康涅狄格州大学：《普通物理》课程PPT教学课件（英文版）Lecture 17 Today's Topics

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

Statics Objects are at rest (static)when ∑F=0 AND 0 No translation No rotation When choosing axes about which to calculate torque, we can be clever and make the problem easy Physics 121: Lecture 17, Pg 6

Physics 121: Lecture 17, Pg 6 Statics: Objects are at rest (Static) when : F = 0  = 0 When choosing axes about which to calculate torque, we can be clever and make the problem easy.... No translation No rotation AND

Statics: Using Torque Now consider a plank of mass M suspended by two strings as shown. We want to find the tension in each string First use∑F=0 71+72=Mg cm M This is no longer enough to L2 solve the problem 14 1 equation, 2 unknowns Mg We need more information ! Physics 121: Lecture 17, Pg 7

Physics 121: Lecture 17, Pg 7 Statics: Using Torque Now consider a plank of mass M suspended by two strings as shown. We want to find the tension in each string: L/2 L/4 x cm M T1 T2 Mg y x First use F = 0 T1 + T2 = Mg This is no longer enough to solve the problem ! 1 equation, 2 unknowns. We need more information !!

Using Torque. We do have more information We know the plank is not rotating TOT=0 T2 cm M L2 The sum of all torques is zero 14 This is true about any axis Mg we choose Physics 121: Lecture 17, Pg 8

Physics 121: Lecture 17, Pg 8 Using Torque... We do have more information: We know the plank is not rotating. TOT = 0 The sum of all torques is zero. This is true about any axis we choose ! L/2 L/4 x cm M T1 T2 Mg y x  = 0

Using Torque Choose the rotation axis to be along the z direction (out of the page )through the cm: The torque due to the string T1 T2 on the right about this axis is cm M L2 14 The torque due to the string on the left about this axis is Mg Gravity exerts no torque about CM Physics 121: Lecture 17, Pg 9

Physics 121: Lecture 17, Pg 9 Using Torque... Choose the rotation axis to be along the z direction (out of the page) through the CM: L/2 L/4 x cm M T1 T2 Mg y x  2 2 4 =T L The torque due to the string on the right about this axis is: 1 1 2 = −T L The torque due to the string on the left about this axis is: Gravity exerts no torque about CM

Using Torque Since the sum of all torques must be 0 0 T2 T2=27 cm M Ve already found that L2 Mg g Mg 9 Physics 121: Lecture 17, Pg 10

Physics 121: Lecture 17, Pg 10 Using Torque... Since the sum of all torques must be 0: L/2 L/4 x cm M T1 T2 T L T L 2 1 4 2 − = 0 Mg y x T2 = 2T1 We already found that T1 + T2 = Mg T1 Mg 1 3 = T2 Mg 2 3 =