Slide 1Conditions for EquilibriumMore on Conditions for EquilibriumCenter of Gravity RevisitedExample for Mechanical EquilibriumExample for Mech. Equilibrium Cont’dExample 12 – 8Example 12 – 8 cont’dSlide 9How do we solve equilibrium problems?Elastic Properties of SolidsYoung’s ModulusBulk ModulusExample for Solid’s Elastic PropertyWednesday, Nov. 17, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu11. Conditions for Equilibrium2. Mechanical Equilibrium3. How to solve equilibrium problems?4. Elastic properties of solids5. Fluid and PressurePHYS 1443 – Section 003Lecture #21Wednesday, Nov. 17, 2004Dr. Jaehoon YuQuiz #3 next Monday, Nov. 22!!Today’s Homework is #11 due on Wednesday, Nov. 24, 2003!!Wednesday, Nov. 17, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu2Conditions 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. 00CMv0Wednesday, Nov. 17, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu3More 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.ANDWednesday, Nov. 17, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu4Center of Gravity RevisitedWhen is the center of gravity of a rigid body the same as the center of mass?CMxUnder 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 CoGxgmgm 2211If g is uniform throughout the bodyGeneralized expression for different g throughout the bodyiiimxmMxmiiCMyiiimymMymii222111xgmxgm CoGgxmm 21 gxmxm 2211CoGxiiimxmCMxWednesday, Nov. 17, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu5Example for Mechanical EquilibriumA uniform 40.0 N board supports the father and the daughter each 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 sitxxF0yFBM g-0FM g-DM g-n=mgMgMDF00.1mm 29.200.135080000.1 gMFxgMD0N11903508000.40 MBgMDgMFg0 gMB0n+ �Wednesday, Nov. 17, 2004 PHYS 1443-003, Fall 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 29.200.1350800Wednesday, Nov. 17, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu7Example 12 – 8 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= - =Wednesday, Nov. 17, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu8Example 12 – 8 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-� �= =� �� �oWednesday, Nov. 17, 2004 PHYS 1443-003, Fall 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
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