Statics: Torque EquilibriumTorque EquilibriumHow to solveExampleWhiteboardsTorque and forceExampleWhiteboards

Torque equilibrium - the sum of all torques is zero

How to set up torque equilibrium:1. Pick a point to torque about.2. Express all torques:3. +rF +rF+rF… = 0

• + is CW, - is ACW• r is distance from pivot

4. Do math

•Clockwise torques are positive (+), anti-clockwise are negative (-) = rFsin

How to set up torque equilibrium:1. Pick a point to torque about.2. Express all torques:3. +rF +rF+rF… = 0

• + is CW, - is ACW• r is distance from pivot

4. Do math

5.25 NF = ?

2.15 m 5.82 m

1. Torque about the pivot point2 and 3. (2.15 m)(5.25 N) - (5.82 m)F = 0F = 1.94 N (mech. adv.)

WhiteboardsSimple Torque Equilibrium

1 | 2 | 3 | 4

315 N 87.5 N

12 m r = ?

Find the missing distance. Torque about the pivot point.

43 m

(315 N)(12 m) - (87.5 N)r = 0r = 43.2 m = 43 m

W

34 N

F = ?

1.5 m 6.7 m

Find the missing force. Torque about the pivot point. (Be careful of the way the distances are marked)

6.2 N

(34 N)(1.5 m) - (8.2 N)F = 0F = 6.2 N

W

512 N

F = ?

2.0 m 4.5 m

Find the missing Force. Torque about the pivot point.

360 N

-(512 N)(2.0 m) - (481 N)(5.1 m) + F(9.6 m) = 0F = 362 N = 360 N

W

481 N

3.1 m

F = ?

12.2 N

35.0 cm

186 cm

Find the missing force. Torque about the pivot point.

-15.3 N (down, not up as we guessed

-F(.35 m) - (27.5 N)(1.02 m) + (12.2 N)(1.86 m) = 0F = -15.3 N (it would be downward, not upward as we guessed)

W

102 cm

27.5 N

How to set up torque equilibrium:1. Pick a point to torque about.2. Express all torques

• CW is +• ACW is -

3. Set <torques> = 0, solve if you can

350 kg

85 kg

Find the tension in the cable

58o

12.0 m

1.0 m5.0 m

1. Let’s choose the left side to torque about.Four forces - hinge (up?), weight of box down , weight of beam down and the tension in the cable up @58o

Force Equilibrium:1. Draw picture2. Calculate weights3. Draw arrows for forces.

(weights of beams act at their center of gravity)

4. Make components5. Set up sum Fx = 0, sum Fy = 0

Torque Equilibrium:1. Pick a Pivot Point

(at location of unknown force)2. Express all torques:3. +rF +rF+rF… = 0

+ is CW, - is ACWr is distance from pivot

Do Math

350 kg

85 kg

Find the tension in the cable

58o

12.0 m

1.0 m5.0 m

2a. The hinge acts at r = 0, and so exerts no torque(torque = rF, r = 0)

F

How to set up torque equilibrium:1. Pick a point to torque about.2. Express all torques

• CW is +• ACW is -

3. Set <torques> = 0, solve if you can

350 kg

85 kg

Find the tension in the cable

58o

12.0 m

1.0 m5.0 m

F

Torques: Hinge = 0 Nm (torque = 0*F)

2b. The box weight (85*9.8) = 833 N, at a distance of 5.0 m torque = rF = (5.0 m)(833 N) = +4165 Nm (CW)

833 N

350 kg

85 kg

Find the tension in the cable

58o

12.0 m

1.0 m5.0 m

F

Torques: Hinge = 0 NmBox = +4165 Nm = (5.0 m)(833 N)

833 N

2c. The beam weight (350*9.8) = 3430 N, at its center of mass, 6.0 m from the left side torque = rF = (6.0 m)(3430 N) = +20,580 Nm (CW)

3430 N

350 kg

85 kg

Find the tension in the cable

58o

12.0 m

1.0 m5.0 m

F

Torques: Hinge = 0 NmBox = +4165 Nm = (5.0 m)(833 N)Beam = +20,580 Nm = (6.0 m)(3430 N)

833 N

3430 N

2d. The cable tension T, at 11.0 m, 58o angletorque = rFsin = (11.0 m)Tsin(58o) = -9.329T m (ACW)

TTsin(58o)

350 kg

85 kg

Find the tension in the cable

58o

12.0 m

1.0 m5.0 m

F

3. Set up your torque equation:0 Nm + 4165 Nm + 20,580 Nm - 9.329T m = 0

833 N

3430 N

TTsin(58o)

Torques: Hinge = 0 NmBox = +4165 Nm = (5.0 m)(833 N)Beam = +20,580 Nm = (6.0 m)(3430 N)Cable = - 9.329T m = (11.0 m)Tsin(58o)

