Monday, Dec. 6, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 1 PHYS 1441 – Section 002 Lecture #23 Monday, Dec. 6, 2010 Dr. Jaehoon Yu • Similarities Between Linear and Rotational Quantities • Conditions for Equilibrium • How to Solve Equilibrium Problems? • A Few Examples of Mechanical Equilibrium • Elastic Properties of Solids • Density and Specific GravityAnnouncements • The Final Exam – Date and time: 11am – 1:30pm, Monday Dec. 13 – Place: SH103 – Comprehensive exam • Covers from CH1.1 – what we finish Wednesday, Dec. 8 • Plus appendices A.1 – A.8 • Combination of multiple choice and free response problems • Bring your Planetarium extra credit sheet to the class next Wednesday, Dec. 8, with your name clearly marked on the sheet! • Reading assignments – Ch9.3 – 9.7 Monday, Dec. 6, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 2Monday, Dec. 6, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 3 Similarity Between Linear and Rotational Motions All physical quantities in linear and rotational motions show striking similarity. Quantities Linear Rotational Mass Mass Moment of Inertia Length of motion Distance Angle (Radian) Speed Acceleration Force Force Torque Work Work Work Power Momentum Kinetic Energy Kinetic Rotational I F= ma τ= Iα P = F⋅ v p= mv L= IωMonday, Dec. 6, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 4 Conditions for Equilibrium What do you think the term “An object is at its equilibrium” means? F∑=The object is either at rest (Static Equilibrium) or its center of mass is moving at 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 object to be at its translational equilibrium. However for an object with size this is not sufficient. One more condition is needed. What is it? Let’s consider two forces equal in magnitude but in opposite direction acting on a rigid object as shown in the figure. What do you think will happen? CM d d F -F The object will rotate about the CM. Thus 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.Monday, Dec. 6, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 5 More on Conditions for Equilibrium To 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 an object? O F3 r5 O’ 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. Why is this true? ANDMonday, Dec. 6, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 6 How do we solve static equilibrium problems? 1. Select the object to which the equations for equilibrium are to be applied. 2. Identify all the forces and draw a free-body diagram with them indicated on it with their directions and locations properly indicated 3. Choose a convenient set of x and y axes and write down force equation for each x and y component with correct signs. 4. Apply the equations that specify the balance of forces at equilibrium. Set the net force in the x and y directions equal to 0. 5. Select the most optimal rotational axis for torque calculations Selecting the axis such that the torque of one of the unknown forces become 0 makes the problem easier to solve. 6. Write down the torque equation with proper signs. 7. Solve the force and torque equations for the desired unknown quantities.Monday, Dec. 6, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 7 Example for Mechanical Equilibrium A uniform 40.0 N board supports the father and the daughter each weighing 800 N and 350 N, respectively, and is not moving. 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 the normal force n exerted on the board by the support? Since there is no linear motion, this system is in its translational equilibrium F D n 1m x Therefore the magnitude of the normal force Determine 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 sit Fx∑MBg MDg MFgMonday, Dec. 6, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 8 Example 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 Therefore The net torque can be rewritten What do we learn? No matter where the rotation axis is, net effect of the torque is identical. F D n MBg MFg MFg 1m x x/2!Rotational axisMonday, Dec. 6, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 9 Example 9 – 7 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. FBD First the translational equilibrium, using components Thus, the y component of the force by the ground is mg FW FGx FGy O The length x0 is, from Pythagorian theoremMonday, Dec. 6, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 10 Example 9 – 7 cont’d From the rotational equilibrium Thus the force exerted on the ladder by the wall is Thus the force exerted on the ladder by the ground is The x component of the force by the ground is Solve for FGx The angle between the ground force to the floorMonday, Dec. 6, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu 11 Ex. 9.8 for Mechanical Equilibrium A 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. Since the system
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