Rigid Bodies in Equilibrium To analyze situation or solve problem involving body in equilibrium Isolate the body Draw Forces Choose pivot for calculating torques Determine torques in terms of forces and distances Solve Examples: A. Simple see-saw For balance: Torque (clockwise)= Torque (counter clock wise) B. Q: Will see-saw be still in balance if both people move towards pivot, to ½ of their original distance? Yes Check torques: T(cw) = 200x2= 400 T (ccw)= 100x 4=400 C. Q: If they move towards pivot by 2 feet each? A: Small guy falls down Check: T(cw)= 200x(4-2)=400 T(ccw)= 100x(8-2)=600 D. Q: If both people are handed a 50 lb pumpkin would they stay the same? A. Small guy falls T(cw)= (200+50)4=1000 T(ccw)= (100+50)8=1200 F. Example with Demos Finding CM for long uniform body 15 pound board, length L Review Conditions for translational equilibrium Fnet = 0 ( without force, body will not accelerate. Will not start to travel) Condition for rotational equilibrium Tnet=0 or Tcw=Tccw (Torque is what makes bodies turn. Without torque, no rotation can begin) Levers (like see-saws, but for doing lifting rather than playing.) First class lever(support on the middle) For balance Tcw=Tccw Fapp x d2= W x d2 Note: -Fapp can be reduced by making D1 smaller or D2 larger or both -Work done on lifting with a lever= work done on lifting without lever - Direction of Fapp –advantages (it’s easier to push down) - Examples crowbar, autojack, bolt cutter - Assuming lever vs. light Fapp= W x d1/d2= 500 x 1/8 = 625 lbs’ If lever itself weighs 50 lbs For balance Tcw=Tccw Fapp x 8+50 x 3.5 = 500 x 1 8 Fapp = 325 Fapp=40.6 lbs Second Class Lever From balance- Fapp= W x d1/d2 Comment: Including weight of the plank would make Fapp larger Examples: wheelbarrow, dolly For balance- Tccw = Tcw Fapp x 6 = 300 x 2 Fapp = 100 lbs Third Class Lever (Fapp in the middle) Note: d1>d2 so Fapp >W Lever does not give advantage of smaller force Examples: Snow shovel, Human Arm, Problem: Lever representation of an arm. weighing 5 lbs For balance: Tcw=Tccw = 10+40+5 x 20 10lb object Fapp=225 lbs! Stability of Equilibrium Difference: In first case, equilibrium is more stable Stability of equilibrium can be measured by magnitude of external torque or force needed to tip body over Examples Refridgerator, weighing 100lb Pivot is at the bottom corner. It will start to rotate when torque CW produced by Fapp equalizes Tccw produced by force of gravity. B. 200 lb, 6 foot tall student, what force is needed to tip him over pushing 3 ft up? His feet are 1 foot apart. About ready to go when: Fapp x 3 = 200 x 0.5 Fapp=33lbs Same, but feet 2 ft apart. Fapp x 3 = 200 x1. Fapp =66.7 lb (more stable, harder to tip over) Like a but Fapp pushing higher at 5 feet. Fapp x 5 = 200 x .5 Fapp = 20 lbs Torque and Angular Acceleration As long as net torque is = to 0, body is in rotational equilibrium and has no angular acceleration (will not start rotating) What happens when net torque (Tnet) is not = to 0? A: Rotation will start, body will change its angular velocity (will have angular acceleration) Relation between torque and angular acceleration. Recall Newtons 2nd Law for linear motion, Fnet= ma or a= Fnet/m By analogy for angular motion unbalanced torque and angular acceleration (review) rotational analog and 2nd law Recall (for translation) : Fnet = ma By angular motion: Tnet=I x angular acceleration T=torque- rotational analog of force( force produces linear acceleration, torque- angular acceleration) Angular acceleration-rotational analog of linear acceleration Moment of inertia- rotational analog of mass m-resistance against acceleration, sluggishness resistance against rotation Depends on body’s mass and how the mass is disturbed Note: There is no one simple formula for evaluation of I for all bodies. Each case must be considered individually. See book for specific cases I=mr^2 - Point mass going circles Solid cylinder= I=1/2 mr^2 Hollow cylinder- I=mr^2 Long rod about the middle- I=1/12ml^2 For combined bodies, moments of inertia add. Find I for composite bodies. I=Ibar+Im1+Im2 Bike Wheel R=.3m m=5kg. Find torque necessary to give it angular acceleration of 10 rad/sec^2 T=I x Ang. Acceleration. I is needed first I= mr^2=5x(.3)^2= .45kgm^2 Bike wheel=empty cylinder I=mr^2=5x(0.3)=.45km Then T=Ix ang. Acc. = .45x10=4.5 Angular Momentum(L) L=angular momentum – rot. Analog of momentum Recall linear momentum- P=mv P-measures bodies persistence in motion. Bodies with high momentum are hard to stop. By analogy, angular momentum: L= I x W L-measures persistence of angular motion W- angular velocity Angular Momentum Recall: Linear momentum – P=mv Linear momentum Measure body’s motion persistence Bodies with large P are hard to stop Define angular momentum: L=I x W Angular momentum is L. L- Persistence of rotation Bodies with great L rotate and their