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Lecture 21

Lecture 21. Goals:. Discussion conditions for static equilibrium Use Free Body Diagrams prior to problem solving Introduce Young’s, Shear and Bulk modulus. Exam 3: Wednesday, April, 18 th 7:15-8:45 PM. Statics. Equilibrium is established when.

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Lecture 21

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  1. Lecture 21 Goals: • Discussion conditions for static equilibrium • Use Free Body Diagrams prior to problem solving • Introduce Young’s, Shear and Bulk modulus Exam 3: Wednesday, April, 18th 7:15-8:45 PM

  2. Statics Equilibrium is established when In 3D this implies SIX expressions (x, y & z)

  3. Three force and three torque expressions

  4. Free Body Diagram Rules • For forces zero acceleration reflects the motion of the center of mass • For torques the position of the force vector matters • We are free to choose the axis of rotation (usually one that simplifies the algebra)

  5. Statics: Using Torque T1 T2 M x CM L/2 L/4 y Mg x • Now consider a plank of mass M suspended by two strings as shown. • We want to find the tension in each string:

  6. Statics: Using Torque T1 T2 M x CM L/2 L/4 y Mg x • Now consider a plank of mass M suspended by two strings as shown. • We want to find the tension in each string:

  7. Statics: Using Torque T1 T2 M x CM L/2 L/4 y Mg x • Now consider a plank of mass M suspended by two strings as shown. • Using a different torque position

  8. Another Example: See-saw 60 kg 30 kg 30 kg • Two children (60 kg and 30 kg) sit on a 3.0 m long uniform horizontal teeter-totter of mass 30 kg. The larger child is 1.0 m from the pivot point while the smaller child is trying to figure out where to sit so that the teeter-totter remains motionless. • Position of the small boy?

  9. N Example: Soln. 60 kg 30 kg 0.5 m 1 m 300 N 300 N 600 N 30 kg • Draw a Free Body diagram (assume g = 10 m/s2) • 0 = 300 d + 300 x 0.5 + N x 0 – 600 x 1.0 0= 2d + 1 – 4 d = 1.5 m from pivot point d

  10. Exercise: Statics • A 1.0 kg ball is hung at the end of a uniform rod exactly 1.0 m long. The system balances at a point on the rod 0.25mfrom the end holding the mass. • What is the mass of the rod? (a) 0.5 kg (b) 1 kg (c) 2 kg 1m 1kg

  11. The classic ladder problem m=0 N’ N mg f • A uniform ladder of length L and mass m is propped up against a frictionless wall. There is friction with the ground with static coefficient m. The angle the ladder makes with the ground is q. • What is the minimum value of m to keep the latter from slipping?

  12. The classic ladder problem m=0 N’ N mg f • A uniform ladder of length L and mass m is propped up against a frictionless wall. There is friction with the ground with static coefficient m. If the angle the ladder makes with the ground is q. • What is the minimum value of m that keeps the ladder from slipping?

  13. Example:Statics 3 1 2 • A box is placed on a ramp in the configurations shown below. Friction prevents it from sliding. The center of mass of the box is indicated by a white dot in each case. • In which case(s) does the box tip over ? (a)all (b)2 & 3(c) 3 only

  14. A stable equilibrium 1 • Potential energy with tipping (left and right) • x-axis corresponds to the x-position of the center-of mass

  15. Balancing Acts • So, where is the center of mass for this construction? • Between the arrows!

  16. Example problem • The board in the picture has an average length of 1.4 m, mass 2.0 kg and width 0.14 m. The angle of the board is 45°. The bottle is 1.5 m long, a center of mass 2/5 of the way from the bottom and is mounted 1/5 of way from the end. If the sketch represents the physics of the problem. • What are the minimum and maximum masses of the bottle that will allow this system to remain in balance? -0.1 m -0.8 m 0.0 m -0.5 m 0.1 m 0.4 m

  17. Example problem -0.1 m -0.8 m 0.0 m -0.5 m 0.1 m 0.4 m • What are the minimum and maximum masses of the bottle that will allow this system to remain in balance? • X center of mass, x, -0.1 m < x < 0.1 m 2.0 kg

  18. Example problem -0.1 m -0.8 m 0.0 m -0.5 m 0.1 m 0.4 m • What are the minimum and maximum masses of the bottle that will allow this system to remain in balance? • Check torque condition (g =10 m/s2) • x, -0.1 m < x < 0.1 m 2.0 kg

  19. Real Physical Objects • Representations of matter • Particles: No size, no shape  can only translate along a path • Extended objects are collections of point-like particles They have size and shape so, to better reflect their motion, we introduce the center of mass Rigid objects: Translation + Rotation Deformable objects: Regular solids: Shape/size changes under stress (applied forces) Reversible and irreversible deformation Liquids: They do not have fixed shape Size (aka volume) will change under stress

  20. For Thursday • All of chapter 12

  21. Some definitions F L0 L V0 F V0 - V • Elastic properties of solids : • Young’s modulus: measures the resistance of a solid to a change in its length. • Bulk modulus: measures the resistance of solids or liquids to changes in their volume. elasticity in length volume elasticity

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