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UT Arlington PHYS 1443 - Mechanical Equilibrium

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PHYS 1443 – Section 003 Lecture #19Conditions for EquilibriumMore on Conditions for EquilibriumCenter of Gravity RevisitedExample for Mechanical EquilibriumExample for Mech. Equilibrium Cont’dSlide 7Slide 8Slide 9How did we solve equilibrium problems?Elastic Properties of SolidsYoung’s ModulusBulk ModulusExample for Solid’s Elastic PropertyWednesday, Nov. 12, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu1PHYS 1443 – Section 003Lecture #19Wednesday, Nov. 12, 2003Dr. Jaehoon Yu1. Conditions for Mechanical Equilibrium2. Center of Gravity in Mechanical Equilibrium 3. Elastic Properties of Solids•Young’s Modulus•Shear Modulus•Bulk ModulusToday’s homework is homework #10 due noon, Wednesday, Nov. 19!!Wednesday, Nov. 12, 2003 PHYS 1443-003, Fall 2003Dr. 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. 00CMv0Wednesday, Nov. 12, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu3More 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 O0F00xF0zOF1F4F3F2F5r5O’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ONet Force exerting on the object''rrriiPosition of force Fi about O’Net torque about O’'O'O0yF332211FrFrFriiFr0   2211'' FrrFrr iiiFrFr '2211'' FrFr 0'rFriiO0Wednesday, Nov. 12, 2003 PHYS 1443-003, Fall 2003Dr. 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 bodyiiimxmMxmiiCMyiiimymMymii222111xgmxgm CoGgxmm 21 gxmxm 2211CoGxiiimxmCMxWednesday, Nov. 12, 2003 PHYS 1443-003, Fall 2003Dr. 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 sitxxF0yFgMB0gMFgMDnmgMgMDF00.1mm 29.200.13508000 gMB00.1 gMFxgMD0N11903508000.40 MBgMFgMFgWednesday, Nov. 12, 2003 PHYS 1443-003, Fall 2003Dr. 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/xgMB0gMgMgMDFB 2/00.1 xgMF2/xn 2/xgMD2/xgMB 2/00.1 xgMF 2/xgMgMgMDFB2/xgMDxgMgMDF 00.10mgMgMDF00.1mm 29.200.1350800Wednesday, Nov. 12, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu7Example 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 conditionOFBFUmgdlThus, the force exerted by the biceps muscle isdFBForce exerted by the upper arm isUF0yFmgFFUB0lmgdFFBU 00lmg BFdlmg N58300.30.350.50mgFBN5330.50583 Wednesday, Nov. 12, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu8Example for Mechanical EquilibriumA 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.xFFrom the rotational equilibriumUsing the translational


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UT Arlington PHYS 1443 - Mechanical Equilibrium

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