rotation is hard to stop. P can be changed by external force only. L can be changed only by external torque only When Fnet. Ext=0 then Pi=Pf Conservation of momentum principle When Tnet =0 then Lf=Li i.e. IfWf=IiWi Examples Qualitative Rotating platform + man Because L=Iw must be conserved Ice skater, ballerina. Ex. DCOM Ice Princess High Diving Second propeller in helicopter Bike wheel and platform Polar caps melting Review L=IW L-angular momentum- persistence of rotation- bodies with large L rotate and their rotation is hard to stop. When Tnet,ext=0 then L must be conserved, that is IfWf=IiWi ^Conservation of angular momentum principal Rotational Kinetic Energy Rotating bodies have energy associated with their motion (Don’t get too close to rotating carnival wheel- it can do some pushing on you.) Formula for kinetic energy associated with rotation. Recal: KElin=1/2 mv^2 By analogy: KErot= ½ IW^2 Rolling bodies have both kids of KE it travels and spins. Race down an incline: 1. Sphere 2. Solid cylinder 3. Hallow Cylinder Review CM-Best representation of extended body location CM=Center of gravity=where gravity force is effectively applied Equilibrium= Rest, No translation and no rotation For no translation Fnet =0 For no rotation Tnet=0 Calculation of torques T=F parallel x distance Levers- Always use Tcw=Tccw for balance With unbalanced torques-Ang acceleration 1. Solids and Fluids Solid- Body with well defined (hard to change) size and shape Fluid- Body not opposing changes to its shape Parameters for fluid description Mass (Quantity of matter) Measured in KG or Slugs Symbol: m Weight (Force of Gravity) Measured in N or Lb. (1lb=4.45N) Symbol W W=m x g Density (quantity of mass per unit of volume) d= m/v (mass over volume) measured in kg/m^3 Weight Density (weight per unit volume) D=w/v Measured in N/m^3, lb/ ft^3 Example: Mercury is a very heavy fluid. It has a large weight density (849 lb/ft^3) Compare to water (62lb/ft^3) a. How many kg/m^3 are equivalent to the density of water 1 kg/cm^3 x (1kg/1ooog) x (10^6cm /1m^3)= 1,000 kg/m^3 1m^3=1m x 1m x 1m= b. Some common densities Dwater= 1g/cm^3=1,000 kg/m^3 Diron = 7.8 g/cm^3= 7800 kg/ m^3 Dhuman body = 1g/ cm^3 c. Determine average density of a baseball. M=145 g. R=3.6 cm d=m/v Volume of a sphere= 4/3 pi radius cubed= 4/3 x 3.14 x (3.6)^3 V=195 cm ^3 d=m/v = 145/195 d=.74 d. Human body has a density of 1g/ cm^3 which is about 1000kg/m^3 Estimate your volume (160lb person) mass=70 kg (d=m/v) 1000=70/V. V=.07m^3= 70 liters Pascal’s Principle Consider solid first. You have two solid blocks next to each other. You apply pressure on one block and it moves the other. Force exerted on M1 is transmitted to M2 Transmitted force is the same direction as Fapp Now, examine fluid… In a balloon, pressure is transmitted to every point Pascal’s Principle- Pressure applied to a fluid is transmitted to every point within the fluid. Example: Barber’s chair (Hydrolic Pressure) By Pascal’s principle, for balance- pressures under both pistons must be equal. W/A2= Fapp/A1 Fapp=A1/A2 W Numerical example W=250 lbs A2= 16in^2 A1=2in^2 Then= Fapp = 2/16 x 250 Fapp = 31.25 lbs 2. Archimedes Principle Consider an object immersed in fluid . The pressure on the bottom will be greater than pressure on the top resulting in upward net force, called buoyant force Fbottom= Pbottom x A= dgh^2 A Ftop= Ptop x A = dgh^1 A Net force by fluid= BF (buoyant) =dgh^2 A –dgh^1 A =dgA (h^2-h^1) = dgV=gdV =BF = m x g = W (weight of displaced fluid) BF=Weight of displaced fluid - Archimedes- Body immersed in a fluid is buoyed up by force equal to the weight of displaced fluid. Archimedes Principle “Body immersed in a fluid is buoyed up by a force equal to the weight of fluid it displaces” or BF = W of displaced fluid Consequences, examples Bodies feel lighter when immersed in water. One needs to apply force upward equal to W to prevent body from falling Density of Water 1G/cm^3 = 1 m=350g=.35kg 1. Fluid Flowing Region 1 Large cross section Low speed Flow lines widely separated High pressure Region 2 Small cross section High speed Flow lines crowded Low pressure Bernovlli’s principle Fluid Resistance Intro Aristoltle: Heavier bodies fall to the ground faster Galileo: Free fall looks the same for all bodies Truth: Free fall is accelerated motion with a~10m/s for all bodies if air resistance can be neglected Moving bodies experience resistance of the medium. Ex. Try to run in a pool Surface Drag Primary source of fluid resistnace comes from the fact that fluid is set in motion my moving body and that requires force More fluid being move, more violent its motion the more force is needed and bigger surface drag Factor influencing surface drag Cross section area of the object Smoothness of the object Velocity of the object Viscosity of the fluid Form Drag Secondary source of fluid resistance