Chapter 9 - Static Equilibrium

Chapter 9 - Static Equilibrium - Chapter 9 Static...

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Unformatted text preview: Chapter 9 Static Equilibrium; Elasticity and Fracture Static Equilibrium Static Equilibrium A special case of mechanics − When the net force and net torque on an object, When or system of objects, are both zero. or − Both linear acceleration and the angular Both acceleration of the object or system is zero. acceleration − The object is either at rest or its center of mass is The moving at a constant velocity. moving Static Equilibrium Static Equilibrium Statics − The calculation of the forces acting on and within The structures that are in equilibrium. structures The Conditions of The Conditions of Equilibrium Equilibrium − Objects in daily life have at least one force acting Objects on them….gravity. on − If they are at rest, then there are other forces If acting on them as well so that the net force is zero. zero. − If all the net forces acting on an object are zero, If then the object is in equilibrium. equilibrium. The Conditions of The Conditions of Equilibrium First Condition for Equilibrium − For an object to be at rest, Newton’s second law For says that the forces acting on it must be zero. says − Since force is a vector, the components of the net Since force must also be zero. force The Conditions of The Conditions of Equilibrium Second Condition for Equilibrium − Although the net force on it is zero, the ruler will Although move (rotate). − A pair of equal forces acting in opposite pair directions but at different points on an object is referred to as a couple. couple. The Conditions of The Conditions of Equilibrium Second Condition for Equilibrium − The sum of the torques acting on an object, The calculated about any axis, must be zero. calculated − This ensures that the angular acceleration is zero. − If the object is not rotating, then it will not start If rotating. rotating. Stability and Balance Stability and Balance Equilibrium − An object in static equilibrium, if left An undisturbed, will undergo no translational or rotational acceleration since the sum of all the forces and the sum of all the torques acting on it is zero. is Stability and Balance Stability and Balance Equilibrium − If an object is displaced slightly, it can: − Return to its original position = stable equilibrium. Return stable − Move even farther from its original position = Move unstable equilibrium. unstable − Remain in its new position = neutral equilibrium. Remain neutral Stability and Balance Stability and Balance Balance − An object that is in stable equilibrium. An stable − An object whose center of gravity is above its base of An support will be stable if a vertical line projected downward from the center of gravity falls within the base of support. of Elasticity; Stress and Elasticity; Stress and Strain Elasticity − Any object changes shape under the action of Any applied forces. applied − If the forces are great enough, the objects will If break, or fracture. fracture. Elasticity; Stress and Elasticity; Stress and Strain Elasticity and Hooke’s Law − If the amount of change Δl is small relative to the If length of the object, then Δl is proportional to the force exerted on the object defined by Hooke’s Law: Law: Elasticity; Stress and Elasticity; Stress and Strain Elasticity and Hooke’s Law − Hooke’s Law is valid for almost any solid up to a Hooke’s certain point. certain − If the force is too great, then the object will If break. break. Elasticity; Stress and Elasticity; Stress and Strain Elasticity and Hooke’s Law − Applied force versus elongation will be linear up Applied to a point called the proportional limit. to − Past the proportional limit, the relationship Past becomes more complicated. becomes − Slightly past the proportional limit, up to the Slightly elastic limit, the object will return to its original shape if the force is removed. shape Elasticity; Stress and Elasticity; Stress and Strain Elasticity and Hooke’s Law − The region between the origin and the elastic The limit is called the elastic region. limit − Beyond the elastic region, the object enters the Beyond plastic region where it will NOT return to its original shape after the force is removed. original − The maximum change in length is reached at the The breaking point. breaking − The maximum force that can be applied without The breaking the object is called the ultimate strength. strength. Elasticity; Stress and Elasticity; Stress and Strain Elasticity and Hooke’s Law Elasticity; Stress and Elasticity; Stress and Strain Stress = force applied to an object per area Stress force force F stress = = area A Strain = the change in length of the object as a Strain result of the stress result change in length ∆l strain = = original length