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Equilibrium and Torque

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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. We’ll 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 Let’s 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. 60° 30°

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Condition 1: No net Force ∑F x = 0and∑F y = 0 mg = 20 N N = 20 N ∑F y = 0 N - mg = 0 N = mg = 20 N

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Condition 1: No net Force ∑F x = 0and∑F y = 0 ∑F y = 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 F x - f = 0 f = F x = mgsin 60 f = 17.4 N N = mgcos 60 N = 10 N

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Condition 1: No net Force ∑F x = 0and∑F y = 0 30° mg = 20 N T1 = 20 NT2 = 20 N ∑F y = 0 T1 y + T2 y - mg = 0 2T y = mg = 20 N T y = mg/2 = 10 N T2 = 20 N T2 x T2 y = 10 N 30° T y /T = sin 30 T = T y /sin 30 T = (10 N)/sin30 T = 20 N ∑F x = 0 T2x - T1x = 0 T1 x = T2 x Equal angles ==> T1 = T2 Note: unequal angles ==> T1 ≠ T2

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Condition 1: No net Force ∑F x = 0and∑F y = 0 Note: The y-components cancel, so T1 y and T2 y both equal 10 N 30° mg = 20 N T1 = 20 NT2 = 20 N

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Condition 1: No net Force ∑F x = 0and∑F y = 0 30° 20 N T1 = 40 N T2 = 35 N ∑F y = 0 T1 y - mg = 0 T1 y = mg = 20 N T1 y /T1 = sin 30 T1 = T y /sin 30 = 40 N ∑F x = 0 T2 - T1 x = 0 T2 = T1 x = T1cos30 T2 = (40 N)cos30 T2 = 35 N

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Condition 1: No net Force ∑F x = 0and∑F y = 0 Note: The x-components cancel The y-components cancel 30° 20 N T1 = 40 N T2 = 35 N

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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?

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a. Which string has the greater tension? ∑F x = 0 so T1 x = T2 x 30°60° T1 T2 T1 must be greater in order to have the same x-component as T2.

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∑F y = 0 T1 y + T2 y - mg = 0 T1sin60 + T2sin30 - mg = 0 T1sin60 + T2sin30 = 20 N Solve simultaneous equations! ∑F x = 0 T2x-T1x = 0 T1 x = T2 x T1cos60 = T2cos30 30°60° T1 T2 Note: unequal angles ==> T1 ≠ T2 What is the tension in each string?

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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.

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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° = 2π radians ex: π radians = 180° ex: π/2 radians = 90° s r

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Linear vs. Rotational Motion Linear DefinitionsRotational 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 400π radians in 20 seconds: a. Find the avg angular velocity

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Linear vs. Rotational Net Force ==> linear acceleration The linear velocity changes Net Torque ==> 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”? 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 ø F ø

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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 “r” that the force is applied 2.The component of force perpendicular to the r-vector r ø F ø

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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 ø F ø Torque = Frsinø

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Torque is like “twisting force” Imagine a bicycle wheel that can only spin about its axle. r ø F ø

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r ø F ø F r ø Cross “r” with “F” and choose any angle to plug into the equation for torque

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Torque = (Fsinø)(r) Two different ways of looking at torque r ø F ø Torque = (F)(rsinø) rø F ø r F

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Imagine a bicycle wheel that can only spin about its axle. Torque = (F)(rsinø) rø F ø F

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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.

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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 TorqueNegative 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 lever at rest. Determine the unknown weight below 20 N ??

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Condition 2: No net Torque F 1 = 20 N r1 = 4 mr2 = 4 m F 2 =?? ∑T’s = 0 T2 - T1 = 0 T2 = T1 F 2 r 2 sinø 2 = F 1 r 1 sinø 1 (F 2 )(4)(sin90) = (20)(4)(sin90) F 2 = 20 N … same as F 1 Upward force from the fulcrum produces no torque (since r = 0)

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Condition 2: No net Torque 20 N

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Condition 2: No net Torque 20 N?? 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 ∑T’s = 0 T2 - T1 = 0 T2 = T1 F 2 r 2 sinø 2 = F 1 r 1 sinø 1 (F 2 )(2)(sin90) = (20)(4)(sin90) F 2 = 40 N F 2 =?? F 1 = 20 N r1 = 4 mr2 = 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 lever at rest. Determine the unknown weight below

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Condition 2: No net Torque F 2 =?? F 1 = 20 N r1 = 3 mr2 = 2 m ∑T’s = 0 T2 - T1 = 0 T2 = T1 F 2 r 2 sinø 2 = F 1 r 1 sinø 1 (F 2 )(2)(sin90) = (20)(3)(sin90) F 2 = 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 ø 1 = ø 2 F 1 R 1 sinø 1 = F 2 R 2 sinø 2 F 1 R 1 = F 2 R 2 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 lever at 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)(R cm ) T cm =(mg)(r cm ) = (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.

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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. F1 = 20 N CM R1 = 6 m R2 = 2 m F2 = ?? Rcm = 2 m Fcm = 100 N ∑T’s = 0 T2 - T1 - Tcm = 0 T2 = T1 + Tcm F 2 r 2 sinø 2 = F 1 r 1 sinø 1 + F cm R cm sinø cm (F 2 )(2)(sin90)=(20)(6)(sin90)+(100)(2)(sin90) F 2 = 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 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 T Ty = mg = 200N mg = 200N T y /T = sin30 T = T y /sin30 = 400N We don’t 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 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 “F T ” 30° Pivot point FTFT mg = 200N F cm = 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 kg and is supported by a cable at an angle of 30° as shown. Determine the tension in the cable. 30° Pivot point FTFT mg = 200N F cm = 100N ∑T = 0 T FT = T cm + T mg F T (2)sin30 =100(1)sin90 + (200)(2)sin90 F T = 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 R cm = 1 R 1 = 1 F cm = 400 N F bolt = 400 N ∑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. a.Determine the downward force of the bolt. (Balance Torques) b. Determine the upward force applied by the fulcrum. (Balance Forces) bolt F cm = 400 NF bolt = 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

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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° F hands F feet CM Answer: F hands = 667 N F feet = 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 T H = T cm F H (1.5)sin60 =1000(1)sin60 F H = 667 N ∑F = 0 F feet + F hands = mg = 1,000 N F feet = 1,000 N - F hands = 1000 N N F Feet = 333 N 1 m 0.5 m 30° F hands F feet 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° F hands CM 90°

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Push-ups #2 A 100 kg man does push-ups as shown ∑T = 0 T H = T cm F H (1.5)sin90 =1000(1)sin60 F H = 577 N 1 m 0.5 m 30° F hands 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 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° F T = 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 FW x = FT x = 500N(cos30) FW x = 433N FW y + FT y = Fcm +mg FW y = Fcm + mg - FT y FW y = 300N FW y = 50N (forces and angles NOT drawn to scale!) Eat at Joe’s 30° FT = 500N mg = 200N Fcm = 100N FW FW y = 50N FW x = 433N FW= 436N

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