Equilibrium and Torque

Equilibrium An object is in “Equilibrium” when: There is no net force acting on the object 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

An object is in “Static Equilibrium” when it is NOT MOVING.

Dynamic Equilibrium An object is in “Dynamic Equilibrium” when it is MOVING with constant linear velocity and/or rotating with constant angular velocity.

Equilibrium Let’s focus on condition 1: net force = 0 The x components of force cancel The y components of force cancel

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°

Condition 1: No net Force ∑Fx = 0 and ∑Fy = 0 mg = 20 N N = 20 N ∑Fy = 0 N - mg = 0 N = mg = 20 N

Condition 1: No net Force ∑Fx = 0 and ∑Fy = 0 T1 = 10 N 20 N T2 = 10 N ∑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 60° mg =20 N f = 17.4 N N = 10 N Fx - f = 0 f = Fx = mgsin 60 f = 17.4 N

Condition 1: No net Force ∑Fx = 0 and ∑Fy = 0 30° mg = 20 N T1 = 20 N T2 = 20 N T2 = 20 N T2x T2y = 10 N 30° ∑Fx = 0 T2x - T1x = 0 T1x = T2x Equal angles ==> T1 = T2 ∑Fy = 0 T1y + T2y - mg = 0 2Ty = mg = 20 N Ty = mg/2 = 10 N Ty/T = sin 30 T = Ty/sin 30 T = (10 N)/sin30 T = 20 N Note: unequal angles ==> T1 ≠ T2

Condition 1: No net Force ∑Fx = 0 and ∑Fy = 0 30° mg = 20 N T1 = 20 N T2 = 20 N Note: The y-components cancel, so T1y and T2y both equal 10 N

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

Condition 1: No net Force A Harder Problem! a. Which string has the greater tension? b. What is the tension in each string? 30° 60°

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.

What is the tension in each string? 30° 60° T1 T2 ∑Fx = 0 T2x-T1x = 0 T1x = T2x T1cos60 = T2cos30 ∑Fy = 0 T1y + T2y - mg = 0 T1sin60 + T2sin30 - mg = 0 T1sin60 + T2sin30 = 20 N Solve simultaneous equations! Note: unequal angles ==> T1 ≠ T2

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.

Torque is like “twisting force” 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: Definition of angle in “radians” One revolution = 360° = 2π radians ex: π radians = 180° ex: π/2 radians = 90° s r

Linear vs. Rotational Motion Linear Definitions Rotational Definitions

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

Linear vs. Rotational Net Force ==> linear acceleration The linear velocity changes Net Torque ==> angular acceleration The angular velocity changes (the rate of spin changes)

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? The place where the force is applied: the distance “r” The strength of the force The angle of the force r ø F

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 r ø F

Torque = Frsinø Imagine a bicycle wheel that can only spin about its axle. Torque = (the component of force perpendicular to r)(r) r ø F

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

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

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

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

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.

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 ??

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) ∑T’s = 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 =??

Condition 2: No net Torque

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 F2 =?? F1 = 20 N r1 = 4 m r2 = 2 m ∑T’s = 0 T2 - T1 = 0 T2 = T1 F2r2sinø2 = F1r1sinø1 (F2)(2)(sin90) = (20)(4)(sin90) F2 = 40 N (force at the fulcrum is not shown)

Condition 2: No net Torque

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 F2 =?? F1 = 20 N r1 = 3 m r2 = 2 m ∑T’s = 0 T2 - T1 = 0 T2 = T1 F2r2sinø2 = F1r1sinø1 (F2)(2)(sin90) = (20)(3)(sin90) F2 = 30 N (force at the fulcrum is not shown)

Condition 2: No net Torque

In this special case where - the pivot point is in the middle of the lever, and ø1 = ø2 F1R1sinø1 = F2R2sinø2 F1R1= F2R2 20 N 40 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

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 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 = 0 T2 = T1 + Tcm F2r2sinø2 = 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 Ty/T = sin30 T = Ty/sin30 = 400N 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. Eat at Joe’s 30° (force at the pivot point is not shown) 30° Pivot point T Ty = mg = 200N mg = 200N We don’t need to use torque if the rod is “massless”! 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” Eat at Joe’s 30° 30° Pivot point FT mg = 200N Fcm = 100N (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. 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)

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. Determine the downward force of the bolt. (Balance Torques) ∑T = 0 bolt Rcm = 1 R1 = 1 Fcm = 400 N Fbolt = 400 N (force at the fulcrum is not shown)

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 bolt Fcm = 400 N Fbolt = 400 N F = 800 N

Remember: An object is in “Equilibrium” when: There is no net Torque b. There is no net force acting on the object

Push-ups #1 A 100 kg man does push-ups as shown Fhands Ffeet 30° Fhands Ffeet CM 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 30° Fhands Ffeet CM mg = 1000 N 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

Push-ups #2 A 100 kg man does push-ups as shown Fhands 30° Fhands CM 90° Find the force acting on his hands

Push-ups #2 A 100 kg man does push-ups as shown Fhands mg = 1000 N 30° Fhands CM 90° mg = 1000 N (force at the feet is not shown) Force on hands: ∑T = 0 TH = Tcm FH(1.5)sin90 =1000(1)sin60 FH = 577 N

Sign on a wall, again Find the force exerted by the wall on the rod 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° FT = 500N mg = 200N Fcm = 100N FW = ? (forces and angles NOT drawn to scale!

Find the force exerted by the wall on the rod Eat at Joe’s 30° FT = 500N mg = 200N Fcm = 100N FW FWx = FTx = 500N(cos30) FWx= 433N FWy= 50N FWx= 433N FW= 436N FWy + FTy = Fcm +mg FWy = Fcm + mg - FTy FWy= 300N - 250 FWy= 50N (forces and angles NOT drawn to scale!)