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Lecture 7: Equilibrium I - Basics SJ 7Th Ed.: Chap 12.1 to 3

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Lecture 7: Equilibrium I - Basics SJ 7Th Ed.: Chap 12.1 to 3 Physics 106 Lecture 7: Equilibrium I - Basics SJ 7th Ed.: Chap 12.1 to 3 • The equilibrium conditions • Types of mechanical equilibrium • Center of gravity – Definition – Methods for finding CG – CG versus mass center – Examples • Solving equilibrium problems – Some useful theorems – Problem solving rules and methods – Sample problems The Equilibrium Conditions • No new physics here … apply Newton’s second law • Special case of 2nd Law with accelerations equal to zero • So… net forces and torques are zero as well G G ⎧p = constant ⎫ G G dp G ⎪⎪G IF F .EQ. 0 THEN a = 0 ∑Fi,ext ≡ Fnet = = ma { net,ext } ⎨ ⎬ dt ⎪ ⎪ ⎩⎭first condition met G “First” condition for equilibrium: ∑Fi,ext = 0 Identify forces, write equations setting their sum = 0 G G ⎧⎫L=constantL = constant dL G ⎪ G ⎪ G G IF τα .EQ. 0 THEN = 0 τi,ext ≡ τnet = = Ια {}net,ext ⎨ ⎬ ∑ dt ⎪ ⎪ ⎩⎭second condition met G “Second” condition for equilibrium ∑ τi,ext = 0 Identify torques, write equations setting their sum = 0 1 MECHANICAL EQUILIBRIUM ISOLATED SYSTEM No change in translational or rotational “state” constant momentum, zero net force, constant angular momentum, zero net torque zero linear acceleration of particle or cm of rigid body, zero angular acceleration of rigid body MECHANICAL EQUILIBRIUM: DYNAMIC VERSUS STATIC Dynamic: ΣF = 0 and Στ = 0....BUT... p and/or L are not = 0 Static: In addition, p = 0 and L = 0 STABILITY OF STATIC EQUILIBRIUM Imagine displacing system a small amount: does it return? stable unstable neutral EXAMPLES: book on table static neutral puck sliding on ice dynamic -- ceiling fan – on dynamic -- ceiling fan – off static neutral car on straight road dynamic -- car on curve not equilibrium -- ladder leaning against wall static unstable (friction) stable (foot in groove) Definition: The center of gravity of a body is the point about which all of the torques due to gravitational forces sum to zero • CG is about torques – gravitational torques • A body is “balanced” if it is in rotational equilibrium • Gravitational torques cancel about the CG • A single support force through CG creates zero torque (moment arm = 0)… ... but ensures zero net force on system (linear equilibrium) • CG often coincides with mass center (CM) ... but may not if gravity varies across the body balance a ruler see-saw a box about to tip L L1 L2 m m1 m2 < m1 A single non-gravitational normal force applied at the CG can produce translational equilibrium 2 What’s the Difference Between CG and CM? Definition of Mass Center for point particles: G ˆ ˆ ˆ rcm ≡ (xcm,ycm,zcm ) ≡ xcmi + ycm j + zcmk • Divide object into many small particles – Each particle has a specific mass and specific coordinates • The x coordinate of the center of mass is a mass-weighted average of the position coordinates: mx ∑ ii G x = i CM rcm ∑ mi • Similar expressionsi for the y and z coordinates xm zm ∑ ii ∑ ymii ∑ ii x = y = z = cm m cm cm m ∑ i ∑mi ∑ i Depends only on mass distribution No consideration of gravity needed What’s the Difference Between CG and CM? CM is determined by mass distribution alone, i.e. by geometry & density function of body CM would be the same on any planet CG also depends on uniformity of local gravity field CG might be shifted away from CM when g is not uniform If density and gravity are both uniform - CG, CM, and geometrical center all coincide Uniform Gravitational Field CG Near left end CG coincides with CM CM still at midpoint CM CM X X Super dense material From neutron star 3 G The CG and CM coincide when g is uniform across a body Proof: Consider a set of particles and a rotation axis at the origin Find net torque due to gravitational forces alone G K K G G G τg,net = ∑ τg,i = ∑ ri ×Fg,i = ∑ ri × migi If gravitation has uniform magnitude and direction, the g ’s factor out G i G G G m r G τ = [ m r ] × g = ∑ i i × Mg g,net ∑ i i M Definition of mass center Total weight Fg acting relative to origin at the CM G K K G ∴ τg,net = rcm × Fg (when g is uniform) ∴ To ptput he body into equ ilibrium (balance), su pport it w ith a force equ al G to F applied at the center of gravity r g cg G ...the torque would be equal and opposite to τnet above. K ...the externalK force cancels the net g force ∴ rcm is the same as rcg - the CG location. Also, the net gravitational torque on a rigid body about any axis equals that produced if the weight were a point mass at the center of gravity – see below Example: Find the balance point (CG) for a massless see-saw L FBD N m1 x m2 sL -s s balance point = CG m1gm2g Balance means: put the support force N (= total weight) at a point