statics: torque equilibrium ttorque equilibrium how to solve example whiteboards torque and force...
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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