Equilibrium and Torque

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1 Equilibrium and Torque

2 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 v t 0 t 0

3 What is Torque? Torque is like twisting force The more torque you apply to a wheel the more quickly its rate of spin changes

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

5 Linear vs. Rotational Motion Linear Definitions x v x t a v t Rotational Definitions in radians t t in radians/sec or rev/min in radians/sec/sec f x i x x f i

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

7 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 F net net

8 Torque Torque is like twisting force The more torque you apply to a wheel, the more quickly its rate of spin changes Torque = Frsinø

9 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!

10 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

11 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

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

13 Torque is like twisting force Imagine a bicycle wheel that can only spin about its axle. F F sinø ø F (F )(r) (F sinø)(r) Fr sinø r ø

14 Cross r with F and choose any angle to plug into the equation for torque ø F r (F )(r) Fr sinø Since and ø are supplementary angles (ie : + ø = 180 ) sin = sinø F F sinø ø F r ø

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

16 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

17 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 v t 0 t 0

18 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

19 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??

20 Condition 2: No net Torque Weights are attached to an 8 meter long lever at rest. Determine the unknown weight below 20 N??

21 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 =??

22 Condition 2: No net Torque 20 N 20 N

23 Condition 2: No net Torque Weights are attached to an 8 meter long lever at rest. Determine the unknown weight below 20 N??

24 Condition 2: No net Torque r1 = 4 m r2 = 2 m T s = 0 F 1 = 20 N 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 =?? F 2 = 40 N (force at the fulcrum is not shown)

25 Condition 2: No net Torque 20 N 40 N

26 Condition 2: No net Torque Weights are attached to an 8 meter long lever at rest. Determine the unknown weight below 20 N??

27 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)

28 Condition 2: No net Torque 20 N 30 N

29 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

30 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??

31 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.

32 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)

33 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

34 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

35 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 F T 30 Eat at Joe s F cm = 100N Pivot point mg = 200N (force at the pivot point is not shown)

36 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)

37 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

38 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)

39 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) F = 800 N F = 0 bolt F bolt = 400 N F cm = 400 N

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

41 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

42 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 N F Feet = 333 N

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

44 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 mg = 1000 N Force on hands: T = 0 T H = T cm F H (1.5)sin90 =1000(1)sin60 F H = 577 N

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

46 Find the force exerted by the wall on the rod FW 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 FW y = 50N FW= 436N FW y = 50N FW x = 433N FW= 436N (forces and angles NOT drawn to scale!)

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