Chapter 12 Equilibrium and Elasticity. Questions: 3, 5 Problems: 3, 7, 11, 18, 19, 25, 31, 34, 43, 49


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1 Chapter 12 Equilibrium and Elasticity Questions: 3, 5 Problems: 3, 7, 11, 18, 19, 25, 31, 34, 43, 49
2 Equilibrium Four situations: 1) book sitting on a table 2) hockey puck sliding with constant velocity on frictionless surface 3) rotating blades of ceiling fan 4) wheel of bicycle that is traveling in straight path at constant speed All have two conditions: 1) linear momentum about center of mass is constant 2) angular momentum about any point is constant Objects that fit these two requirements are said to be in equilibrium.
3 For the first situation, the book on a table, not only are the linear and angular momentums constant, they are zero. The book is not moving. The book is in static equilibrium. Objects in static equilibrium will not move or fall over (rotate) as long as no more outside forces are applied.
4 Stable equilibrium if an object in equilibrium is displaced by a small amount by a force and it returns to the initial state, the object is in stable static equilibrium. A small force does not end the equilibrium. If you push the ball in either direction, it will return to the bottom of the trough. When you push the ball in either direction, you increase the potential energy of the ball. The ball wants to minimize the potential energy so it returns to the original position. This is related to the potential energy curve.
5 Unstable equilibrium if you give a small displacement to an object and the displacement increases afterwards, the object is in unstable equilibrium. A small force can end the equilibrium. Push the ball slightly in either direction and it will roll down the side.
6 When the center of mass is directly above a supportive edge, the object is unstable. When the center of mass is between the supportive edges of an object it is not at unstable. This tells us that the size of the base of an object determines how stable the object is. (see pictures with dominos and block)
7 Requirements for Equilibrium All the external forces acting on an object balance out. dp Fnet dt F 0 net All the external torques acting on an object balance. dl net dt 0 net
8 These two conditions are vector equations, so we can write them for each direction. F net,x =0 F net,y =0 F net,z =0 F 0 0 net net,x=0 net,y=0 net,z=0 net In 3D the object can rotate about all 3 axis. You end up with a system of 6 equations and 6 unknowns. If we stick to 2D (objects in the xy plane, they can only rotate about the z axis). We now will see systems of 3 equations and 3 unknowns.
9 Sample problem 121 Examples A uniform beam of length L and mass m=1.8 kg is at rest on two scales. A block of mass 2.7 kg with is on the beam with its center L/4 from the left end. What do the two scales read? Problem number 74 A beam of length 12 m is supported by a cable and a hinge. The tension in the cable is 400 N. The beam is 50 0 from the vertical. What is the gravitational force on the beam and what is the force that the hinge exerts on the beam?
10 Indeterminate structures We will solve problems in 2D by using a system of 3 equations and 3 unknowns. Two of the equations come from the sum of the forces and the third comes from taking the torque about a point. Sometimes there are more than 3 unknowns. Theses problems are indeterminate.
11 Elasticity Rigid bodies are made up of a 3D lattice of atoms. The atoms are held together by interatomic forces that can be modeled as tiny springs.. When forces act to stretch, compress, or twist a rigid object, the springs that connect the atom are deformed. Just like springs, if you deform an objet too much you can permanently deform or break the object.
12 Book gives example of hanging cars from a steel rod. Hang a 1 m long, 1 cm diameter steel rod from ceiling. Attach a small car to one end. The rod will stretch about 0.5 mm or 0.05 %. When you remove the car, the rod will return to its original length. Hang two small cars. The rod will stretch about 1 mm but will not return to its original length after removing the car. Hang 3 small cars. The rod will stretch less than 2 mm but will then break.
13 3 ways to deform a solid Elasticity in length (tension and compression) Elasticity in shape (shearing) Elasticity in volume (hydraulic stress)
14 Pressure Pressure is force per area. SI unit for pressure is the Pascal. 1 Pa = 1N/m 2
15 Definitions Stress is the force per unit area causing a deformation. (Stress acts like pressure) Strain is a measure of the amount of the deformation. The stress will be proportional to the strain. stress = elastic modulus X strain Similar to Hooke s Law for springs F = k x
16 Young s Modulus Elasticity in length Take a long rod. Internal forces keep the rod together. Pull on the rod a bit. Internal forces resist the force of the pull. So even if the rod is stretched a tiny bit, the bar is in equilibrium. The internal forces balance out the external force. The rod is stressed. This is an example of tensile stress.
17 Tensile Strain tensile strain = ratio of the change in length ( L) to the original length (L 0 ). stress = modulus x strain F A E L L 0 F = k L E = Young s Modulus k = AE/L 0 E describes how easy it is for a solid to be stretched. Big E = Hard to stretch Steel: E = 20 x PA Small E = Easier to stretch Rubber: E = 0.1x10 7 Pa Table with some values on page 317
18 Yield strength the stress needed to be produced to permanently deform the object. Ultimate strength if the stress reached this amount, the object will rupture. Stress (F/A) Ultimate strength Yield strength rupture Range of permanent deformation Linear (elastic) range Important examples: 1) steel rod example 2) tendons and ligaments Strain( L/L)
19 Shear Modulus Elasticity in shape. See figure 1211b on page 316 Example: shearing a book Shear strain = x/h, where x is the horizontal distance the shear force moves, and h is the height of the object. shear stress = F A G = shear modulus G x h
20 Again we can make the analogy to Hooke s Law. F = (A G/h) x = k x k = (A G/h) Big G = difficult to bend Small G = easy to bend
21 Bulk Modulus volume elasticity (hydraulic stress) relates to the response of an object to uniform squeezing. A uniform squeezing could be produced by burying or submerging an object. The hydraulic stress, p, is defined as the ratio of the magnitude of the change in the applied Force F to the surface area A. volume strain = V/V p = B V/V
22 Note: I m use to seeing the equation for hydraulic stress with a negative and written like this.) This book takes the absolute value of V/V P = B V/V Notice the minus sign. If V is negative P is positive. If you squeeze an object, the pressure increases. B = bulk modulus. Tells how easy it is to compress an object. The reciprocal of B is defined as the compressibility ( ). B = 1/ Small B easy to compress Big easy to compress
23 Problems: 6, 10, 24, 50
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