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Equilibrium and Torque

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.

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Equilibrium and Torque

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  1. Equilibrium and Torque

  2. 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 linearacceleration or rotational acceleration. We’ll get to this later

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

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

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

  6. 30° 30° 30° 60° 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.

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

  8. T1 = 10 N T2 = 10 N 20 N 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

  9. f = 17.4 N N = 10 N 60° mg =20 N Condition 1: No net Force N = mgcos 60 N = 10 N Fx - f = 0 f = Fx = mgsin 60 f =17.4 N

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

  11. 30° 30° T1 = 20 N T2 = 20 N mg = 20 N Condition 1: No net Force ∑Fx = 0 and ∑Fy = 0 Note: The y-components cancel, so T1y and T2y both equal 10 N

  12. 30° T1 = 40 N T2 = 35 N 20 N 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

  13. 30° T1 = 40 N T2 = 35 N 20 N Condition 1: No net Force ∑Fx = 0 and ∑Fy = 0 Note: The x-components cancel The y-components cancel

  14. 60° 30° Condition 1: No net Force A Harder Problem! a. Which string has the greater tension? b. What is the tension in each string?

  15. T1 T2 60° 30° 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.

  16. T1 T2 60° 30° What is the tension in each string? ∑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

  17. 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 accelerationor rotational acceleration.

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

  19. s r Math Review: Definition of angle in “radians” One revolution = 360° = 2π radians ex: π radians = 180° ex: π/2 radians = 90°

  20. Linear Definitions Rotational Definitions Linear vs. Rotational Motion

  21. 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 Velocity

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

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

  24. This one! 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”?

  25. F ø ø r 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

  26. F ø ø r 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

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

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

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

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

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

  32. 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 accelerationor rotational acceleration.

  33. 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

  34. ?? 20 N ?? 20 N ?? 20 N Condition 2: No net Torque Weights are attached to 8 meter long levers at rest. Determine the unknown weights below

  35. ?? 20 N Condition 2: No net Torque Weights are attached to an 8 meter long lever at rest. Determine the unknown weight below

  36. Condition 2: No net Torque Upward force from the fulcrumproduces 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 =??

  37. 20 N 20 N Condition 2: No net Torque

  38. 20 N ?? Condition 2: No net Torque Weights are attached to an 8 meter long lever at rest. Determine the unknown weight below

  39. r1 = 4 m r2 = 2 m F1 = 20 N F2 =?? Condition 2: No net Torque ∑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)

  40. 20 N 40 N Condition 2: No net Torque

  41. ?? 20 N Condition 2: No net Torque Weights are attached to an 8 meter long lever at rest. Determine the unknown weight below

  42. r1 = 3 m r2 = 2 m F1 = 20 N F2 =?? Condition 2: No net Torque ∑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)

  43. 30 N 20 N Condition 2: No net Torque

  44. (20)(4) = (20)(4) 20 N 20 N (20)(4) = (40)(2) 40 N 20 N (20)(3) = (30)(2) 30 N 20 N • In this special case where • - the pivot point is in the middle of the lever, • and ø1 = ø2 • F1R1sinø1 = F2R2sinø2 • F1R1= F2R2

  45. CM 20 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.

  46. CM ?? 20 N Weight of lever 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.

  47. CM Rcm = 2 m R1 = 6 m R2 = 2 m Fcm = 100 N F1 = 20 N F2 = ?? Masses are attached to an 8 meter long lever at rest. The lever has a mass of 10 kg. Determine the unknown weight below. ∑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)

  48. 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

  49. Ty = mg = 200N T 30° 30° Eat at Joe’s Pivot point mg = 200N Sign on a wall #1 A 20 kg sign hangs from a 2 meter long massless rodsupported 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 don’t need to use torque if the rod is “massless”! Ty/T = sin30 T = Ty/sin30 = 400N

  50. FT 30° 30° Fcm = 100N Eat at Joe’s Pivot point mg = 200N 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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