Lecture02

Forces: Equilibrium Examples ,[object Object],Physics 101: Lecture 02 Phys 101 URL: http://online.physics.uiuc.edu/courses/phys101 Read the course description & FAQ! Office hours start Monday; see web page for locations & times. Exam I

Overview ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

Free Body Diagrams ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Physics hand Gravity Normal friction y x

Book Pushed Across Table ,[object Object],Physics hand Gravity Normal friction x-direction:  F=0 F hand -F friction = 0 F hand =F friction F hand =  k F Normal y-direction:  F=0 F Normal -F Gravity = 0 F Normal = F Gravity F Normal =  x 9.8=9.8 N Combine: F hand =  k F Normal F hand =0.75 x 9.8 N F hand =   ewtons Constant Speed =>  F=0 y x

Contact Force: Springs ,[object Object],[object Object],[object Object],F 1 +F 2 - F gravity = 0 F 1 +F 2 = F gravity k 1 x 1 + k 2 x 2 = m g 2 k x = mg x = m g / (2k) = (½ m) g / k We know mg/k = 8 cm. So: ½ mg /k = 4 cm. y x F 2 F gravity F 1

Contact Force: Springs ,[object Object],[object Object],[object Object],y x

Contact Force: Tension ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],F = mg = 440 Newtons From Prelecture

Pulley ACT ,[object Object],[object Object],[object Object],[object Object],1 2  F = m a 1) T - m 1 g = 0 2) T – m 2 g = 0 => m 1 = m 2 1 T m 1 g 2 T m 2 g

Tension Example: ,[object Object], F = m a T + T – W = 0 2 T = W T = m g / 2 = (45 kg x (9.8 m/s 2 )/ 2 = 220 N ,[object Object],A) 220 N B) 440 N C) 660 N D) 880 N E) 1100 N y x T T W

Tension ACT II ,[object Object], F = m a F c -T - T – T = 0 F c = 3 T F c = 3 x 220 N = 660 N ,[object Object],A) 220 N B) 440 N C) 660 N D) 880 N E) 1100 N y x F c T

Springs Preflight ,[object Object],[object Object],1 The magnitude of tension in a ideal string is equal everywhere.

Springs ACT ,[object Object],[object Object],In both cases the NET FORCE on the spring is zero, and the force to the right is 550N. Therefore the force to the left is also 550 N. 1 2

Forces in 2 Dimensions: Ramp ,[object Object],[object Object],[object Object],Note, weight is not in x or y direction! Need to decompose it!  y x T N W

Forces in 2 Dimensions: Ramp ,[object Object],  W sin  W cos  y x W T N W

Forces in 2 Dimensions: Ramp ,[object Object],x- direction W sin  – T = 0 T = W sin   m g sin   Newtons  y x T N W

Normal Force ACT ,[object Object],[object Object],In “y” direction  F = ma N – W cos  = 0 N = W cos  N = m g cos   T N W y x W  W sin  W cos 

Force at Angle Example ,[object Object],x- direction:  F x = ma x F push cos(  ) – F friction = 0 F push cos(  ) –  F Normal = 0 F Normal = F push cos(  ) /  y- direction:  F y = ma y F Normal –F weight – F Push sin(  ) = 0 F Normal –mg – F Push sin(  ) = 0 Combine: (F push cos(  ) /  –mg – F Push sin(  ) = 0 F push ( cos(  ) /  - sin(  )) = mg F push = m g / ( cos(  )/  – sin(  )) F push = 80 N y x Normal Weight Pushing  Friction 

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

Lecture02

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