Equilibrium and Torque Equilibrium An object is in
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) 3. In other words, the object is NOT experiencing 4. linear acceleration or rotational acceleration. We’ll get to this later
Static Equilibrium 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° 30°
Condition 1: No net Force ∑Fx = 0 N = 20 N mg = 20 N and ∑Fy = 0 N - mg = 0 N = mg = 20 N
Condition 1: No net Force ∑Fx = 0 T 1 = 10 N T 2 = 10 N 20 N and ∑Fy = 0 T 1 + T 2 - mg = 0 T 1 = T 2 = T T + T = mg 2 T = 20 N T = 10 N
Condition 1: No net Force N = mgcos 60 N = 10 N Fx - f = 0 f = Fx = mgsin 60 f = 17. 4 N 60° mg =20 N N = 10 N
Condition 1: No net Force ∑Fx = 0 30° and ∑Fy = 0 30° T 2 = 20 N T 1 = 20 N T 2 y = 10 N T 2 = 20 N 30° T 2 x mg = 20 N ∑Fx = 0 T 2 x - T 1 x = 0 T 1 x = T 2 x ∑Fy = 0 T 1 y + T 2 y - mg = 0 2 Ty = mg = 20 N Ty/T = sin 30 T = Ty/sin 30 T = (10 N)/sin 30 Equal angles ==> T 1 = T 2 Ty = mg/2 = 10 N T = 20 N Note: unequal angles ==> T 1 ≠ T 2
Condition 1: No net Force ∑Fx = 0 30° T 1 = 20 N and 30° T 2 = 20 N mg = 20 N Note: The y-components cancel, so T 1 y and T 2 y both equal 10 N ∑Fy = 0
Condition 1: No net Force ∑Fx = 0 and ∑Fy = 0 30° T 1 = 40 N T 2 = 35 N 20 N ∑Fy = 0 T 1 y - mg = 0 T 1 y = mg = 20 N T 1 y/T 1 = sin 30 T 1 = Ty/sin 30 = 40 N ∑Fx = 0 T 2 - T 1 x = 0 T 2 = T 1 x= T 1 cos 30 T 2 = (40 N)cos 30 T 2 = 35 N
Condition 1: No net Force ∑Fx = 0 and 30° T 1 = 40 N T 2 = 35 N 20 N Note: The x-components cancel The y-components cancel ∑Fy = 0
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? ∑Fx = 0 so T 1 x = T 2 x 60° 30° T 1 T 2 T 1 must be greater in order to have the same x-component as T 2.
What is the tension in each string? 60° 30° T 1 ∑Fx = 0 T 2 x-T 1 x = 0 T 1 x = T 2 x T 1 cos 60 = T 2 cos 30 T 2 ∑Fy = 0 T 1 y + T 2 y - mg = 0 T 1 sin 60 + T 2 sin 30 = 20 N Note: unequal angles ==> T 1 ≠ T 2 Solve simultaneous equations!
Equilibrium An object is in “Equilibrium” when: 1. There is no net force acting on the object 2. There is no net Torque 3. In other words, the object is NOT experiencing 4. linear acceleration or rotational acceleration.
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” s r 2. One revolution = 360° = 2π radians 3. ex: π radians = 180° 4. ex: π/2 radians = 90°
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? 1. 2. 3. The place where the force is applied: the distance “r” The strength of the force The angle of the force ø ø r F
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 ø ø r F
Torque = Frsinø Imagine a bicycle wheel that can only spin about its axle. Torque = (the component of force perpendicular to r)(r) ø ø r F
Torque is like “twisting force” Imagine a bicycle wheel that can only spin about its axle. ø ø r F
Cross “r” with “F” and choose any angle to plug into the equation for torque F ø r ø ø r F
Two different ways of looking at torque Torque = (Fsinø)(r) Torque = (F)(rsinø) F r ø ø r F F r ø ø
Imagine a bicycle wheel that can only spin about its axle. Torque = (F)(rsinø) F r ø ø F
Equilibrium An object is in “Equilibrium” when: 1. There is no net force acting on the object 2. There is no net Torque 3. In other words, the object is NOT experiencing 4. 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) r 1 = 4 m ∑T’s = 0 T 2 - T 1 = 0 T 2 = T 1 F 2 r 2 sinø 2 = F 1 r 1 sinø 1 (F 2)(4)(sin 90) = (20)(4)(sin 90) r 2 = 4 m F 2 = 20 N … same as F 1 = 20 N F 2 =? ?
