Ch 9 Rotational Forces Torques Static Dynamic Equilibrium

Ch. 9 Rotational Forces: Torques & Static & Dynamic Equilibrium 1

Physics 2 Spring 2017 • HW 1: Ch. 8 Due Mon 2/6 • pg. 282 Practice 8 A # 1, 2, 3 • pg. 282 Section Rev # 1, 2, 4 • pg. 288 Practice 8 B # 2, 4 • pg. 289 Section Review # 1, 5 2

Rotating Objects • If forces are not applied to the center of an object’s mass, that object will rotate. 3

Torque = Rotational Force Symbol: (tau), Unit: Nm When An Applied Force Causes Rotation = r Fsin clockwise rotation negative • counter-clockwise rotation, positive • Lever arm, r, points from the pivot to where the force is applied • • 4

Examples of Torque 5

Torque Factors • Torque Greatest When Applied Force Perpendicular to Lever Arm, F r • Torque Greatest When Force Applied Furthest from Pivot (big r) B Easier to open a door when A you push away from the hinge. C 6

Example Problem Luis uses a 20 -cm-long wrench to turn a nut. The wrench handle is tilted 30 above the horizontal, and Luis pulls straight down on the end with a force of 100 N. How much torque does Luis exert on the nut? = r Fsin = (0. 2 m)(100 N)sin(30°) = 10 Nm 7

Static Equilibrium • Fnet = 0 (so a=0) • All Forces are Balanced • v = 0 (no motion) 8

Dynamic Equilibrium • Fnet = 0 (a = 0) • Forces are Balanced • v = constant (object IS moving) 9

Objects Are NOT in Equilibrium If They Rotate! Or if they accelerate 10

Rotating Objects: Pivot Point • Objects Rotate around a hinge or place the Object is CONSTRAINED to Rotate around • Center of Mass (aka center of gravity) if not constrained. 11

Center of Gravity • When there’s no set pivot point, objects will want to pivot around the center of gravity. 12

Center of Gravity • When there is a set pivot point, the object tends to fall so that the center of gravity will be just below the pivot point (keep in mind that it may be unstably stable). FN pivot Ff Fg 13

Even when Linear Forces are Balanced, An Object is NOT in Equilibrium if it Rotates 14

Second Condition for Equilibrium • 15

True Equilibrium • Net Force = 0 F 1 + F 2 + Fp= 0 • Net Torque = 0 + + = 0 1 2 p F 2 F 1 Fp 16

Example Problem • The two children shown below are balanced on a seesaw of negligible mass. The first child has a mass of 26. 0 kg and sits 1. 60 m from the pivot. a. If the second child has a mass of 32. 0 kg, how far is she from the pivot? b. What is supporting force, Fp, exerted by the pivot?

Example 9. 1 = r. Fsin • m 1 = 26. 0 kg r 1 = 1. 6 m a. m 2 = 32. 0 kg, how far is she from the pivot? (find r 2) System is is Equilibrium (“BALANCED”), and 1 + 2 + p= 0 so F 1 + F 2 + Fp= 0 r 1 F 1 + r 2 F 2 + rp. Fp= 0 F 1 = (26 kg)(9. 8 m/s 2) =254. 8 N +(1. 6 m)(254. 8 N) - r 2(313. 6 N) + (0)Fp= 0 F 2 = (32 kg)(9. 8 m/s 2) (1. 6 m)(254. 8 N) = r 2 =313. 6 N (313. 6 N) r 2 =1. 3 m

Example 9. 1 = r. Fsin • m 1 = 26. 0 kg r 1 = 1. 6 m a. m 2 = 32. 0 kg, how far is she from the pivot? (find r 2) System in Equilibrium (“BALANCED”), and 1 + 2 + p= 0 so F 1 + F 2 + Fp= 0 F 1 = 254. 8 N F 2 = 313. 6 N r 2 =1. 3 m b. What is the Fp, - m g + -m g + F = 0 1 2 p the supporting F = (26 kg)(9. 8 m/s 2)+(32 kg)(9. 8 m/s 2) p force exerted Fp= 568 N by the pivot? Fp= 568. 4 N

Example Problem: 8 B #3 pg 288 A bridge 20. 0 m long weighing 5 4. 00 x 10 N is supported by two pillars located 3. 00 m from each 4 end. If a 1. 96 x 10 N car is parked 8. 00 m from one end of the bridge, how much force does each pillar exert? 20