Physics 201, Lecture 21 Today s Topics q Static Equilibrium of Rigid Objects(Ch. 12.1-3) Review: Rotational and Translational Motion Conditions for Translational and Rotational Equilibrium Demos and Exercises q Stable and Instable Equilibrium (Section 7.9) Hope you have previewed!
Review: Torque τ = r F q τ Fsinφ r ( ) Torque depends on F, r, and sinφ No torque If F=0, or φ=0 o /180 o, or r = 0 q The torque acting on the object is proportional to its angular acceleration Στ = I α Moment if inertia
Review: Rotational Dynamics Στ = I α Rotational Dynamics compared to 1-D Dynamics Angular Linear τ F α = a = I m ω = ω 0 + αt v v + at θ = θ + ω 2 1 2 0 + ω0t 2 αt ω 2 + αδθ 0 = 2 = 0 x = x + v 2 1 2 0 + v0t 2 at 2 = v + 2aΔx 0 KE = 1 2 Iω 2 L = I ω KE = 1 2 mv2 p = m v
Review: Motion of Rigid Object: Translation + Rotation = +
Review: Dynamics For Rigid Objects q Translational Motion Σ F external = M a CM q Rotational Motion Σ τ = dl /dt (= Iα for object around fixed axis)
Mechanical Equilibrium (Static)
q Mechanic Equilibrium: Mechanic Equilibrium Translational acceleration a=0, AND Rotational acceleration α=0 q Conditions of Mechanic Equilibrium: Ø a=0 The net external force must be zero (think a=f/m) Σ F = 0 Ø α=0 The net external torque must be zero (think α = τ/i) Σ τ = 0 F x =0 F y =0, F Z =0 q Static and Dynamic Equilibrium Note in equilibrium, translational velocity does not have to be 0 Static equilibrium: v=0 Dynamic equilibrium: v =constant (!=0)
Mechanical Equilibrium (Case Analyses) N mg mg N center of gravity for board+bottle N m 1 g m 2 g Mg
Solving Equilibrium Problems q Draw Free Body Diagram (FBD) with exact acting point of each force q Establish convenient coordinate axes for each object. Ø Apply the First Condition of Equilibrium F net = 0. (Note in 2D this gives you F net,x = 0 and F net,y = 0) q Choose a convenient rotational axis for calculating net torque on object. and then apply τ net = 0. v Note: In the situation of static equilibrium, the choice of rotational axis is arbitrary. Just choose a convenient one such as a pivot (some choices are better, choose a simple one cleverly.) q Solve the resulting simultaneous equations for all of the unknowns.
The Acting point of Gravity: Center of Gravity q The force of gravity acting on an object must be considered in determining equilibrium q In finding the torque produced by the force of gravity, all of the weight of the object can be considered to be concentrated at one point called center of gravity (cg) q Effectively, assuming gravitation field is uniform, the CG of an object is the same as its CM (that is usually true at the Phy103 level) x cg = Σm ix i Σm i and y cg = Σm iy i Σm i See demo: finding CG
Example of a Free Body Diagram Fig 8.12, p.228 Slide 17 Important: The direction and acting point of each force must be drawn correctly.
Another Example of FBD
Exercise: Seesaw q Masses of father and daughter are m f =70 kg and m d =45 kg, respectively. The seesaw bar has a mass of M=50 kg. Ø where the father has to sit to balance the bar? Solution: Identify 4 forces as in FBD. Choose the pivot point as origin for torque τ daughter = - m d g L τ seesaw = Mg x 0m =0 τ N = N x 0m =0 τ father = m f g d Στ =0 d= (m d /m f )L = 1.3m Ø How much is N? ΣF=0 N= m f g+m d g+mg = (70+45+50)g =1617 N N L= 2m
Exercise: Leaning On the Wall (I) q A ladder of mass m=1kg is leaning without slipping on the wall at an angle φ=60 o as shown. Assume there is no friction on the wall. Ø What is the friction from ground. A: 2.8N, B:4.9N, C:9.8N, D: not sure as length is not given Solution: Identify forces (N F,N W, mg, F friction ), draw FBD Take P as the origin for torque. τ NW = N W sinφ L out of page τ mg =- mgsin(90 o -φ)l/2 into page τ NF = τ friction =0 Στ=0 N W sinφ L mgcosφl/2=0 N W =mg/(2tanφ) = 2.83N also: 0= ΣF x = F friction -N W F friction = N W = 2.83 N to the right
Exercise: Leaning On the Wall(2) q A ladder of mass m=1kg is leaning without slipping on the wall at an angle φ=60 o as shown. Assume there is no friction on the wall. Ø What is the friction from ground. Solution: Identify forces (N F,N W, mg, F friction ), draw FBD Take CM of the ladder as the origin for torque. (Spend 1-2 min. now, more after class) Repeat steps on previous slide And verify that you get the same answer
Exercise: Leaning On the Wall (3) q A ladder of mass m=1kg is leaning without slipping on the wall at an angle φ as shown. There is no friction on the wall and the µ s between the ladder and the ground is 0.35. Ø What is the minimum angle φ such that ladder does not sliding down? Solution: Identify forces (N F,N W, mg, F friction ), draw FBD Take P as the origin for torque. τ NW = N W sinφ L out of page τ mg =- mgsin(90 o -φ)l/2 into page τ NF = τ friction =0 Στ=0 N W sinφ L mgcosφl/2=0 N W =mg/(2tanφ) also: in y: N F -mg=0 N F =mg in x 0= ΣF x = F friction -N W F friction = N W = mg/(2tanφ) < µ s N F tanφ > 1/(2µ s ) f > 55 o
Stable and Unstable Equilibriums (Section. 7.9, conceptual only) q It can be shown that when a system is in equilibrium, the first derivative of its potential energy must be zero. (du/dx=0) This is consistent with conditions F=0, and τ=0 q There are 2 class of equilibrium: Stable and Unstable du/dx=0 and d 2 U/dx 2 >0 du/dx=0 but d 2 U/dx 2 <0 stable equilibrium at x=0 (lowest energy principle) unstable equilibrium at x=0