Equilibrium and Torque

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. a = Dv Dt = 0 a = Dw Dt = 0 We ll get to this later

Static Equilibrium An object is in Static Equilibrium when it is NOT MOVING. v = Dx Dt = 0 a = Dv Dt = 0

Dynamic Equilibrium An object is in Dynamic Equilibrium when it is MOVING with constant linear velocity and/or rotating with constant angular velocity. a = Dv Dt = 0 a = Dw Dt = 0

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

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 30 30

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

Condition 1: No net Force F x = 0 and F y = 0 T1 = 10 N 20 N T2 = 10 N F y = 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 f = 17.4 N N = 10 N F x - f = 0 f = F x = mgsin 60 f = 17.4 N 60 mg =20 N

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

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

Condition 1: No net Force F x = 0 and F y = 0 30 T1 = 40 N T2 = 35 N 20 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

Condition 1: No net Force F x = 0 and F y = 0 T1 = 40 N 30 T2 = 35 N 20 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? 60 30

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

What is the tension in each string? 60 T1 T2 30 F x = 0 T2x-T1x = 0 T1 x = T2 x T1cos60 = T2cos30 F y = 0 T1 y + T2 y - 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: 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. a = Dv Dt = 0 a = Dw Dt = 0

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: 1. Definition of angle in radians Dq = s/r Dq = arc length radius Dq r s 2. One revolution = 360 = 2π radians ex: π radians = 180 ex: π/2 radians = 90

Linear vs. Rotational Motion Linear Definitions Dx v = Dx Dt a = Dv Dt Rotational Definitions Dq in radians w = Dq Dt a = Dw Dt in radians/sec or rev/min in radians/sec/sec q f Dx x i x f Dq q i

Linear vs. Rotational Velocity A car drives 400 m in 20 seconds: a. Find the avg linear velocity v = Dx Dt = 400m 20s = 20m /s w A wheel spins thru an angle of 400π radians in 20 seconds: a. Find the avg angular velocity = Dq 400p radians = Dt 20s w = 20p rad/sec =10 rev/sec = 600 rev/min Dx x i x f Dq

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

Torque 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? 1. The place where the force is applied: the distance r 2. The strength of the force 3. The angle of the force F^ to r ø F r ø F // to r

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 F^ to r ø F r ø

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

Torque is like twisting force Imagine a bicycle wheel that can only spin about its axle. t = (F^)(r) t = (F sinø)(r) t = Frsinø F^ = F sinø r ø ø F

Cross r with F and choose any angle to plug into the equation for torque J ø F r t = (F^)(r) t = Frsinø Since J and ø are supplementary angles (ie : J + ø = 180 ) sinj = sinø F^ = F sinø ø F r ø

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

Imagine a bicycle wheel that can only spin about its axle. Torque = (F)(rsinø) r^ is called the "moment arm" or "moment" F F r^ r ø ø r^

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 r1 = 4 m Upward force from the fulcrum produces no torque (since r = 0) r2 = 4 m 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 F 1 = 20 N F 2 =??

Condition 2: No net Torque 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 F 1 = 20 N r1 = 4 m r2 = 2 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 )(2)(sin90) = (20)(4)(sin90) F 2 = 40 N (force at the fulcrum is not shown)

Condition 2: No net Torque 20 N 40 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 r1 = 3 m r2 = 2 m T s = 0 F 1 = 20 N F 2 =?? 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)

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 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)(4) = (20)(4) 20 N 20 N (20)(4) = (40)(2) 20 N 40 N (20)(3) = (30)(2) 20 N 30 N

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. CM 20 N??

More interesting problems (the pivot is not at the center of mass) 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 CM 20 N?? 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. CM R1 = 6 m Rcm = 2 m R2 = 2 m T s = 0 F1 = 20 N Fcm = 100 N F2 =?? 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)

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 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) 30 T 30 Ty = mg = 200N Eat at Joe s Pivot point mg = 200N We don t need to use torque if the rod is massless! T y /T = sin30 T = T y /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 F T F T 30 30 Eat at Joe s F cm = 100N Pivot point mg = 200N (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. F T 30 T = 0 T FT = T cm + T mg F T (2)sin30 =100(1)sin90 + (200)(2)sin90 F T = 500 N F cm = 100N Pivot point mg = 200N (force at the pivot point is not shown)

bolt 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.

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 T = 0 R 1 = 1 R cm = 1 F bolt = 400 N F cm = 400 N (force at the fulcrum 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. (Balance Torques) b. Determine the upward force applied by the fulcrum. (Balance Forces) bolt F = 800 N F = 0 F bolt = 400 N F cm = 400 N

Remember: An object is in Equilibrium when: a. There is no net Torque St = 0 b. There is no net force acting on the object SF = 0

Push-ups #1 A 100 kg man does push-ups as shown F hands F feet 0.5 m 1 m CM 30 Find the force on his hands and his feet Answer: F hands = 667 N F feet = 333 N

A 100 kg man does push-ups as shown F hands F feet 0.5 m 1 m CM 30 mg = 1000 N 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 - 667 N F Feet = 333 N

Push-ups #2 A 100 kg man does push-ups as shown 0.5 m F hands 1 m CM 90 30 Find the force acting on his hands

Push-ups #2 A 100 kg man does push-ups as shown (force at the feet is not shown) 0.5 m F hands 1 m CM 90 30 mg = 1000 N Force on hands: T = 0 T H = T cm F H (1.5)sin90 =1000(1)sin60 F H = 577 N

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 FW =? 30 30 F T = 500N Eat at Joe s Fcm = 100N mg = 200N (forces and angles NOT drawn to scale!

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