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1 Introduction • Statics: a special case of motion. • Net force and the net torque on an object, or system of objects, are both zero. • The system is not accelerating—it is either are rest or its CM is moving at a constant velocity. 2 Statics—The Study of Forces in Equilibrium • Gravity acts on all objects except those is deep space, where it still acts, but can be considered negligible for most purposes. • If the net force on an object is zero, then other forces must be acting on it to counteract gravity. 3 The book is in equilibrium. 4 The Conditions for Equilibrium First condition for equilibrium: F = 0 5 Example 1 Pull-ups on a scale. A 90-kg weakling cannot do even one pull-up. By standing on a scale, he can determine how close he gets. His best effort results in a scale reading of 23 kg. What force is he exerting? 6 7 Example 2 Chandelier cord tension. Calculate the tensions F1 and F2 in the two cords, which are connected to the cord supporting the 200kg chandelier. 8 9 •Net force is zero. •Net torque is not, so body will rotate. 10 The Conditions for Equilibrium • Second condition for equilibrium: = 0 • If the forces that cause the torque act in the xy plane, then the torque is calculated about the z axis. 11 Conceptual Example 3 A lever. The bar in the figure is being used as a lever to pry up a rock. The small rock acts as a fulcrum. The force FP required a the long end of the bar can be quite a bit smaller than the rock’s weight Mg, since it is the torques that balance in the rotation about the fulcrum. If, however, the leverage isn’t quite good enough, and the rock isn’t budged, what are the two ways to increase leverage? 12 13 3 Solving Statics Problems 1. Choose a body at a time for consideration. Draw a free-body diagram showing all force acting on that body and the points at which these forces act. 2. Choose a convenient coordinate system and resolve the forces into their components. 3. Using letters to represent the unknowns, write down equations for SFx = 0, SFy = 0, S = 0. 4. For the S = 0 equation, choose any axis perpendicular to the xy plane. Give each torque a + or – sign. 5. Solve these equation for the unknowns. If an unknown force comes out negative, it means the direction you originally choose for that force is actually the opposite. 14 Example 4 Tower crane. A tower crane must always be carefully balanced so that 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. The crane’s dimensions are given in the figure. (a) Where must the crane’s 9500-kg counterweight be placed when the load is lifted from the ground? (b) Determine the maximum load that can be lifted with this counterweight when it is placed at its full extent. 15 16 17 Example 5 Forces on a beam and supports. A uniform 1500-kg beam, 20.0 m long, supports a 15,000-kg printing press 5.0 m from the right support column. Calculate the force on each of the vertical support columns. 18 19 A cantilever 20 Example 6 Beam supported by a pin and cable. A uniform beam, 2.20 m long with a mass m = 25.0 kg, is mounted by a pin on a wall. The beam is held in a horizontal position by a cable that makes an angle q = 30.0o. The beam supports a sign of mass M = 280 kg suspended from its end. Determine the components of the force FP that the pin exerts on the beam, and the tension FT in the supporting cable. 21 22 Example 7 Force exerted by biceps muscle. How much force must the biceps muscle exert when a 5.0-kg mass is held in the hand (a) with the arm horizontal? (b) when the arm is at a 30o? Assume that the mass of forearm and hand together is 2.0 kg and their CG is as shown. 23 24 Example 8 Ladder. A ladder of length L=12 m leans against a wall at a point 9.3 m above the ground. The ladder has a mass m = 45.0 kg and its center of mass is L/3 from the lower end. A firefighter of mass M=72 kg climbs the ladder until her center of mass id L/2 from the lower end. What then are the magnitudes of the forces on the ladder from the wall and the pavement? 25 26 27 Stability and Balance • A body in static equilibrium, if left undisturbed, will undergo no translational or rotational acceleration since the sum of all forces and the sum of all torques acting on it are zero. • If object displaced slightly, three possible outcomes: Stable Equilibrium Unstable Equilibrium Neutral Equilibrium 28 Equilibrium • In general, an object whose CG is below its point of support will be in stable equilibrium • A body whose CG is above its base of support will be stable if a vertical line projected downward from the CG falls within the base of support. 29 30 31 32 Levers and Pulley • Mechanical Advantage of a lever Fl = F’l’ 33 Levers and Pulley • Mechanical Advantage of a pulley F’/F = x/x’ 34 Elasticity and Elastic Modulii; Stress and Strain • Study effects of forces on objects in equilibrium. • If such forces are strong enough, the object will break, or fracture. • If the amount of elongation, DL, is small compared to the length of the object, experiment shows that DL is proportional to the weight or force exerted on the object. Sometimes called F = k DL Hook’s law. 35 36 Elastic Limit • This law is found to be valid for almost any solid material. • But valid only up to a point. • If the force is too great, the object stretches excessively and eventually breaks. • The maximum force that can be applied without breaking is called the ultimate strength. 37 38 Elastic Modulus The amount of elongation depends not only on the force, but also on the material. Experiment shows: DL = 1 F L0 YA where L0 is the original length of the object, A is the cross-sectional area, and DL is the change in length due to the applied force, and Y is a constant of proportionality known as the elastic modulus, or Young’s modulus. 39 Stress and Strain force F stress = area = A DL change in length strain = = original length L0 stress Y= strain 40 Example 9 Tension in a piano wire. A 1.60-m-long steel piano wire has a diameter of 0.20 cm. How great is the tension in the wire if it stretches 0.30 cm when tightened? 41 Tensile Stress • Tensile stress: the rod in the figure is under tension or tensile stress. • Not only is the force pulling down on the rod at its lower end, but since the rod is in equilibrium the support at the top is exerting an equal upward force. • Tensile stress exists throughout the material. 42 43 Compressive Stress Compressive stress is the exact opposite. Instead of being stretched, the material is compressed: the forces act inwardly on the body. 44 45 Shear Stress • An object under shear stress has equal and opposite forces applied across its opposite faces. 46 47 Calculating Shear Strain 1 F Dx = h S A where S is called the shear modulus. 48 Bulk Stress • If an object is subject to forces from all sides, its volume will decrease. • A common example is a body submersed in a liquid. DV 1 = - DP V0 B B = - DP DV/V0 where pressure is force per unit area, and B is the bulk modulus. or 49 50 Fracture • If the stress on a solid object is too great, the object fractures or breaks. 51 52 Example 10 When water freezes, it expands by 9.00%. What pressure increase would occur inside your automobile engine block if the water froze? The bulk mo9dulus for ice is 2.00 x 109 N/m2. 53 54