not shown)
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Diving board

• A 4 meter long diving board with a mass of 40 kg.
• Determine the downward force of the
  bolt.(Balance Torques)
• b. Determine the upward force applied by the
  fulcrum.(Balance Forces)

?F 0
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Remember

• An object is in Equilibrium when
• There is no net Torque
• b. There is no net force acting on the object

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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
Answer Fhands 667 N Ffeet 333 N
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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
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Push-ups 2
A 100 kg man does push-ups as shown
Find the force acting on his hands
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Push-ups 2
A 100 kg man does push-ups as shown
(force at the feet is not shown)
Force on hands
?T 0 TH Tcm FH(1.5)sin90 1000(1)sin60 FH
577 N
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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 kg and is supported by a cable
at an angle of 30 as shown
Find the force exerted by the wall on the rod
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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 FWy 50N
(forces and angles NOT drawn to scale!)