Physics 121, Sections 9, 10, 11, and 12 Lecture 17 Today's Topics Homework 7: due Friday Nov 4@ 6: 00 PM Chap7:#3,11,20,21,25,27,30,40, 46,47,52,and68 Chapter 8 Torque Static equilibrium C.M. motion Rotation Moment of inertia Physics 121: Lecture 17, Pg 1
Physics 121: Lecture 17, Pg 1 Physics 121, Sections 9, 10, 11, and 12 Lecture 17 Today’s Topics: Homework 7: due Friday Nov. 4 @ 6:00 PM. Chap. 7: # 3, 11, 20, 21, 25, 27, 30, 40, 46, 47, 52, and 68. Chapter 8: Torque Static equilibrium C.M. motion Rotation Moment of inertia
Torque The lever arm is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force t=Fd= FL sin Fsinφ F F Fcosφ The sign of the torque is positive if its turning tendency is counterclockwise and negative if its turning tendency is clockwise(right-hand rule) Physics 121: Lecture 17, Pg 2
Physics 121: Lecture 17, Pg 2 Torque The lever arm is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force = Fd = FL sin d L F L F F cos F sin The sign of the torque is positive if its turning tendency is counterclockwise and negative if its turning tendency is clockwise (right-hand rule)
Conditions for Equilibrium An object in mechanical equilibrium must satisfy 1. The net external force must be Zero ∑F=0 2. The net external torque must be zero ∑T=0 The first condition is translational equilibrium and the second is rotational equilibrium If the object is in equilibrium, then the choice of axis of rotation does not influence the calculation of the net torque The location of the axis is arbitrary Physics 121: Lecture 17, Pg 3
Physics 121: Lecture 17, Pg 3 Conditions for Equilibrium An object in mechanical equilibrium must satisfy: 1. The net external force must be Zero: SF = 0 2. The net external torque must be Zero: S = 0 The first condition is translational equilibrium and the second is rotational equilibrium If the object is in equilibrium, then the choice of axis of rotation does not influence the calculation of the net torque; The location of the axis is arbitrary
Objects in Equilibrium Walking a horizontal beam A Uniform horizontal 300 N beam, 5.00 m long, is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by cable that makes an angle of 53 00. If a 600 N person stand 1.5 m from the wall. find the tension in the cable and the force exerted by the wall on the beam 600N 53.0 5.00 Physics 121: Lecture 17, Pg 4
Physics 121: Lecture 17, Pg 4 Objects in Equilibrium Walking a Horizontal Beam A Uniform horizontal 300 N beam, 5.00 m long, is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by cable that makes an angle of 53.00 . If a 600 N person stand 1.5 m from the wall, find the tension in the cable and the force exerted by the wall on the beam. 53.00 5.00 m 600 N
Walking a Horizontal Beam F=R.-Tcos5300=0 F=Ry-Tsin5300-600N-300N=0 Tsin53.09)(500m) (300N)(2.5m) (600N)(1.5m)=0 T=413N.R.=249N.R.=570N 53.0 600N 300N 600N 300N Physics 121: Lecture 17, Pg 5
Physics 121: Lecture 17, Pg 5 Walking a Horizontal Beam Fx = Rx - T cos 53.00 = 0 Fy = Ry - T sin 53.00 – 600 N – 300 N = 0 0 = (T sin 53.00 )(5.00 m) - (300 N)(2.5 m) - (600 N)(1.5 m) = 0 T = 413 N, Rx = 249 N, Ry = 570 N 53.00 600 N 300 N o 600 N 300 N o