PHYS 1441 – Section 501 Lecture #15AnnouncementsConditions for EquilibriumMore on Conditions for EquilibriumExample for Mechanical EquilibriumExample for Mech. Equilibrium Cont’dExample 9 – 9Example 9 – 9 cont’dSlide 9How do we solve equilibrium problems?Elastic Properties of SolidsYoung’s ModulusBulk ModulusExample for Solid’s Elastic PropertyDensity and Specific GravityFluid and PressureExample for PressureMonday, July 26, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu1PHYS 1441 – Section 501Lecture #15Monday, July 26, 2004Dr. Jaehoon Yu•Conditions for Mechanical Equilibrium•Elastic Property of SolidsYoung’s ModulusBulk Modulus•Density and Specific Gravity•Fluid and PressureRemember the quiz in the class, Monday, Aug. 2!!Monday, July 26, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu2Announcements•Term2 exam results–Average: 66.4–Top score: 97•2nd quiz next Monday, Aug. 2 –Section 8.6 – whatever we cover till next Wednesday•Final term exam Wednesday, Aug. 11–Covers: Section 8.6 – wherever we cover by Aug. 4–Review on Aug. 11•Dr. Osturk will give the lecture this Wednesday.Monday, July 26, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu3Conditions 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. 00CMv0Monday, July 26, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu4More on Conditions for EquilibriumTo simplify the problem, 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?0F00xF0zOF1F4F3F2F5r5O’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.0yFWhy is this true?Because the object is not movingnot moving, no matter what the rotational axis is, there should not be a motion. It is simply a matter of mathematical calculation.Monday, July 26, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu5Example for Mechanical EquilibriumA 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?Since there is no linear motion, this system is in its translational equilibriumFDn1m xTherefore the magnitude of the normal force nDetermine where the child should sit to balance the system.The net torque about the fulcrum by the three forces are Therefore to balance the system the daughter must sitxxF0yFgMB0gMFgMDnmgMgMDF00.1mm 29.200.13508000 gMB00.1 gMFxgMD0N11903508000.40 MBgMFgMFgMonday, July 26, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu6Example for Mech. Equilibrium Cont’d Determine 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 ThereforexnThe net torque can be rewritten What do we learn?No matter where the rotation axis is, net effect of the torque is identical.FDnMBgMFg MFg1m xx/2Rotational axis2/xgMB0gMgMgMDFB 2/00.1 xgMF2/xn 2/xgMD2/xgMB 2/00.1 xgMF 2/xgMgMgMDFB2/xgMDxgMgMDF 00.10mgMgMDF00.1mm 29.200.1350800Monday, July 26, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu7Example 9 – 9 A 5.0 m long ladder leans against a wall at a point 4.0m above the ground. The ladder is uniform and has mass 12.0kg. Assuming the wall is frictionless (but ground is not), determine the forces exerted on the ladder by the ground and the wall. xFFBDFirst the translational equilibrium, using componentsThus, the y component of the force by the ground ismgFWFGxFGyOGyFGx WF F= -0yFGymg F=- +0mg12.0 9.8 118N N= � =The length x0 is, from Pythagorian theorem2 205.0 4.0 3.0x m= - =Monday, July 26, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu8Example 9 – 9 cont’dOFrom the rotational equilibrium02 4.0Wmg x F=- +0Thus the force exerted on the ladder by the wall isWFThus the force exerted on the ladder by the ground isTx component of the force by the ground is44Gx WF F N= =GF024.0mg x=118 1.5444.0N�= =0x Gx WF F F= - =�Solve for FGx2 2Gx GyF F= +2 244 118 130N= + �The angle between the ladder and the wall isq1tanGyGxFF-� �=� �� �1118tan 7044-� �= =� �� �oMonday, July 26, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu9Example for Mechanical EquilibriumA 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.xFSince the system is in equilibrium, from the translational equilibrium conditionFrom the rotational equilibrium conditionOFBFUmgdlThus, the force exerted by the biceps muscle isdFBForce exerted by the upper arm isUF0yFmgFFUB0lmgdFFBU 00lmg BFdlmg N58300.30.350.50mgFBN5330.50583 Monday, July 26, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon
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