Lecture14

Physics 101: Lecture 14 Torque and Equilibrium ,[object Object],Exam II

Review ,[object Object],[object Object],[object Object],[object Object],[object Object]

You Know Torque! ,[object Object],[object Object],[object Object],Balance Demo 1 kg 2kg 50 cm 0 25 50 75 100

Torque ,[object Object],[object Object],[object Object],[object Object]

ACT ,[object Object],[object Object],[object Object],CORRECT

ACT 2 ,[object Object],[object Object],[object Object],CORRECT

Torque Example and ACT ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],= 65 N m 30 F=89 N r = 0.84 m

Equilibrium Acts ,[object Object],[object Object],[object Object],F F Y direction:  F y = ma y +F – F = 0 Yes, it rotates! y x

Equilibrium ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Balance Demo  = 0 9.8 (0.5) – (19.6)d = 0 d = 25 9.8 N 19.6 N 50 cm d

Static Equilibrium and Center of Mass ,[object Object],[object Object],[object Object],[object Object],Center of mass pivot d W=mg

Static Equilibrium A method to find center of mass of an irregular object Not in equilibrium Equilibrium Center of mass pivot d W=mg Torque about pivot  0 Center of mass pivot Torque about pivot = 0

Preflight ,[object Object],[object Object],[object Object],[object Object],“ assume r for F1(person on the left to the log) is 1/4R, then after calculation i get F1=2F2” “ The guy on the left has to support more weight because he is further in and has to support more of the board. ” CORRECT

Preflight ,[object Object],[object Object],[object Object],[object Object],Look at torque about center: F R L – F L L/2 = 0 F R = ½ F L CORRECT mg F L F R L/2 L

Preflight ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

Homework 8 Hints ,[object Object],A B 1 meter 1 meter 1) Draw FBD 2)  F = 0 3) Choose pivot 4)  = 0 F A + F B – mg – Mg = 0 -F A (1) sin(90)+ F B (0) sin(90) + mg (0.5)sin(90) + Mg(0.5) sin(90) = 0 F A = 0.5 mg + 0.5 Mg = 612.5 Newtons F A F B Mg mg 1 meter 0.5meter

Homework 8 Hints ,[object Object],[object Object],A B 1 meter 1 meter

Homework 8 Hints ,[object Object],A B 1 meter 1 meter Just before board tips, force from A becomes zero 1) Draw FBD 2)  F = 0 3) Choose pivot 4)  = 0 F B – mg – Mg = 0 F B (0) sin(90) + mg (0.5)sin(90) – Mg(x) sin(90) = 0 0.5 m = x M F B Mg mg 0.5meter x

Equilibrium Example ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],mg F 1 F 2

Homework 8 Hints ,[object Object],Using F TOT = 0 : T = m 1 g + m 2 g + Mg allows you to solve for m 1 T Mg m 2 g m 1 g

[object Object], (m 1 g) = 0 since lever arm is 0 x T Mg m 2 g m 1 g

 (Mg ) = -Mg L/2  (m 1 g) = 0 since lever arm is 0 L/2 T Mg m 2 g m 1 g

 (Mg ) = -Mg L/2  (m 1 g) = 0 since lever arm is 0 x  (T ) = T x T Mg m 2 g m 1 g

Work Done by Torque ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

Summary ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

Lecture14

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