Wednesday, May 9 noon-1:50PM Remaining Schedule Today , April 27 Static Equilibrium (12-1, 12-2, 12-3, 12-4) Quiz 4 Review Monday, April 30 Quiz 4 / 50 points / Crib Sheet Wednesday, May 2 Results of Quiz 4 Final Review Wednesday, May 9 noon-1:50PM Final / 100 points / Open book 12/5/2018 Physics 253

Static Equilibrium Static equilibrium is special case of motion – net forces and net torque on an object are zero. Furthermore they are usually at rest. This is the preferred state for almost every structure and requires a detailed knowledge of statics. By analyzing the forces of an object in equilibrium the internal forces can be calculated and the ability of the construction materials to support equilibrium. This may seem trivial, but it’s not! Consider the Golden Gate Bridge: the forces of gravity, the load, and weather must be canceled completely by the countervailing forces of the supports, the trusses, and the suspension cables. It’s an incredibly detailed problem. We’ll content ourselves with some examples. 12/5/2018 Physics 253

Conditions of Equilibrium The first condition for equilibrium of an object is that it not be accelerating or that the vector SUM of the external forces acting on the body be zero: SF=0. This also means the three components are zero: SFx=0, SFy=0, SFz=0. The second condition for equilibrium requires that the vector sum of all external torques add to zero: St=0 and in components Stx=0, Sty=0, Stz=0. The second condition implies the selection of an origin and an axis. If all the forces act in a single plane then there can only be a torque in the perpendicular direction. The conditions then reduce to SFx=0, SFy=0, Stz=0 12/5/2018 Physics 253

A Tower Crane A crane must be in static equilibrium with respect to forces and torques or it’ll tip! The crane at right is about to lift a 2800kg load. When the load is first lifted where must the 9500kg counterweight be located? What is the maximum load that can be lifted when the counter weight is fully extended? 12/5/2018 Physics 253

The toques must also sum to 0 for static equilibrium: First we go to a free-body diagram for the BEAM and find forces and locations. The load + counterweight act downward and the normal force from the tower upward. These three forces must cancel to avoid acceleration. That is:SFy=FN-Mg-mg= 0 FN=(M+m)g=1.2x105N The toques must also sum to 0 for static equilibrium: St=Mg(x)-mg(7.7m)=0 x=m(7.7m)/M= (2800kg)(7.7m)/9500kg =2.3m The maximum load must have the counterweight at it’s maximum extent, the torque equation is then: St=Mg(3.4m)-mg(7.7m)=0 m=M(3.4m)/(7.7m)= (9800kg)(3.7m)/7.7m=4200kg. 12/5/2018 Physics 253

The Cantilever A very similar system is the cantilever, where a structure overhangs its supports. For the cantilever at right suppose the beam has a mass of 1200kg and the center of mass is 5m past the central support. What are the forces F1 and F2? Well we have two unknown forces so we need two equations which we have in the two equilibrium conditions. 12/5/2018 Physics 253

Hint: similar problem on quiz! 12/5/2018 Physics 253

Supports with a Pin and Cable. A uniform 25.0kg beam, 2.20m long is mounted with a pin and held horizontally with a cable at an angle of 30 degrees. The beam also supports a sign of mass M=280 kg. Determine the force Fp the pin exerts on the beam and the tension FT in the cable. 12/5/2018 Physics 253

12/5/2018 Physics 253

12/5/2018 Physics 253

A Final Example: The Ladder! A 5.0m ladder leans against a wall 4.0m above the ground. The uniform ladder has a mass of 12.0kg. Assume the wall is frictionless and the floor is not. What are the forces exerted on the ladder by the ground and wall? If the coefficient of static friction is ms=0.40, how far horizontally can a 58kg painter be from the point where the ladder contacts the ground before the ladder slips? 12/5/2018 Physics 253

12/5/2018 Physics 253

12/5/2018 Physics 253

12/5/2018 Physics 253

Stability and Balance If in static equilibrium and undisturbed a system will undergo no linear or rotational acceleration. However if perturbed or displaced slightly three possible outcomes can occur A return to original position (stable equilibrium) The object continues to move farther from the original position (unstable equilibrium) The object remains in the same position (neutral equilibrium) 12/5/2018 Physics 253

Stable Support above CM Support below CM Unstable 12/5/2018 Physics 253

Equilibrium more complicated if CG above support. For this fridge if tipped slightly torques act to restore stability. If CG passes point of support torques topple the fridge. Rule-of-thumb: a body with CG above it’s base will be stable if a vertical line downward from the CG falls within the base. 12/5/2018 Physics 253

Concepts on the Quiz Definition and use of the vector product, calculations using unit vectors Definitions of torque and calculation of torque for a simple systems. Definition of the moment of inertia and calculation of the moment for simple systems. Kinetic energies of a rotating disk Mechanical energy of rotating system in a gravitational field Application of the relationship between angular momentum and torque. Solution of a simple statics problem (just discussed). 12/5/2018 Physics 253

The Vector Cross Product 12/5/2018 Physics 253

12/5/2018 Physics 253

Rotational + Translational Kinetic Energy An object in translation has kinetic energy which can be described by the total mass M and the velocity of the CM vCM: KE= (1/2)M vCM2 An object rotating about an axis through it’s center of mass has rotational kinetic energy given by KE= (1/2)ICM w2 An object such as a rolling wheel has both! KE= (1/2)M vCM2+(1/2)ICM w2 12/5/2018 Physics 253

Mechanical Energy for a Rolling System 12/5/2018 Physics 253

Conservation of Angular Momentum In Chapter 9 we saw that Newton’s 2nd Law SF=ma could be rewritten as SF=dP/dt for both a single particle or a system of particles. When the net forces are zero, dP/dt=0 or P=constant, this is the Conservation Law of Momentum. Similarly St=dL/dt is true for a system of particles in an inertial frame or about the center of mass. If the net torques are zero then dL/dt=0 and L=constant, which is the powerful Conservation Law of Angular Momentum. 12/5/2018 Physics 253

See you on Monday! 12/5/2018 Physics 253