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12.4(B) Zero Torque and Static Equilibrium
AP PHYSICS
March 22, 2019
PRACTICE LABS TESTSUnit 12
Problems (123)
Paper Car Crash
Balancing Act
Unit 12 Test Tuesday (3/26/19)
bit.ly/bshprom19 Prom Royalty
Zero Torque and Static Equilibrium
12.4I can describe, interpret, and solve problems involving static equilibrium.
The center of mass of an object is the point on the object that moves in the same way that a point particle would move.
Center Of Mass Conditions For Equilibrium
Net Force = 0
Net Torque = 0
1.) TranslationalEquilibrium
2.) RotationalEquilibrium
12.4(B) Zero Torque and Static Equilibrium
AP PHYSICS
March 22, 2019
Static EquilibriumA 5.8 kg ladder, 1.80 m long, is resting on two sawhorses. Sawhorse A is 0.60 m from one end of the ladder, and sawhorse B is 0.15 m from the other end of the ladder. What force does each sawhorse exert on the ladder?
Scaffolding
SCAFFOLDS
160 lb. 140 lb.
UNIT 12 IN CLASS PROBLEMSA 5.00 m long diving board of negligible mass is supported by two pillars. One pillar is at the left end of the diving board and the other is 1.50 m away. Find the forces, in Newton's, exerted by the pillars when a 90.0 kg diver stands at the far end of the board.
22.
A banner is suspended from a horizontal, pivoted pole, as shown in Figure 830. The pole is 2.10 m long and weighs 175 N. The banner, which weighs 110 N, is suspended 1.80 m from the pivot point or axis of rotation. What is the tension in the cable supporting the pole?
23.
UNIT 12 IN CLASS PROBLEMSA 5.00 m long diving board of negligible mass is supported by two pillars. One pillar is at the left end of the diving board and the other is 1.50 m away. Find the forces, in Newton's, exerted by the pillars when a 90.0 kg diver stands at the far end of the board.
22.
1 2
UNIT 12 IN CLASS PROBLEMSA 5.00 m long diving board of negligible mass is supported by two pillars. One pillar is at the left end of the diving board and the other is 1.50 m away. Find the forces, in Newton's, exerted by the pillars when a 90.0 kg diver stands at the far end of the board.
22.
1 2
τcc = τc F2 r2 sin θ = Fm rm sin θ
F2 r2 = mm g rm
F2 = 2940 N F2 = mmg + F1F1 = 2060 N
12.4(B) Zero Torque and Static Equilibrium
AP PHYSICS
March 22, 2019
UNIT 12 IN CLASS PROBLEMSA banner is suspended from a horizontal, pivoted pole, as shown in Figure 830. The pole is 2.10 m long and weighs 175 N. The banner, which weighs 110 N, is suspended 1.80 m from the pivot point or axis of rotation. What is the tension in the cable supporting the pole?
23.
UNIT 12 IN CLASS PROBLEMSA banner is suspended from a horizontal, pivoted pole, as shown in Figure 830. The pole is 2.10 m long and weighs 175 N. The banner, which weighs 110 N, is suspended 1.80 m from the pivot point or axis of rotation. What is the tension in the cable supporting the pole?
23.
τcc = τc Tc rc sin θ = Fp rp sin θ + FB rB sin θ
Tc = 430 N
Tc (2.10 m) sin 25o = (175 N)(1.05 m) + (110 N)(1.80 m)
UNIT 12 PROBLEMS(2428)
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