170 likes | 229 Views
0. 0. ?. ?. 30. 30. 10. 10. 20. 20. 1.0 m. 0.5m. 1.0 kg. Predict the readings on the scales:. M bar = 1.0 kg. 10 N. 10 N. 1.0 m. 0.5m. Predict the readings on the scales:. 0. 0. 30. 10. 30. 10. 20. 20. M bar = 1.0 kg. 1.0 kg. 0. 0. ?. ?. 30. 30. 10. 10. 20.
E N D
0 0 ? ? 30 30 10 10 20 20 1.0 m 0.5m 1.0kg Predict the readings on the scales: Mbar = 1.0 kg
10 N 10 N 1.0 m 0.5m Predict the readings on the scales: 0 0 30 10 30 10 20 20 Mbar = 1.0 kg 1.0kg
0 0 ? ? 30 30 10 10 20 20 1.0 m 1.0kg Now predict the readings: 0.8m Mbar = 1.0 kg a=0
F2 F1 standard meter stick standard pivot point standard frictionless pond Which force is most effective in causing rotation?
F2 F1 standard meter stick standard pivot point standard frictionless pond Now which force is most effective in causing rotation?
F = 200 N r = 3 m Pivot In practice…. t = rF = (3 m)(200 N) = 600 m - N
q F sin Useful force method: only the component Fperp causes torque. F q q r t q) = r (F sin = r Fperp Fixed Pivot Point
Torque • A torque is produced when a force is applied with “leverage”. • net torque causes angular acceleration • Only the perpendicular force contributes to the Torque.
Torque • net torque causes angular acceleration if the force is applied “off center” • If net torque = 0, angular velocity() = constant. • Doubling the net torque doubles angular acceleration ().
0 ? 30 10 20 1.0 m Let’s practice: 200 N pivot 200 N
0 ? 30 10 20 1.0 m 0.5m Let’s practice: 400 N pivot 200 N
100 N ??? N 150 N Linear equilibrium: a = 0 Fleft = Fright
0 ? 30 10 20 1.0 m 0.25m Rotational equilibrium a = 0 800 N t1 = t2 pivot 200 N
0 0 ? ? 30 30 10 10 20 20 1.0 m 1.0kg Let’s solve the problem: 0.8m Mbar = 1.0 kg a = 0 Fnet = 0 = 0 net = 0
TR 0.5m TL 0.8m Pivot 10 N 10 N t1 = t2 (10 N)(.5 m) + (10 N)(.8 m) = TR(1 m) TL= 7 N TR= 13 N
0 0 ? ? 30 30 10 10 20 20 1.0 m 1.0kg Let’s solve the problem: 0.6m Mbar = 1.0 kg a = 0 Fnet = 0 = 0 net = 0