1443-501 Spring 2002 Lecture #17Conditions for EquilibriumMore on Conditions for EquilibriumCenter of Gravity RevisitedExample 12.1Example 12.1 ContinuedExample 12.2Example 12.3Example 12.4Elastic Properties of SolidsYoung’s ModulusShear ModulusBulk ModulusExample 12.71443-501 Spring 2002Lecture #17Dr. Jaehoon Yu1. Conditions for Equilibrium2. Center of Gravity3. Elastic Properties of Solids•Young’s Modulus•Shear Modulus•Bulk ModulusToday’s Homework Assignment is the Homework #8!!!2nd term exam on Wednesday, Apr. 10. Will cover chapters 10 -13.Apr. 1, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #172Conditions for EquilibriumWhat do you think does the term “An object is at its equilibrium” mean?0FThe object is either at rest (Static Equilibrium) or its center of mass is moving with a constant velocity (Dynamic Equilibrium). Is this it? When do you think an object is at its equilibrium?Translational Equilibrium: Equilibrium in linear motion The above condition is sufficient for a point-like particle to be at its static equilibrium. However for object with size this is not sufficient. One more condition is needed. What is it? Let’s consider two forces equal magnitude but opposite direction acting on a rigid object as shown in the figure. What do you think will happen?CMddF-FThe object will rotate about the CM. The net torque acting on the object about any axis must be 0. For an object to be at its static equilibrium, the object should not have linear or angular speed. 00CMv0Apr. 1, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #173More on Conditions for EquilibriumTo simplify the problems, we will only deal with forces acting on x-y plane, giving torque only along z-axis. What do you think the conditions for equilibrium be in this case? The six possible equations from the two vector equations turns to three equations.What happens if there are many forces exerting on the object?Net torque about O0F000yxFF0zOF1F4F3F2F5r5O’r’If an object is at its translational static equilibrium, and if the net torque acting on the object is 0 about one axis, the net torque must be 0 about any arbitrary axis. 0321FFFF 0332211 iiOFrFrFrFrNet Force exerting on the object''rrriiPosition of force Fi about O’Net torque about O’ iiiOFrFrFrrFrrFrFr '''''22112211' 00'' OiiOrFrApr. 1, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #174Center of Gravity RevisitedWhen is the center of gravity of a rigid body the same as the center of mass?MymmymyMxmmxmxiiiiiCMiiiiiCMUnder the uniform gravitational field throughout the body of the object.CMm1g1CoGm2g2m3g3migiLet’s consider an arbitrary shaped objectThe center of mass of this object isLet’s now examine the case with gravitational acceleration on each point is giSince the CoG is the point as if all the gravitational force is exerted on, the torque due to this force becomes 2221112211xgmxgmxgmgmCoGIf g is uniform throughout the body CMiiiCoGCoGxmxmxgxmxmgxmm221121Generalized expression for different g throughout the bodyApr. 1, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #175Example 12.1A uniform 40.0 N board supports a father and daughter weighing 800 N and 350 N, respectively. If the support (or fulcrum) is under the center of gravity of the board and the father is 1.00 m from CoG, what is the magnitude of normal force n exerted on the board by the support?00ngMgMgMFFDFByxSince there is no linear motion, this system is in its translational equilibriumFDnMBgMFg MFg1m xTherefore the magnitude of the normal force Nn 11903508000.40 Determine where the child should sit to balance the system.The net torque about the fulcrum by the three forces are 000.10 xgMgMgMDFBTherefore to balance the system the daughter must sitmmmgMgMxDF29.200.135080000.1 Apr. 1, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #176Example 12.1 ContinuedDetermine the position of the child to balance the system for different position of axis of rotation.Since the normal force is The net torque about the axis of rotation by all the forces are 02/2/2/00.12/ xgMxnxgMxgMDFBThereforemmmgMgMxDF29.200.135080000.1 gMgMgMnDFBThe net torque can be rewritten 000.12/2/2/00.12/xgMgMxgMxgMgMgMxgMxgMDFDDFBFBWhat do we learn?No matter where the rotation axis is, net effect of the torque is identical.FDnMBgMFg MFg1m xx/2Rotational axisApr. 1, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #177Example 12.2A person holds a 50.0N sphere in his hand. The forearm is horizontal. The biceps muscle is attached 3.00 cm from the joint, and the sphere is 35.0cm from the joint. Find the upward force exerted by the biceps on the forearm and the downward force exerted by the upper arm on the forearm and acting at the joint. Neglect the weight of forearm.00mgFFFFUByxSince the system is in equilibrium, from the translational equilibrium conditionFrom the rotational equilibrium conditionOFBFUmgdl00 lmgdFFBUThus, the force exerted by the biceps muscle isNdlmgFlmgdFBB58300.30.350.50Force exerted by the upper arm isNmgFFBU5330.50583 Apr. 1, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #178Example 12.3A uniform horizontal beam with a length of 8.00m and a weight of 200N is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle of 53.0o with the horizontal. If 600N person stands 2.00m from the wall, find the tension in the cable, as well as the magnitude and direction of the force exerted by the wall on the beam.02006000.53sinsin00.53coscosNNTRFTRFyxFrom the rotational equilibriumUsing the translational equilibrium 8m53.0o2mFBDRT02053.0oTsin53Tcos53RsinRcosFirst the translational equilibrium, using
View Full Document
Unlocking...