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 =