lo Elasticity; Stress and Elasticity; Stress and Strain Young’s Modulus − The amount of elongation also depends on the The type of material from which an object is made. type − It is independent of the object’s original size and It shape. shape. Elasticity; Stress and Elasticity; Stress and Strain Tension, Compression, and Shear Stress − An object in tension has tensile stress. An tensile − A force is pulling one down and an equal force is force pulling up to balance. pulling − Tensile stress extends objects. Elasticity; Stress and Elasticity; Stress and Strain Tension, Compression, and Shear Stress − When forces act inwardly on an object, it is When compressed. compressed. − Compressive stresses shorted objects. Elasticity; Stress and Elasticity; Stress and Strain Tension, Compression, and Shear Stress − An objects under shear stress has equal and An opposite forces applied across its opposite faces. opposite Elasticity; Stress and Elasticity; Stress and Strain Tension, Compression, and Shear Stress − Shear stress changes the shape of an object. − The shear stress can be calculated by: − G is the Shear modulus. Elasticity; Stress and Elasticity; Stress and Strain Volume Change – Bulk Modulus − If an object is subject to inward forces from all If sides, its volume decreases. sides, − The force per area on an object is called pressure The and is equivalent to stress. and Elasticity; Stress and Elasticity; Stress and Strain Volume Change – Bulk Modulus − The change is volume as a result of pressure is The proportional to the original volume and to the change in pressure. change − B is the proportionally constant called the bulk is modulus. modulus. Elasticity; Stress and Elasticity; Stress and Strain Elasticity; Stress and Elasticity; Stress and Strain Fracture − If the stress on an object is too great, then is breaks, or If fractures. fractures. − A horizontal beam will be under both tensile and horizontal compressive stress due to its own weight. Therefore, it must be made of a material that is strong under both compression and tension. compression Elasticity; Stress and Elasticity; Stress and Strain Problems Example 1: A tower crane must always be carefully balanced so that tower there is no net torque tending to tip it. A particular crane at a building site is about to lift a 2800-kg air-conditioning unit. (a) Where must the crane’s 9500-kg counterweight be unit. Where placed when the load is lifted from the ground? (Note that the counterweight is usually moved automatically via sensors and motors to precisely compensate for the load.) (b) Determine the maximum load that can be lifted with this Determine counterweight when it is placed at its full extent. Ignore the mass of the beam. mass Problems Example 1: Problems Problems Example 1: • (a) See the free-body diagram. Calculate See torques about the pivot point P labeled in the diagram. The upward force at the pivot will not have any torque. r P x r Mg FP d r mg Problems Example 1: • (a) The total torque is zero since the crane is The in equilibrium. in ∑τ = Mgx − mgd = 0 x= md M → ( 2800 kg ) ( 7.7 m ) = ( 9500 kg ) = 2.3m Problems Example 1: • (b) Again we sum torques about the pivot Again point. Mass m is the unknown in this case, and the counterweight is at its maximum distance from the pivot. distance ∑τ = Mgxmax − mmax gd = 0 → mmax = Mxmax d = ( 9500 kg ) ( 3.4 m ) ( 7.7 kg ) = 4200 kg Problems Problems Example 2: Find the tension in the two wires supporting Find the traffic light. Problems Example 2: • Draw a free-body diagram of the junction of Draw the three wires. the 37 o r FT1 53o r FT2 r mg Problems Example 2: • The tensions can be found from the The conditions for force equilibrium. conditions ∑F = FT1 cos 37° − FT2 cos 53° = 0 → FT2 = ∑F = FT1 sin 37° + FT2 sin 53° − mg = 0 x y cos 37° cos 53° FT1 Problems Example 2: FT1 sin 37° + FT1 = FT2 = cos 37° cos 53° FT1 sin 53° − mg = 0 → ( 33 kg ) ( 9.80 m sin 37° + cos 37° cos 53° FT1 = cos 37° cos 53° cos 37° s2 ) = 194.6 N ≈ 190 N sin 53° ( cos 53° ) 1.946 × 102 N = 258.3 N ≈ 260 N Problems Problems Example 3: A refrigerator is approximately a uniform refrigerator rectangular solid 1.9 m tall, 1.0 m wide, and 0.75 m deep. If it sits upright on a truck with its 1.0-m dimension in the direction of travel, and if the refrigerator cannot slide on the truck, how rapidly can the truck accelerate without tipping the refrigerator over? Problems Problems Example 3: • Assume the truck is accelerating to the right. Assume We want the refrigerator to not tip in the non-inertial reference frame of the truck. Accordingly, to analyze the refrigerator in the non-inertial reference frame, add a pseudoforce in the opposite direction of the actual acceleration. Problems Problems Example 3: • The free-body diagram is for a side view of the The refrigerator, just ready