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Last Time : Torque, Conditions for Equilibrium Today : Center of Gravity, Equilibrium Problems

Last Time : Torque, Conditions for Equilibrium Today : Center of Gravity, Equilibrium Problems I will be away at a conference next week (Tues-Fri). Lectures will be given by: Tuesday: Dr. Wolfgang Korsch (006 recitation instructor) Thursday: Dr. Tim Gorringe (other PHY 211 lecturer)

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Last Time : Torque, Conditions for Equilibrium Today : Center of Gravity, Equilibrium Problems

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  1. Last Time: Torque, Conditions for Equilibrium • Today: Center of Gravity, Equilibrium Problems • I will be away at a conference next week (Tues-Fri). Lectures will be given by: • Tuesday: Dr. Wolfgang Korsch (006 recitation instructor) • Thursday: Dr. Tim Gorringe (other PHY 211 lecturer) • Will be in email contact.

  2. The See-Saw: Revisited L/2 L/2 mass m Question : Something we have neglected to mention so far … The see-saw itself has a mass. So there is a downward force due to gravity on the see-saw itself. Why did we not include this force in our consideration of the torques on the see-saw ?

  3. The See-Saw: Revisited L/2 L/2 mass m mg Answer : If the see-saw is homogeneous (i.e., its mass is distributed uniformly along its length), we can think of the see-saw’s entire mass as being located at its very center. So if all of its mass is located at its center, and if the pivot is at its center, the torque due to its weight mg about the center is: A “perfectly uniform” see-saw would balance on a pivot located exactly at its center point.

  4. Center of Gravity Consider this weird-shaped object. We can think of it as being made up of a LARGE NUMBER of small masses: M = m1 + m2 + m3 + ... We could, in principle, calculate the gravitational torque due to the weight of all of these small masses about the origin O. The CENTER OF GRAVITY of the object is that point where: If the entire weight Mg = (m1 + m2 + m3 + …)g was located at the c.g., the torque about O due to M would be equal to The torque about O due to all the individual weights.

  5. Center of Gravity Consider these 3 masses along a massless rod in the x-direction. x1 x2 x3 O x m1g m2g m3g all CW torques about O The net torque about O due to the individual torques is: Suppose there is a Center of Gravity located at xcg with mass M = m1 + m2 + m3. The torque about O due to M at the c.g. would also be in the CW direction, and would be:

  6. Center of Gravity Let’s now equate these two torques (minus sign cancels!) : Solving for xcg gives us: Having the entire weight located at the center of gravity yields the same torque about point A as the individual weights ! xcg O x Mg = (m1 + m2 + m3)g

  7. Center of Gravity In general, the formulas for the x- and y-coordinates of the center of gravity are: Sometimes also called the “center of mass”.

  8. Example Find the center of gravity of these 3 masses : y (m) 2 kg 5 kg x (m) 6 kg

  9. What About Symmetric Objects ? For a symmetric, homogeneous (i.e., uniform) object, it is often easy to guess where the center of gravity is … Example: Solid, homogenous piece of wood c.g. located right at the center h/2 h/2 L/2 L/2 Example: Solid, homogeneous sphere For solid, homogenous objects, the c.g. is usually located at the geometric center c.g. located right at the center

  10. Objects in Equilibrium An object in equilibrium must satisfy these conditions : Net force on the object must be zero. #1 Net torque on the object must be zero. #2 Steps to solving equilibrium problems : Draw a diagram (including a coordinate system) and then draw a free-body diagram for the object. #0 Apply ΣFx = 0 and ΣFy = 0 [net force must be zero !!] #1 in both x- and y-directions Apply Στi = 0 [net torque must be zero !!] #2 sum of CW (-) and CCW (+) torques

  11. Example: 8.8 A uniform beam of length 7.60 m and weight 450 N is carried by two people. (a) Determine the forces that each person exerts on the beam. (b) Qualitatively, how would the answers change if Sam moved closer to the midpoint ? (c) What would happen if Sam moved beyond the midpoint ?

  12. Example: 8.10 A meter-stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 50.0-gram mass is attached at the 10.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter-stick ?

  13. Example: 8.20 A window washer weighing 700 N is standing on a scaffold supported by a vertical rope at each end. The scaffold weighs 200 N and is 3.0 m long. What is the tension in each rope when the window washer stands 1.0 m from one end? 1.0 m 3.0 m

  14. Next Class • 8.5 : Moment of Inertia, Torque and Angular Acceleration

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