such that • system does not accelerate linearly (static translational equilibrium) • angular accelerat ion is also zero (rotati onal equilibilibirium ) To locate CG coordinate “s” • Evaluate torques around CG, including only gravitational forces • Assume support force N is applied at CG, exerting zero torque 1) ∑ τcg = 0 ⇒ m1gs − m2g(L − s) + N x 0 m2 Ex: Let m1 = 2m2 ∴ s = L locates cg then s = 1/3 L (m1 + m2 ) 2) ∑Fy = 0 ⇒ (m1 + m2 )a = 0 = N − m1g − m2g ∴ N = (m1 + m2 ) g total weight Actually, if an object is in equilibrium, then the net torque = 0 for any axis chosen (more below). For an axis away from CG, torque due to N is not zero but still cancels the gravitational torques 4 Center of gravity for a sledge hammer 7.1. The center of gravity for a sledge hammer lies on the centerline of the handle, close to the head, at the X mark. Suppose you saw across the handle through the center of gravity, cutting the ax in two pieces. You then weigh both pieces. Which of the following will you find? x A) The handle piece is heavier than the head piece B) The head piece is heavier than the handle piece C) The two pieces are equally heavy D) The comparative weights depend on more information Definition: The center of gravity of a body is the point about which all of the torques due to gravitational forces sum to zero The torque on a body due to gravity equals the torque gravity would exert on a point mass with same weight (magnitude and direction) located at the CG Proof: Assume a collection of particles { mi } Arbitrary rotation axis at “O” cg K K r ' Include only gravitational forces r i G G G G cg ' O K ri = ri + rcg where... rcg = CG location mi G ri ri ≡ particle coordinate relative to O G' r ≡ particle coordinate relative to cg Gi K G G G G G K G τg,net,O = ∑ ri ×Fg,i ≡ ∑ ri × migi = ∑ ri' × migi + rcg × ∑ migi K Total weight Net torque of gravitational forces Center of ≡ Fg about CG = 0, by definition of CG point gravity location of body G K G Prescription: for torque about any axis ∴ τnet,O = rcg × Fg due to rigid body’s weight replace the body by a point mass at CG with the same weight. This works even if g varies. 5 Finding CG’s of rigid bodies FINDING T CG! -Pick body up (external force T) and let it swing freely, come to equilibrium position, and stop (friction). previous - Body comes to rest with CG below or above suspension suspension point, o/w there would be non-zero torque point on it and body would not be in equilibrium. - Mark a vertical line at suspension point - Repeat process for a different suspension point - The CG is where the lines intersect. If gravitation field is uniform, CG & CM coincide but may not be at geometrical symmetry points unless density is uniform. • For uniform sphere, cube, disc, rectangle,.. CG at center • For cone. cylinder, bar, etc CG along axis of symmetry • For composite objects, break into shapes with symmetry STABILITY Block tips if CG is to the left of WEIGHT all support points in base. W ACTS AT Nete torque then can not = 0 CG N Center of gravity: summary • All the various gravitational forces acting on all the various mass elements are equivalent to a single gravitational force (the total weight) acting through a single point called the center of gravity (CG) • The net torque due to gravitational forces on an object of mass M equals the force Mg acting at the center of gravity of the object. • If g is uniform over the object, then the center of gravity of the object coincides with its center of mass. • If the object has constant density and is symmetrical, the center of gravity coincides with its geometric center. 6 Does it matter which axis you choose for calculating torques? No. Use any rotation axis to solve equilibrium problems. IF Fnet = 0 THEN τnet is the same around every parallel axis ..so... IF τnet = 0 for one axis THEN it is zero for every other axis Proof: Compare rotation axis at O with one shifted to O’ for a system of point particles { m } O’ K' G i K r i K K a LO = ∑ ri ×pi angular momentum about axis O G G G G O K mi r ri = ri' + a shift axis by a to O' origin G i K' G K K' K G K LO = ∑[ri + a] ×pi = ∑ ri ×pi + a × ∑ pi G K angular momentum angular momentum linear momentum = Lo' ≡ P relative to axis O relative to axis O’ of mass center cm Take the time Gderivative Gof each term, useG second law (rotational and linear) dL dL G dP Rightmost term vanishes O = O' + a × cm if F = ΣF = 0 K dt dt K dt G net i = τnet, o = τnet, o' = Fnet, ext K K ∴ If ∑Fi = 0 then τnet is the same about any axes O or O' Note also: if Pcm = 0, the angular momentum is the same around all parallel axes Rules for solving equilibrium problems G First Condition: Fnet = 0 G Second Condition: τnet = 0 Choose anyyq convenient axis for torque calculation.
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