to tip so that the normal force and frictional force are at the lower back corner of the refrigerator. The center of mass is in the geometric center of the refrigerator. w r ma truck r FN r Ffr r mg h Problems Example 3: • Write the conditions for equilibrium, taking Write torques about an axis through the center of mass, perpendicular to the plane of the paper. The normal force and frictional force cause no torque about that axis. cause Problems Example 3: ∑F ∑F = Ffr − ma truck = 0 → Ffr = matruck horiz vert = FN − mg = 0 → FN = mg ∑τ = F ( N FN Ffr = h w = 1 2 w ) − Ffr ( h ) = 0 → 1 2 mg ma truck → atruck = g w h FN Ffr ( = h w = 9.80 m s 2 1.0 m ) 1.9 m = 5.2 m s 2 Problems Example 4: How much pressure is needed to compress How the volume of an iron block by 0.10%? Express your answer in N/m2 and compare it Express to atmospheric pressure (1 X 105 N/m2) to (1 Problems Example 4: • The relationship between pressure change The and volume change is given by: and ∆V = −V0 ∆P Patm = ∆P B → ∆P = − 9.0 × 107 N m 2 1.0 × 105 N m 2 ∆V V0 ( B = − 0.10 × 10−2 ) ( 90 × 10 = 9.0 × 102 , or 900 atmospheres 9 ) N m 2 = 9.0 × 107 N m 2 Problems Example 5: A steel cable is to support an elevator whose steel total (loaded) mass is not to exceed 3100 kg. If the maximum acceleration of the elevator is 1.2m/s2 calculate the diameter of cable is required. Assume a safety factor of 8.0. required. Problems Problems Example 5: • From the free-body diagram, write Newton’s From second law for the vertical direction. Solve for the maximum tension required in the cable, which will occur for an upwards acceleration. ∑F y = FT − mg = ma → FT = m ( g + a ) r F T r mg Problems Example 5: • The maximum stress is to be 1/8th of the tensile strength for steel. The maximum stress will occur for the minimum area, and thus the minimum diameter. thus stress max = d= FT Amin = tensile strength 4 ( 8.0 ) m ( g + a ) 8.0 π ( tensile strength ) = → A1 = π ( 1 d ) = 2 2 ( 32 ( 3100 kg ) 11.0 m s 2 π ( 500 × 106 N m 2 ) ) 8.0 FT tensile strength → = 2.6 × 10−2 m = 2.6 cm Problems Problems Example 6: A 50-story building is being planned. It is to be 50-story 180.0 m high with a base 46.0 m by 76.0 m. Its total mass will be about 1.8 X 107 kg and its weight mass therefore about 1.8 X 107 N. Suppose a 200 km/h therefore wind exerts a force of 950 N/m2 over the 76.0 m wind wide face. Calculate the torque about the potential pivot point, the rear edge of the building, and determine whether the building will topple. Assume the total force of the wind acts at the midpoint of the building’s face, and that the building is not anchored in bedrock. Problems Example 6: Problems Example 6: • Assume that the building has just begun to Assume tip, so that it is essentially vertical, but that all of the force on the building due to contact with the Earth is at the lower left corner. with r FA 23.0 m r mg 90.0 m r FE y r FE x Problems Example 6: • Take torques about that corner, with counterclockwise Take torques as positive. torques ∑τ = F ( 90.0 m ) − mg ( 23.0 m ) = ( 950 N m ) ( 180.0 m ) ( 76.0 m ) ( 90.0 m ) − ( 1.8 × 10 kg ) ( 9.80 m s ) ( 23.0 m ) A 2 7 2 = −2.9 × 109 mg N • Since this is a negative torque, the building will tend to rotate clockwise, which means it will rotate back down to the ground. Thus the building will not topple. Problems Problems Example 7: The roof over a 9.0 m X 10.0 m room in a The school has a total mass of 13,600 kg. The roof is to be supported by vertical “2 X 4’s” (actually about 4 cm X 9 cm ) equally spaced along the 10.0 m sides. How many supports are required on each side, and how far apart must they be? Consider only compression, and assume a safety factor of 12. Problems Problems Example 7: • The number of supports can be found from The the compressive strength of the wood. Since the wood will be oriented longitudinally, the stress will be parallel to the grain. Problems Example 7: Compressive Strength Safety Factor ( # supports ) = Load force on supports = Area of supports = Weight of roof ( # supports ) ( area per support ) Weight of roof Safety Factor ( area per support ) Compressive Strength ( 1.36 × 10 kg ) ( 9.80 m s ) = 4 ( 0.040 m ) ( 0.090 m ) 2 12 ( 35 × 10 6 Nm 2 ) = 12.69 supports Problems Problems Example 7: Since there are to be more than 12 supports, Since and to have the same number of supports on each side, there will be 14 supports, or 7 supports on each side. That means there will be 6 support-to-support spans, each of which would be given by would Spacing = 10.0 m 6 gaps = 1.66 m gap End of Chapter 9 End of Chapter 9 ...
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This note was uploaded on 08/10/2011 for the course GOV 102 taught by Professor Smithy during the Spring '11 term at CSU San Marcos.

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