350 kg

85 kg

Find the tension in the cable

58o

12.0 m

1.0 m5.0 m

F

4. Do Math:0 Nm + 4165 Nm + 20,580 Nm - 9.329T Nm = 024,745 Nm = 9.329T Nm T = 2652.62 N = 2700 N

833 N

Torques: Hinge = 0 NmBox = +4165 Nm = (5.0 m)(833 N)Beam = +20,580 Nm = (6.0 m)(3430 N)Cable = - 9.329T m = (11.0 m)Tsin(58o)

Whiteboards: Torque Equilibrium

1a | 1b | 1c | 1d | 1e | 1f

TOC

Blue, camels don’t need much water W

35 kg

15 kgT

7.0m

16.0m

20.0m

(Fulcrum) Find the tension in the cable

Step 1 - Let’s torque about the fulcrum

Green, because of their feet W

35 kg

15 kgT

7.0m

16.0m

20.0m

(Fulcrum) Find the tension in the cable

Step 2 - Express your torques:The fulcrum, the beam, the box, and the cableThe fulcrum is r = 0 from the fulcrum, and so exerts no torque

+1029 Nm (CW) W

35 kg

15 kgT

7.0m

16.0m

20.0m

(Fulcrum) Find the tension in the cable

Step 2a - Calculate the torque exerted by the beam itself. + = CW, - = ACW

r

F

r = (20.0 m)/2 - 7.0 m = 3.0 mF = (35 kg)(9.8 N/kg) = 343 Ntorque = rF = (3.0 m)(343 N) = +1029 Nm (CW)

+1323 Nm (CW) W

35 kg

15 kgT

7.0m

16.0m

20.0m

(Fulcrum) Find the tension in the cable

Step 2b - Calculate the torque exerted by the 15 kg box. + = CW, - = ACW

r

F

r = 16.0 m - 7.0 m = 9.0 mF = (15 kg)(9.8 N/kg) = 147 Ntorque = rF = (9.0 m)(147 N) = +1323 Nm (CW)

-(13.0m)T m (ACW) W

35 kg

15 kg

7.0m

16.0m

20.0m

(Fulcrum) Find the tension in the cable

T

Step 2c - Express the torque exerted by the cable. + = CW, - = ACW

r

r = 20.0 m - 7.0 m = 13.0 mF = Ttorque = rF = (13.0 m)T = -(13.0 m)T Nm (ACW)

180 N W

35 kg

15 kg

7.0m

16.0m

20.0m

(Fulcrum) Find the tension in the cable

T

Step 3, and 4 - Set up your torque equilibrium, and solve for T:Beam = +1029 NmBox = +1323 NmCable = -(13.0 m)T Nm

+1029 Nm + 1323 Nm - (13.0 m)T = 0T = (1029 Nm + 1323 Nm)/(13.0 m) = 180.9 N = 180 N

Whiteboards: Torque and force2a | 2b | 2c | 2d | 2e

TOC

T1 + T2 -509.6 N - 754.6 N = 0

Step 1 - Set up your vertical force equation

52 kg

77 kgT1 T2

Beam is 18.0 m long, person is 5.0 m from the right side. Find the two tensions in the cables at either end.

T1 and T2 are up, and the beam and person weights are down:Beam: -(52 kg)(9.8 N/kg) = -509.6 N (down)Person: -(77 kg)(9.8 N/kg) = -754.6 N (down)T1 + T2 -509.6 N - 754.6 N = 0

+4586.4 Nm + 9809.8 Nm - (18.0 m)(T2) = 0

Step 2 - Let’s torque about the left sideSet up your torque equation: torque = rF

52 kg

77 kgT1 T2

Beam is 18.0 m long, person is 5.0 m from the right side. Find the two tensions in the cables at either end.

T1 = 0 Nm torque, (r = 0)Beam: (9.0 m)(509.6 N) = +4586.4 Nm (CW)Person: (13.0 m)(754.6 N) = +9809.8 Nm (CW)T2: T2 at 18.0 m = -(18.0 m)(T2) (ACW)Finally:+4586.4 Nm + 9809.8 Nm - (18.0 m)(T2) = 0

9.0 m 13.0 m 18.0 m

509.6 N 754.6 N

T2 = 799.8 N T1 =464.4 N

Step 3 - Math time. Solve these equations for T1

and T2:+4586.4 Nm + 9809.8 Nm - (18.0 m)(T2) = 0T1 + T2 -509.6 N - 754.6 N = 0

52 kg

77 kgT1 T2

Beam is 18.0 m long, person is 5.0 m from the right side. Find the two tensions in the cables at either end.

+4586.4 Nm + 9809.8 Nm = (18.0 m)(T2), T2 = 799.8 NT1 + 799.8 N -509.6 N - 754.6 N = 0, T1 = 464.4 N