Condition 2: No net Torque 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 r 1 = 4 m r 2 = 2 m ∑T’s = 0 T 2 - T 1 = 0 T 2 = T 1 F 2 r 2 sinø 2 = F 1 r 1 sinø 1 (F 2)(2)(sin 90) = (20)(4)(sin 90) F 1 = 20 N F 2 =? ? (force at the fulcrum is not shown) F 2 = 40 N
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 r 1 = 3 m r 2 = 2 m ∑T’s = 0 T 2 - T 1 = 0 T 2 = T 1 F 2 r 2 sinø 2 = F 1 r 1 sinø 1 (F 2)(2)(sin 90) = (20)(3)(sin 90) F 1 = 20 N F 2 =? ? (force at the fulcrum is not shown) F 2 = 30 N
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)(4) = (40)(2) 40 N 20 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)(Rcm) Tcm =(mg)(rcm) = (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 R 1 = 6 m Rcm = 2 m R 2 = 2 m ∑T’s = 0 T 2 - T 1 - Tcm = 0 T 2 = T 1 + Tcm F 2 r 2 sinø 2 = F 1 r 1 sinø 1 + Fcm. Rcmsinøcm (F 2)(2)(sin 90)=(20)(6)(sin 90)+(100)(2)(sin 90) Fcm = 100 N F 1 = 20 N F 2 = ? ? (force at the fulcrum is not shown) F 2 = 160 N
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) Ty = mg = 200 N T 30° Eat at Joe’s We don’t need to use torque if the rod is “massless”! 30° Pivot point mg = 200 N Ty/T = sin 30 T = Ty/sin 30 = 400 N
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” FT 30° Eat at Joe’s 30° Fcm = 100 N Pivot point mg = 200 N (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. ∑T = 0 TFT = Tcm + Tmg FT(2)sin 30 =100(1)sin 90 + (200)(2)sin 90 FT = 500 N FT 30° Fcm = 100 N Pivot point mg = 200 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. a. Determine the downward force of the bolt. (Balance Torques) ∑T = 0 bolt R 1 = 1 Fbolt = 400 N Rcm = 1 Fcm = 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. b. Determine the upward force applied by the fulcrum. (Balance Forces) F = 800 N bolt Fbolt = 400 N Fcm = 400 N ∑F = 0
Remember: An object is in “Equilibrium” when: a. There is no net Torque b. b. There is no net force acting on the object
Push-ups #1 A 100 kg man does push-ups as shown Fhands Ffeet 0. 5 m 1 m CM 30° 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 Fhands Ffeet 0. 5 m 1 m CM 30° mg = 1000 N Find the force on his hands and his feet ∑T = 0 TH = Tcm FH(1. 5)sin 60 =1000(1)sin 60 ∑F = 0 Ffeet + Fhands = mg = 1, 000 N Ffeet = 1, 000 N - Fhands = 1000 N - 667 N FH = 667 N FFeet = 333 N
Push-ups #2 A 100 kg man does push-ups as shown Fhands 0. 5 m 1 m 90° CM 30° Find the force acting on his hands
Push-ups #2 A 100 kg man does push-ups as shown Fhands (force at the feet is not shown) 0. 5 m 1 m 30° Force on hands: 90° CM mg = 1000 N ∑T = 0 TH = Tcm FH(1. 5)sin 90 =1000(1)sin 60 FH = 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 FT = 500 N FW = ? 30° Eat at Joe’s 30° Fcm = 100 N mg = 200 N (forces and angles NOT drawn to scale!
Find the force exerted by the wall on the rod FT = 500 N 30° Eat at Joe’s FWx = FTx = 500 N(cos 30) FWx= 433 N FWy + FTy = Fcm +mg FWy = Fcm + mg - FTy FWy= 300 N - 250 FWy= 50 N FW 30° Fcm = 100 N FW= 436 N mg = 200 N FWy= 50 N FWx= 433 N FW= 436 N (forces and angles NOT drawn to scale!)
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